VECTORS AND MOTION IN TWO DIMENSIONS Q3.1.srjcstaff.santarosa.edu/~lwillia2/20/20ch3hw.pdf3-1 3...

3-1

VECTORS AND MOTION IN TWO DIMENSIONS 3 Q3.1. Reason: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if = 0 mxA and = 5 m,yA then the magnitude of the vector is

2 2(0 m) (5 m) 5 mA = + =

(b) A zero magnitude says that the length of the vector is zero, thus each component must be zero. Assess: It stands to reason that a vector can have a nonzero magnitude with one component zero as long as the other one isn’t. It also makes sense that for the magnitude of the vector to be zero all the components must be zero.

Q3.2. Reason: No, it is not possible. A scalar has a magnitude only but a vector has direction as well. Even if each has the same dimensions, the result of the addition of a scalar and a vector is ambiguous. Assess: We already learned in chapter 1 that we can’t add quantities unless they have the same dimensions; here we also point out that two quantities that will be added (or subtracted) must both be scalars or both vectors.

Q3.3. Reason: Consider two vectors A

and .B

Their sum can be found using the method of algebraic addition. In Question 3.2 we found that the components of the zero vector are both zero. The components of the resultant of A

and B

must then be zero also. So

00

x x x

y y y

R A BR A B

= + == + =

Solving for the components of B

in terms of A

gives x xB A= − and .y yB A= − Then the magnitude of B

is 2 2( ) ( )x yB B+ = 2 2 2 2( ) ( ) ( ) ( ) .x y x yA A A A− + − = + So then the magnitude of B

is exactly equal to the

magnitude of .A

Assess: For two vectors to add to zero, the vectors must have exactly the same magnitude and point in opposite directions.

Q3.4. Reason: (a) =C A B+ only if A

and B

are in the same direction. Size does not matter. (b) C = A – B if A

and B

are in the opposite direction to each other. Size matters only in that A > B because C as a magnitude can only be positive. Assess: Visualize the situation with arrows.

Q3.5. Reason: The ones that are constant are vx, ax, and ay. Furthermore, ax is not only constant, it is zero.

3-2 Chapter 3

Assess: There are instants when other quantities can be zero, but not throughout the flight. Remember that ay = –g throughout the flight and that vx is constant; that is, projectile motion is nothing more than the combination of two simple kinds of motion: constant horizontal velocity and constant vertical acceleration.

Q3.6. Reason: The acceleration of the ball is due to gravity, so the acceleration the ball experiences is always straight downward. (a) The velocity vector of the ball always has a component in the horizontal direction since it was thrown at an angle of 40°. The horizontal component of the ball’s velocity is constant throughout its trajectory. Since the velocity vector always has a component in the horizontal direction, it is never pointing entirely straight up or down. So there is no point on the trajectory where the acceleration and velocity are parallel. Note that if the ball were thrown straight up or down instead of at 40°, the ball’s velocity and acceleration would be parallel. (b) The acceleration of the ball is always straight downward. For the velocity vector to be perpendicular to the acceleration vector, the velocity must point entirely in the horizontal direction. The velocity vector always has a component in the horizontal direction, as reasoned in part (a). It has a component in the vertical direction during most of its motion also since the ball travels upward and downward in the vertical direction. There is one point where the ball is not traveling up or down, and that is at the top of its trajectory. This is the only point where the velocity and acceleration are perpendicular. Assess: The acceleration due to gravity always points straight downward. See Figure 3.30, which shows the average velocity vectors and acceleration along the trajectory of a tossed ball.

Q3.7. Reason: By extending their legs forward, the runners increase their time in the air. As you will learn in chapter 7, the “center of mass” of a projectile follows a parabolic path. By raising their feet so that their feet are closer to their center of mass, the runners increase the time it takes for their feet to hit the ground. By increasing their time of flight, they increase their range. Also, having their feet ahead of them means that their feet will land ahead of where they would have landed otherwise. Assess: By simply moving their feet, runners can change their time of flight and change the spot where their feet land.

Q3.8. Reason: Running while throwing a ball increases the distance of the throw because it increases the horizontal component of the ball’s velocity without changing the time of flight. Relative to the ground, the ball’s horizontal velocity component is increased by an amount equal to the speed of the runner. Since the velocity of the runner is purely horizontal, the vertical velocity component of the ball is unchanged and so the time of flight of the ball is unchanged. It is interesting to note that since, relative to the ground, the ball has a greater horizontal component of velocity, the angle of the ball’s velocity is higher relative to the person throwing the ball than relative to the ground. However, neither relative to the ground, nor relative to the person throwing the ball will a launch angle of 45˚ give the maximum range. The angle 45˚ gives the maximum range for a ball launched by someone not moving because it gives the best compromise between having a high time of flight and a large horizontal component of velocity. But when the person is running, the horizontal component of velocity gets an advantage so the time of flight can be greater at the expense of the horizontal component of velocity. Thus the best angle relative to the person is greater than 45˚. Assess: While the best launch angle is no longer 45°, it is still true that running increases the range of the ball.

Q3.9. Reason: The claim is slightly misleading, since the passenger cannot walk at a speed in excess of 500 mph due to her own efforts relative to the walking surface (the floor of the plane in this case); but she can be walking and moving with such a speed relative to the ground thousands of feet below. Assess: It is important to specify the coordinate system when reporting velocities.

Q3.10. Reason: The lower the angle of the slope, the lower the acceleration down the slope since acceleration along a ramp is given by sing θ . Furthermore, the speed at the bottom of a slope will be less if the acceleration is less. Assess: A lower angle gives the skier a lower velocity and, consequently, better control.

Q3.11. Reason: The acceleration is due to gravity, so the acceleration will always act to pull the cars back down the ramp. Since the roller coaster is constrained to move along the ramp, the acceleration must be along the ramp. So in all three cases the acceleration is downward along the ramp. Assess: Gravity always acts, even if motion is constrained by an inclined ramp. See Figure 3.24 in the text.

Vectors and Motion in Two Dimensions 3-3

Q3.12. Reason: The time for an object to hit the ground does not depend on its horizontal speed, but only on its height and initial vertical speed. When the pilot goes twice as fast, all that changes is the horizontal speed of the projectile. Therefore the time of flight will be the same in both cases, 2.0 s. However the distance travelled horizontally will be doubled since the horizontal speed is doubled and the time of flight is the same. At the doubled speed, the weight will travel twice as far, or 200 m. Assess: The above answer assumed no air resistance. Actually the greater speed of the faster projectile will slightly increase the time of flight since a faster object experiences more air resistance.

Q3.13. Reason: The cyclist’s speed may be constant, but the direction of her motion is always changing. Since the direction of her velocity vector is constantly changing, she is always accelerating. Assess: Acceleration is caused by any change in velocity, either in magnitude or direction. In circular motion, the acceleration is always towards the center of the circle and is called centripetal acceleration.

Q3.14. Reason: Because acceleration is defined as the change in the (vector) velocity divided by the corresponding time interval, it, too, is a vector.

= vat

∆∆

A car in uniform circular motion (constant speed) is still accelerating because the direction of the velocity vector is changing. The way to determine the direction of the acceleration vector is to realize that a must always point in the same direction as v∆ because t∆ is not only a scalar, it is a positive scalar (as long as time doesn’t go backward).

f i ,v v v∆ = − so, to determine the direction of ,v∆ look at the velocity vector arrows just before and just after the

car is pointed north, and subtract them using vector subtraction. f i f i= = ( )v v v v v∆ − + −

and, as shown in the figure, v∆ points west.

Assess: It is worth noting that the specific directions were not as important as the conclusion that the v∆ vector and therefore the a vector both point toward the center of the circle in uniform circular motion.

Q3.15. Reason: Since the plane is moving in a circle and constantly changing direction, the plane is constantly accelerating. At any point along its motion, even directly north, the plane is accelerating. For circular motion, the direction of the acceleration is always toward the center of the circle. When the plane is headed directly north, the plane’s acceleration vector points directly east. See the next figure.

3-4 Chapter 3

Assess: In circular motion, there is a centripetal acceleration that always points toward the center of the circle.

Q3.16. Reason: To make a tighter turn with a smaller radius, you need to reduce your speed. If you are traveling at the greatest speed which is safe, then you are accelerating at the highest acceleration which is safe. Now if the radius is reduced, this tends to increase the acceleration above safe values. To bring it back down, your speed should be reduced. If we solve the formula for centripetal equation, 2 / ,a v r= for v, we have:

.v ar= So v is proportional to the square root of r. For example, if we take a turn with a radius which is four times smaller, we need to cut our speed in half. Assess: The maximum safe speed is proportional to the square root of the radius of the turn.

Q3.17. Reason: The longest vector will be obtained by turning C

around or by turning A

around so that the two point in the same direction. The choices A, C, and D all have A

and C

added together. B is the only choice in which the two vectors are subtracted and not added together. Since B

is perpendicular to A

and ,C

we will get a vector of the same length whether we add or subtract B

from the other two. See the figure for the lengths of the different vector combinations.

Assess: A longer vector can be created by adding two vectors which point in the same direction.

Q3.18. Reason: To generate a vector which points to the left, we could add two vectors which point left, one pointing up and the other down. In C, Q

and P−

fit this description so their sum points to the left. The various vector combinations are shown.

Assess: If two vectors have equal and opposite components in a certain direction, say the x direction, then when we add the vectors, the equal and opposite components will cancel and leave us with a vector perpendicular to that direction.

Q3.19. Reason: The gas pedal can be considered an “accelerator” because it can change the velocity of the car. So can any other controls in the car that change the velocity of the car. The brakes certainly can change the velocity by slowing the car.


The steering wheel also can change the velocity by changing the direction of travel. The gear shift can also change the speed. (Shifting to a lower gear can slow the car.) So the answer is D, all of the above. Assess: When driving on a straight road at constant speed the gas pedal is not then acting as an “accelerator.” You have to keep your foot on the gas pedal just to keep the velocity constant, and so, in that scenario, the gas pedal is a nonaccelerator. This question certainly highlights the difference in language between every day usage and the more specific physics usage. It pays to know your audience!

Q3.20. Reason: The car is traveling at constant speed, so the only possible cause for accelerations is a change in direction. (a) At point 1 the car is traveling straight to the right on the diagram, so its velocity is straight to the right. The correct choice is B. (b) At point 1 the car is traveling at constant speed and not changing direction so its acceleration is zero. The correct choice is E. (c) The car’s velocity at this point on the curve is in the direction of its motion, which is in the direction shown at choice C. (d) The car is moving on a portion of a circle. The acceleration of an object moving in a circle is always directly toward the center of the circle. The correct choice is D. (e) The car is moving on a portion of a circle at point 3. The instantaneous velocity vector is directly to the right, which is choice B. (f) The car is accelerating because it is moving on a portion of a circle. The acceleration is toward the center of the circle, which is in direction A. Assess: The instantaneous velocity of a particle is always in the direction of its motion at that point in time. For motion in a circle, the direction of the acceleration is always toward the center of the circle.

Q3.21. Reason: (a) The ball is going along the trajectory, so the best choice for the velocity at position 2 is D. (b) The direction of the acceleration is not related to the direction of the velocity—only the direction of the change in velocity. The ball is in free fall, and so its acceleration is down, just like the acceleration of all other objects in free fall. The best choice is A. This does mean, however, that the change in velocity vectors ( )v∆ must also be down. (c) The ball at position 3 is moving horizontally (the vertical component is zero for an instant), so choice C is best. (d) The same argument for the acceleration at position 2 applies at position 3; the acceleration is down. So the answer is A. Assess: Galileo taught us the law of falling bodies: “All bodies in free fall (projectiles) have the same acceleration.” The ball is in free fall at both position 2 and position 3 so the accelerations are the same, regardless of the fact that the velocities are in different directions.

Q3.22. Reason: The maximum height the ball reaches only depends on the initial velocity it had in the vertical direction. The y-component of the velocity of this ball is

i i( ) sin ( ) (23.0 m/s) sin (37.0 ) 13.8 m/syv v θ= = ° =

In order to reach the same height when being thrown vertically upward the ball’s initial velocity must be 13.8 m/s. The correct choice is A. Assess: Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction.

Q3.23. Reason: The key to projectile motion problems is to realize that the motion in the x-coordinate is independent of the motion in the y-coordinate. We can solve an equation in one of these directions and use the results in an equation for the other direction. For example, t∆ is the same for the horizontal and vertical components of the motion. (a) First find the horizontal component of the velocity, and, realizing it will be constant, find the time to impact.

i i( ) = cos = (89 m/s) cos 40 = 68.2 m/sxv v θ ° distance 300 mtime 4.4 s

speed 68.2 m/s= = =

3-6 Chapter 3

So the correct choice is C. (b) The vertical component of the initial speed is

i i( ) sin (89 m/s) sin 40 57.2 m/syv v θ= = ° =

Now that we know 4.4st∆ = and i( ) 57.2 m/syv = we can solve for h (which is yf).

2 2 2f i i

1 1( ) ( ) 0.0 m (57.2 m/s)(4.4 s) (9.8 m/s )(4.4 s) 157 m 160 m2 2yy y v t g t= + ∆ − ∆ = + − = ≈

So the correct choice is D. Assess: The answers to both parts seem reasonable; in either case if we had been off by a factor of 10 in either direction we would think the result not realistic.

Q3.24. Reason: The car drops in the vertical direction by a distance of 73 m. Since the car drives off the cliff horizontally, its initial velocity in the vertical direction is 0 m/s. Projectile motion is described by Equations 3.25. (a) To find time given distance we can use

2f i i

1( ) ( )2yy y v t g t= + ∆ − ∆

With i 0 m,y = f 73 my = − and i( ) 0 m/s.yv = Substituting these values in and solving for t∆

f2

2 2( 73 m) 3.9 s(9.80 m/s )

ytg

− − −∆ = = =

The correct choice is C. (b) The horizontal distance traveled by the car is found by multiplying xv by t∆ :

(27 m/s)(3.86 s) 104 m.xv t∆ = =

Here we have used three figures in t∆ since it is used in an intermediate calculation. To two significant figures, the car lands 100 m from the base of the cliff, so the answer is C. Assess: Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Note that the initial velocity of the car was not relevant to this problem because it was entirely in the horizontal direction.

Q3.25. Reason: “To the nearest second” means we don’t even need a calculator if we recall that 12sin30 .° = We

can also assume no air resistance. The vertical component of the initial velocity is 1i i 2( ) sin (20 m/s)( ),yv v θ= = so we

can analyze this as if the projectile had gone straight up with an initial velocity of i( ) 10 m/s.yv =

Understanding that ay = –g ≈ –10 m/s2 means that the velocity changes by 10 m/s 2 each second. So after 1 s the ball has lost 10 m/s of vertical speed, which makes its vertical speed zero (and consequently it is at the top of the trajectory). We ignore air resistance and say the motion is symmetric—it will take as long to come down as it did to go up, which is one more second. The total is therefore 2 s and the correct choice is B. Assess: The value of g is within about 2% of 10 m/s2, so when we only want one significant figure we can use that value for very easy calculations. Even when you want more significant figures, it is nice to round g to 10 m/s2 and do a quick mental calculation as an assessment check of your more precise calculator work.

Q3.26. Reason: Let us try to avoid the calculator as in Question 3.25. Since this question asks for range, we need the horizontal component of the ball’s speed. For that you need to know that cos30 3 / 2° = and that

3 1.73.≈ The horizontal component is ( ) ( )( )iicos 20 m/s 3 / 2 17.3 m/sxv v θ= = = . Since xv is constant (there

is no acceleration in the x direction), the range of the football is given by

( ) ( )( )i17.3 m/s 2 s 34.6 m 35 mxR v t= ∆ = = ≈

Because we have used the time from the previous question, this answer only has one or two significant figures. The correct choice is C.


Assess: As Questions 3.25 and 3.26 illustrate, by using 210 m/sg ≈ , we can estimate the main features of a projectile motion problem such as the time of flight and the range without using a calculator.

Q3.27. Reason: The magnitude of the centripetal acceleration is 2

cvar

=

where we want to solve for r: 2 2

2c

(4.0 m/s)= = = 8.0 m2.0 m/s

vra

This is r, but the diameter is twice the radius 16m,d = so the correct choice is D. Assess: Think about Ferris wheels you have been on. One with a radius of 16 m would be reasonable without being overly large.

Problems

P3.1. Prepare: (a) To find ,A B+

we place the tail of vector B

on the tip of vector A

and then connect vector sA′

tail with vector sB′

tip. (b) To find ,A B−

we note that ( ).A B A B− = + −

We place the tail of vector B−

on the tip

of vector A

and then connect vector sA′

tail with the tip of vector .B−

Solve:

P3.2. Prepare: (a) To find ,A B+

we place the tail of vector B


and connect the tail of

vector A

with the tip of vector .B

Solve:

(b) Since ( ),A B A B− = + −

we place the tail of the vector ( )B−


and then connect the tail of

vector A

with the tip of vector ( ).B−

P3.3. Prepare: We can find the positions and velocity and acceleration vectors using a motion diagram.

3-8 Chapter 3

Solve: The figure gives several points along the car’s path. The velocity vectors are obtained by connecting successive dots. The acceleration vectors are obtained by subtracting successive velocity vectors. The acceleration vectors point toward the center of the diagram. Assess: Notice that the acceleration points toward the center of the turn. As you will learn in chapter 4, whenever your car accelerates, you feel like you are being pushed the opposite way. This is why you feel like you are being pushed away from the center of a turn.

P3.4. Prepare: Acceleration is found by the method of Tactics Box 3.2. Solve: (a) Let iv be the velocity vector between points 0 and 1 and fv be the velocity vector between points 1 and 2.

(b) Its average speed between points 1 and 2 is probably greater than its average speed between points 0 and 1. This is because the distance between 1 and 2 is greater than the distance between 0 and 1. However, it is possible that the object traveled a greater distance between 0 and 1 than between 1 and 2 if for example it followed a very curved path between 0 and 1.

P3.5. Prepare: Acceleration is found by the method of Tactics Box 3.2. Solve: The acceleration vector at each location points directly toward the center of the Ferris wheel’s circular motion.


Assess: As we will learn later, this acceleration that is directed toward the center is called centripetal acceleration.

P3.6. Prepare: The position vector d

whose magnitude d is 10 m has an x-component of 6 m. It makes an angle θ with the +x-axis in the first quadrant. We will use trigonometric relations to find the y-component of the position vector.

Solve: Using trigonometry, dx = d cos ,θ or 6 m = (10 m) cos .θ This gives 53.1 .θ = Thus the y-component

of the position vector d

is dy = d sin (10 m)θ = sin 53.1° = 8 m. Assess: The y-component is positive since the position vector is in the first quadrant.

P3.7. Prepare: The figure below shows the components vx and vy, and the angle θ. We will use Tactics Box 3.3 to find the sign attached to the components of a vector.

Solve: We have, sin 40 ,yv v= − ° or 10 m/s sin 40 ,v− = − ° or 15.56 m/s.v =

Thus the x-component is cos 40 (15.56 m/s ) cos 40 12 m/s.xv v= ° = ° = Assess: Note that we had to insert the minus sign manually with vy since the vector is in the fourth quadrant.

3-10 Chapter 3

P3.8. Prepare: The figure below shows the components vx and vy, and the angle θ. We will use Tactics Box 3.3 to find the sign attached to the components of a vector.

Solve: (a) Since cos ,xv v θ= we have 2.5 m/s (3.0 m/s) cos θ= 1cos (2.5 m/s 30 m/s) 33.6 34 .θ −⇒ = = ° = ° (b) The vertical component is sinyv v θ= (3.0 m/s) sin 33.6= ° 1.7 m/s.=

P3.9. Prepare: The figure below shows the components and ,v v⊥

and the angle θ. We will use Tactics Box 3.3 to find the sign attached to the components of a vector.

Solve: We have x yv v v= + .v v⊥= +

Thus, cos v v θ=

(100 m/s) cos 30= ° 87 m/s.= Assess: For the small angle of 30°, the obtained value of 87 m/s for the horizontal component is reasonable.

P3.10. Prepare: Vector E

points to the left and up, so according to the Tactics Box 3.3 the components Ex and Ey are negative and positive, respectively.

Solve: (a) cos xE E θ= − and sin .yE E θ=

(b) sin xE E φ= − and cos .yE E φ= Assess: Note that the role of sine and cosine is reversed because we are using a different angle. θ and φ are complementary angles.


P3.11. Prepare: We will follow rules given in the Tactics Box 3.3.

Solve: (a) Vector d

points to the right and down, so the components xd and yd are positive and negative, respectively:

cos (100 m) cos 45 70.7 mxd d θ= = ° = sin (100 m) sin 45 71 myd d θ= − = − ° = −

(b) Vector v points to the right and up, so the components xv and yv are both positive:

cos (300 m/s) cos 20 280 m/sxv v θ= = ° = sin (300 m/s) sin 20 100 m/syv v θ= = ° =

(c) Vector a has the following components: 2 2 cos (5.0 m/s ) cos 90 0 m/sxa a θ= − = − ° = 2 2 sin (5.0 m/s ) sin 90 5.0 m/sya a θ= − = − ° = −

Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according to the Tactics Box 3.3.

P3.12. Prepare: We will follow rules given in Tactics Box 3.1.

Solve: (a) (2 km) sin 30 1 kmxd = − ° = − (2 km) cos 30 1.7 kmyd = ° =

(b) (5 cm/s) sin 90 5 cm/sxv = − ° = − (5 cm/s) cos 90 0 cm/syv = ° =

(c) 2 2(10 m/s ) sin 40 6.4 m/sxa = − ° = − 2 2(10 m/s ) cos 40 7.7 m/sya = − ° = − Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to the Tactics Box 3.3.

3-12 Chapter 3

P3.13. Prepare: We will draw the vectors to scale as best we can and label the angles from the positive x-axis (positive angles go CCW). We also use Equations 3.11 and 3.12. Make sure your calculator is in degree mode. Solve: (a)

2 2 2 2= ( ) ( ) (20 m/s) (40 m/s) 45 m/sx yv v v+ = + =

1 1 140 m/stan tan tan (2) 6320 m/s

y

x

vv

θ − − − = = = = °

(b)

2 2 2 2 2 2 2( ) ( ) (2.0 m/s ) ( 6.0 m/s ) 6.3 m/sx ya a a= + = + − = 2

1 1 12

6.0 m/stan tan tan ( 3) = 722.0 m/s

y

x

aa

θ − − − −= = = − − °

Assess: In each case the magnitude is longer than either component, as is required for the hypotenuse of a right triangle. The negative angle in part (b) corresponds to a clockwise direction from the positive x-axis.

P3.14. Prepare: We can use Equations 3.11 and 3.12 to find the magnitude and direction of a vector given its components. Solve: (a) See the following diagram.


Using Equation 3.11,

2 2 2 2( ) ( ) (10 m/s) ( 30 m/s) 32 m/sx yv v v= + = + − =


1 1| | 30 m/stan tan 7210 m/s

y

x

vv

θ − − = = = °

(b) See the following diagram.


2 2 2 2 2 2 2( ) ( ) (20 m/s ) (10 m/s ) 22 m/sx ya a a= + = + =

Using Equation 3.12, 2

1 12

10 m/stan tan 2720 m/s

y

x

aa

θ − − = = = °

Assess: Comparing the vector diagrams to the calculations, both these answers make sense.

P3.15. Prepare: Assume you start at the spot labeled home and that Strawberry Fields and Penny Lane are perpendicular. We will not assume that the lengths in the figure are to scale. We will write each vector in component form for easy addition. We will then need to take the sum and compute the magnitude and direction to report the final answer. The Strawberry Fields vector is = ( , ) = (2.0 km,0.0 km).x yS S S

The Penny Lane vector is ( , ) (0.0 km, 1.0 km).x yP P P= = −

The Abbey Road vector is

o o

( , )

( cos , sin )((4.0 km) sin (40 ),(4.0 km) cos (40 ))(2.57 km, 3.06 km)

x yA A AA Aφ φ

=

=

==

(The cos and sin are interchanged from Equation 3.10 because φ is measured from the y-axis. We have used an extra significant figure for extra accuracy.) Solve: Now add the respective components of the three vectors to get the components of the total displacement.

3-14 Chapter 3

S

= (2.0 km, 0.0 km)

P

= (0.0 km, 1.0 km)−

A

= (2.57 km, 3.06 km)

D

= (4.57 km, 2.06 km)

Now use Equations 3.11 and 3.13. 2 2 2 2= ( ) ( ) (4.57 km) (2.06 km) 5.1 kmx yD D D+ = + =

1 1 2.06 kmtan tan 244.57 km

y

x

DD

θ − − = = = °

where θ is measured ccw from the positive x-axis. Assess: Even though the figure may not be precisely to scale, it, or one you draw, would convince you that the answers for the magnitude and direction are both reasonable.

P3.16. Prepare: With air resistance and friction ignored, the acceleration down a slope is given by Equation 3.21. Assume you start from rest. Solve:

Since you are moving down the slope, your acceleration along the slope will be

2 2sin ( ) (9.80 m/s ) sin (15 ) 2.54 m/sa g θ= = ° =

The final velocity can be obtained with the equation

2f i( ) ( ) (2.54 m/s )(10 s)=25.4 m/sx xv v a t= + ∆ =

This should be reported as 25 m/s to two significant figures. From Table 1.3, 1 m/s = 2.24 mph. Converting to miles per hour, you are traveling at

2.24 mph(25.4 m/s) 57 mph1 m/s

=

Assess: This answer is reasonable. Compare to Example 3.7 in the text.

P3.17. Prepare: A visual overview of the car’s motion that includes a pictorial representation, a motion diagram, and a list of values is shown below. We have labeled the x-axis along the incline. Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with acceleration a = –g sin θ = –g sin 5.0° = –0.854 m/s2.


Solve: Constant acceleration kinematics gives

2 22 2 2 if i f i i f f 2

(30 m/s)2 ( ) 0 = 2 530 m2 2( 0.854 m/s )vv v a x x v ax xa

= + − ⇒ + ⇒ = − = − =−

Notice how the two negatives canceled to give a positive value for xf. Assess: We must include the minus sign because the a vector points down the slope, which is in the negative x-direction.

P3.18. Prepare: We can find the acceleration on the ramp and then use the acceleration to find the final velocity of the car.

Solve: (a) The maximum possible acceleration is given by the formula sin .xa g θ= Plugging in the values of

g and ,θ we get 22.2 m/s .xa =

(b) The final velocity can be obtained from the formula 2 2( ) ( ) 2x f x i xv v a x− = ∆ , using ( ) 0 m/sx iv = and

16.8 m,x∆ = the latter being obtained by converting 55 ft: 1 m55 ft 55 ft3.28 ft

=

. 1 m55 ft 55 ft3.28 ft

=

The

solution to the first equation is ( ) 8.6 m/s.x fv =

Assess: As with most ramp problems, this one was best solved by using a rotated coordinate system with the x-axis along the ramp.

P3.19. Prepare: Make a sketch with tilted axes with the x-axis parallel to the ramp and the angle of inclination labeled. We must also make a bold assumption that the piano rolls down as if it were an object sliding down with no friction.

As part of the preparation, compute the length of the ramp in the new tilted x-y coordinates.

o

1.0 m 2.9 msin sin 20

hLθ

= = =

3-16 Chapter 3

The acceleration in the new coordinate system will be 2 2sin (9.8 m/s ) sin 20 3.4 m/s .xa g θ= = ° = Solve: Since this is a case of constant acceleration we can use the second equation from Table 2.4 with xi = 0.0 m and (vx)i = 0.0 m/s.

2f

1= ( )2 xx a t∆

Solve for ,t∆ and use xf = 2.9 m and ax = 3.4 m/s2, which we obtained previously.

f2

2 2(2.9 m) 1.3 s3.4 m/sx

xta

∆ = = =

Assess: They may catch it if they have quick reactions, but the piano will be moving 4.5 m/s when it reaches the bottom.

P3.20. Prepare: The acceleration down a ramp is given by Equation 3.21. The acceleration of each car depends on the angle of the ramp. The final velocity will depend on the acceleration and the length of the ramp. The cars start from rest, so their initial velocities are all 0 m/s. Solve: The acceleration down a ramp is given by a = g sin ( ).θ The final velocity of the cars can be calculated

with Equation 2.13, 2 2f i( ) ( ) 2 ,x x xv v a x= + ∆ with i( ) 0 m/s.xv = Solving for f( ) ,xv f( ) 2 .x xv a x= ∆

For car A, 2 2sin ( ) (9.80 m/s ) sin (15 ) 2.54 m/sa g θ= = ° =

2f( ) 2 2(2.54 m/s )(10 m) 7.1 m/sx xv a x= ∆ = =

For car B, 2 2sin ( ) (9.80 m/s ) sin (20 ) 3.35 m/sa g θ= = ° =

2f( ) 2 2(3.35 m/s )(10 m) 8.2 m/sx xv a x= ∆ = =

For car C, 2 2sin ( ) (9.80 m/s ) sin (20 ) 3.35 m/sa g θ= = ° =

2f( ) 2 2(3.35 m/s )(8.0 m) 7.3 m/sx xv a x= ∆ = =

For car D, 2 2sin ( ) (9.80 m/s ) sin (12 ) 2.04 m/sa g θ= = ° =

2f( ) 2 2(2.04 m/s )(12 m) 7.0 m/sx xv a x= ∆ = =

The car with the greatest speed at the bottom of the ramp is car B. Assess: The answer makes sense. Car B and C are both on the ramp with the highest inclination out of all four cars, while car B travels a longer distance than car C.

P3.21. Prepare: For everyday speeds we can use Equation 3.22 to find relative velocities. We will use a subscript A for Anita and a 1 and a 2 for the respective balls; we also use a subscript G for the ground. We will consider all motion in this problem to be along the x-axis (ignore the vertical motion including the fact that the balls also fall under the influence of gravity) and so we drop the x subscript. It is also worth noting that interchanging the order of the subscripts merely introduces a negative sign. For example,

=AG 5 m s,v so vGA = –5 m/s. “According to Anita” means “relative to Anita.” Solve: For ball 1:

1A 1G G 10 m/s ( 5 m/s) 5 m/sAv v v= + = + − =

For ball 2:

2A 2G G 10 m/s ( 5 m/s) 15 m/sAv v v= + = − + − = −


The speed is the magnitude of the velocity, so the speed of ball 2 is 15 m/s. Assess: You can see that at low speeds velocities simply add or subtract, as the case may be. Mentally put yourself in Anita’s place, and you will confirm that she sees ball 1 catching up to her at only 5 m/s while she sees ball 2 speed past her at 15 m/s.

P3.22. Prepare: We can use the technique of “canceling” subscripts to find relative velocities. Solve: Anita’s friends are standing on the ground, so we can calculate the velocities they threw the balls with by calculating the velocities of the balls relative to the ground. The velocity of ball 1 relative to Anita is (vx)1A = +10 m/s. The velocity of ball 2 relative to Anita is (vx)2A = –10 m/s. Anita’s velocity relative to the ground is (vx)Ag = +5 m/s. Then the velocity of ball 1 relative to the ground is

1g 1A Ag( ) ( ) ( ) 10 m/s 5 m/s 15 m/sx x xv v v= + = + + = +

The velocity of ball 2 relative to the ground is

2g 2A Ag( ) ( ) ( ) 10 m/s 5 m/s 5 m/sx x xv v v= + = − + = −

Assess: The results make sense. The ball to the left of Anita must be traveling faster than Anita, and the ball to the right must be traveling slower than Anita.

P3.23. Prepare: Assume motion along the x-direction. The velocity of the boat relative to the ground is (vx)bg; the velocity of the boat relative to the water is (vx)bw; and the velocity of the water relative to the ground is (vx)wg. We will use the technique of Equation 3.22: bg bw wg( ) ( ) ( ) .x x xv v v= +

Solve: For travel down the river,

bg bw wg30 km( ) ( ) ( ) 10.0 km/hr3.0 hrx x xv v v= + = =

For travel up the river,

bg bw wg30 km( ) ( ) ( ) 6.0 km/hr5.0 hrx x xv v v = − + = − = −

Adding these two equations yields wg( ) 2.0 km/hr.xv = That is, the velocity of the flowing river relative to the earth is 2.0 km/hr. Assess: Note that the speed of the boat relative to the water downstream and upstream are the same. P3.24. Prepare: We can find the relative velocities using the subscript cancellation formula, typified by Equation 3.24. Solve: The first drawing in the figure below gives the first part of the race and the second drawing shows the second part.

(a) For the first part of the trip, Phillippe’s velocity relative to the sidewalk is PS( ) 2.0 m/s,xv = his velocity relative to the floor can be found with subscript cancellation:

3-18 Chapter 3

Pf PS Sf( ) ( ) ( ) 2.0 m/s 1.5 m/s 3.5 m/sx x xv v v= + = + =

For the second part of the trip, Phillippe’s velocity relative to the sidewalk is PS( ) 2.0 m/sxv = − and his velocity relative to the floor is given by:

Pf PS Sf( ) ( ) ( ) 2.0 m/s 1.5 m/s 0.5 m/sx x xv v v= + = − + = −

The time it takes Phillippe to finish the race can be found using the formula /t x v∆ = ∆ for the two parts of the race:

P20 m 20 m 46 s

3.5 m/s 0.5 m/st −

∆ = + =−

The reason for the negative sign in front of the 20 m in the second fraction is that during the second leg of the trip, Phillippe is going back to his original position. Since his velocity relative to the floor is also negative, the time for that portion of the race comes out positive as it should. Renee’s velocity relative to the floor is 20 m/s for the first part of the trip and 20 m/s− for the second part. The time it takes Renee to finish the race is as follows:

P20 m 20 m 20 s

2.0 m/s 2.0 m/st −

∆ = + =−

Renee wins the race. Her time is 10 s to the end and 10 s back. Her total time is 20 s. Phillippe, on the other hand takes only 5.7 s to make it to the end but 40 s to make it back, for a total time of 46 s. (b) Renee wins by 26 s. Assess: This problem is reminiscent of the story of the tortoise and the hare. Renee with her constant speed outperforms Phillippe whose performance was exceptional at first and then less so.

P3.25. Prepare: First we can find the velocity of the skydiver with respect to the ground using the idea of cancelling subscripts. Then, knowing the components of this vector, we can find the angle of the vector with the vertical.

Solve: (a) The velocity of the skydiver relative to the ground, SG ,v can be found since we know his velocity relative to the air: SA (0, 5.0 m/s)v = −

and the velocity of the air relative to the ground, AG ( 2.0 m/s, 0 m/s)v = − :

SG SA AG (0 m/s, 5.0 m/s) + (–2.0 m/s, 0 m/s) = (–2.0 m/s, –5.0 m/s)v v v= + = −

The figure shows the three velocity vectors. The skydiver’s velocity vector relative to the ground forms the angle φ with the vertical where:

1 1 2.0 m/stan tan 225.0 m/s

x

y

vv

φ − − = = =

He falls at an angle of 22 to the vertical.

(b) The time the skydiver takes to fall is given by


1000 m 200 s5.0 m/sy

ytv∆ −

∆ = = =−

The wind causes the skydiver to drift horizontally at a rate of 2.0 m/s and he falls for 200 s. By the time he lands, he has drifted through a distance D, which is a product of drift speed and time:

(2.0 m/s)(200 s) 400 mD = =

He will miss his desired landing spot by 400 m. Assess: Notice that the skydiver fell at a constant speed whereas an object in free fall accelerates. This is due to the presence of air resistance. The constant speed at which an object falls when there is air resistance is called “terminal speed.”

P3.26. Prepare: The object is undergoing projectile motion as illustrated in Figure 3.32. Solve: (a) The components of vi are

i i( ) cos (50 m/s) cos 36.9 40.0 m/sxv v θ= = ° =

i i( ) sin (50 m/s) sin 36.9 30.0 m/syv v θ= = ° =

Since (vx)i is constant, the position x increases by 40.0 m every second. The y-velocity and the y-position are obtained from the following equations:

f i f i( ) ( ) ( )y y yv v a t t= + − 21f i i f i f i2( ) ( ) ( )y yy y v t t a t t= + − + −

Thus, and so on.

The table showing x, y, vx, vy and v (from t = 0 s to t = 6 s) is given as follows.

tf – ti (s) x (m) y (m) vx (m/s) vy (m/s) v (m/s) 0 0 0 40 30.0 50.0 1 40 25.1 40 20.2 44.8 2 80 40.4 40 10.4 41.3 3 120 45.9 40 0.6 40.0 4 160 41.6 40 −9.2 41.0 5 200 27.5 40 −19.0 44.3 6 240 3.6 40 −28.8 49.3

(b)

Assess: As expected the trajectory is a parabola. Given an initial velocity of 50 m/s, a value of x = 240 m for t = 6 s is reasonable.

2f f i f i

2 2 21 1f f i i f i f i2 2

f f i f i

At 1.0 : ( ) ( ) ( ) (30.0 m/s) ( 9.8 m/s )(1 s 0 s) 20.2 m/s

At 1.0 : ( ) ( ) ( ) 0 m (30.0 m/s)(1 s 0 s) ( 9.8 m/s )(1 s 0 s) 25.1 m

At 2.0 : ( ) ( ) ( )

y y y

y y

y y y

t s v v a t t

t s y y v t t a t t

t s v v a t t

= = + − = + − − =

= = + − + − = + − + − − =

= = + − = 2

2 2 21 1f f i i f i f i2 2

(30.0 m/s) ( 9.8 m/s )(2 s 0 s) 10.4 m/s

At 2.0 : ( ) ( ) ( ) 0 (30.0 m/s)(2 s 0 s) ( 9.8 m/s )(2 s 0 s) 40.4 my yt s y y v t t a t t

+ − − =

= = + − + − = + − + − − =

3-20 Chapter 3

P3.27. Prepare: We will assume the ball is in free fall (i.e., we neglect air resistance). The trajectory of a projectile is a parabola because it is a combination of constant horizontal velocity (ax = 0.0 m/s2) combined with constant vertical acceleration (ay = –g). In this case we see only half of the parabola. The initial speed given is all in the horizontal direction, that is, (vx)i = 5.0 m/s and (vy)i = 0.0 m/s. Solve: (a)

(b)

(c)

(d) This is a two-step problem. We first use the vertical direction to determine the time it takes, then plug that result into the equation for the horizontal direction.

21 ( )2 yy a t∆ = ∆

2

2 2( 20 m) 2.0 s9.8 m/sy

yta∆ −

∆ = = =−

We we use the 2.0 s in the equation for the horizontal motion.

(5.0 m/s)(2.0 s) 10 mxx v t∆ = ∆ = =

Assess: The answers seem reasonable, and we would get the same answers to two significant figures in a quick mental calculation using 210m/s .g ≈ In fact, I did this before computing the algebra so I would know how to scale the graphs.

P3.28. Prepare: We can use the vertical part of the motion to calculate the time it takes the ball to hit the floor and the horizontal part of the motion to find out how far from the bench it lands. Solve: Refer to the visual overview shown.

The initial vertical velocity is zero. Take the floor as the origin of coordinates. The ball falls from yi = 1.00 m and lands at yf = 0 m. (a) We can use the vertical-position equation from Equations 3.25 to find the time it takes the ball to reach the floor.

2f i i

1( ) ( )2yy y v t g t= + ∆ − ∆

2 210.00 m 1.00 m (9.80 m/s )( )2

t= − ∆

Solving for ,t∆


2

2( 1.00m) 0.452s9.80m/s

t − −∆ = =

(b) The distance the ball travels horizontally is governed by the horizontal-position equation from Equations 3.25.

f i i( ) (1.25 m/s)(0.452 s) = 0.565 mxx x v t= + ∆ =

Assess: This seems reasonable. Compare to Example 3.11 in the text, which is similar.

P3.29. Prepare: This problem is asking us for a range so we need the horizontal component of velocity, ,xv and the time of flight, .t∆ The time of flight, in turn, depends on the initial vertical component of velocity, ( )y iv and the overall vertical displacement of the rock, .y∆

Solve: We first find xv and ( )y iv using Equations 3.25:

cos (25 m/s) cos 30 21.7 m/s

(v ) sin (25 m/s) sin 30 12.5 m/sx i

y i i

v v

v

θ

θ

= = =

= = =

Using the equation which relates vertical velocity, acceleration and displacement: 2 2( ) ( ) 2y f y i yv v a y− = ∆ , we can find the final vertical component of velocity.

2 2 2( ) (12.5 m/s) 2( 9.8 m/s )( 12 m)y fv − = − −

This equation has two solutions: ( ) 19.8 m/sy fv = ± . We choose the negative solution since the rock is

descending when it lands. Now we can find the time of flight using /y ya v t= ∆ ∆ :

2/ ( 19.8 m/s 12.5 m/s) / ( 9.8 m/s ) 3.29 sy yt v a∆ = ∆ = − − − =

Finally, the range of the rock is given by xx v t∆ = ∆ :

(21.7 m/s)(3.29 s) 71 mxx v t∆ = ∆ = =

The rock lands 71 m from the castle wall. Assess: Notice that the mass of the rock was not involved. So if this had been a baseball thrown from a 12 m tower, it would have gone equally far—about three quarters of a foot ball field.

3-22 Chapter 3

P3.30. Prepare: We will apply the constant-acceleration kinematic equations to the horizontal and vertical motions as described by Equations 3.25.

Solve: Using 21

f A iA y i A fA iA A fA iA2( ) ( ) ( ) ( ) ,yy y v t t a t t= + − + − we get

2 21fA21.0 m 0 m 0 m ( 9.8 m/s )( 0s)t− = + + − − fA f B0.452 st t⇒ = =

Because yfA = y fB, so both take the same time to reach the floor. We are now able to calculate xfA and xfB as follows:

21fA iA i A fA iA A fA iA2

21fB iB i B fB iB B fB iB2

( ) ( ) ( ) ( ) 0 m (5.0 m/s)(0.452 s 0 s) 0 m 2.3 m

( ) ( ) ( ) ( ) 0 m (2.5 m/s)(0.452 s 0 s) 0 m 1.1 mx x

x x

x x v t t a t t

x x v t t a t t

= + − + − = + − + =

= + − + − = + − + =

Assess: Note that tfB = tfA since both the spheres move with the same vertical acceleration and both of them start with zero vertical velocity. The horizontal distance for sphere B is one-half the distance for sphere A because the horizontal velocity of sphere B is one-half that of A.

P3.31. Prepare: We will apply the constant-acceleration kinematic equations to the horizontal and vertical motions as described by Equations 3.25. The effect of air resistance on the motion of the bullet is neglected.

Solve: (a) Using 21f i i f i f i2( ) ( ) ( ) ,y yy y v t t a t t= + − + − we obtain

2 2 21f2( 2.0 10 m) 0 m 0 m ( 9.8 m/s )( 0 s)t−− × = + + − − f 0.0639 st⇒ =

(b) Using 21f i i f i f i2( ) ( ) ( ) ,x xx x v t t a t t= + − + −

i(50 m) 0 m ( ) (0.0639 s 0 s) 0 mxv= + − + i( ) 782 m/sxv⇒ =

Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a value of 782 m/s is not surprising.


P3.32. Prepare: We are asked to find the take-off speed and horizontal speed of the kangaroo given its initial angle, 20 , and its range. Since the horizontal speed is given by cosxv v θ= and the time of flight is given by 2 sin / ,t v gθ∆ = the range of the kangaroo is given by the product of these: 2 sin cos / .x v gθ θ∆ = Solve: (a) We can solve the above formula for v and then plug in the range and angle to find the take-off speed:

( ) ( )2/ 2 sin cos 9.8 m/s (10 m)/(2 sin 20 cos 20 ) 12.3 m/sv g x θ θ= ∆ = =

Its take-off speed is 12 m/s, to two significant figures. (b) Its horizontal speed is given by cos (12.3 m/s) cos 20 11.6 m/sxv v θ= = = or 12 m/s to two significant figures. Assess: The reason the horizontal speed and take-off speed appear the same is that 20 is a small angle and the cosine of a small angle is approximately equal to 1.

P3.33. Prepare: The golf ball is a particle following projectile motion. We will apply the constant-acceleration kinematic equations to the horizontal and vertical motions as described by Equations 3.25.

Solve: (a) The distance traveled is xf = (vi)xtf = vi cos θ × tf. The flight time is found from the y-equation, using the fact that the ball starts and ends at y = 0:

2 i1 1f i i f f f f f2 2

2 sin0 sin ( sin )ivy y v t gt v gt t t

gθθ θ− = = − = − ⇒ =

Thus the distance traveled is 2

i if i

2 sin 2 sin coscos v vx vg g

θ θ θθ= × =

For θ = 30°, the distances are 2 2i

f earth 2earth

2 2 2i i i

f moon f earth1moon earth earth6

2 sin cos 2(25 m/s) sin 30 cos 30( ) 55.2 m9.80 m/s

2 sin cos 2 sin cos 2 sin cos( ) 6 6( ) 331.2 m

vxg

v v vx xg g g

θ θ

θ θ θ θ θ θ

° °= = =

= = = × = =

The flight times are

if earth

earth

i if moon f earth1

moon earth6

2 sin( ) 2.55 s

2 sin 2 sin( ) 6( ) 15.30 s

vtgv vt tg g

θ

θ θ

= =

= = = =

The ball spends 15.30 s – 2.55 s = 12.75 s = 13 s longer in flight on the moon. (b) From part (a), the distance traveled on the moon is 331 m or 330 m to two significant figures. (c) From part (a), the golf ball travels 331.2 m – 55.2 m = 276 m farther on the moon than on earth.

P3.34. Prepare: We can use Equation 3.26. Solve: (a) Converting revolutions per minute to revolutions per second

1 revolutions 1 minute33 0.56 rev/s3 minute 60 s

=

3-24 Chapter 3

(b) Using Equation 3.26

1 1 1.8 s0.56 rev/s

Tf

= = =

Assess: This seems reasonable, if you’re old enough to remember LPs.

P3.35. Prepare: We need to convert the 5400 rpm to different units and then find the period which is the inverse of frequency. Solve: (a) The hard disk’s frequency can be converted as follows:

rev rev 1 min rev5400 5400 90min min 60 sec sec

= =

Its frequency is 90 rev/s. (b) Rewriting Equation 3.26, we have the following:

1 1 11 ms90 rev/s

Tf

= = =

Its period is 11 ms. Assess: This is about the rate that the engine in a car turns if it is straining. So an automobile engine completes a cycle every 10 or 20 ms.

P3.36. Prepare: We can use the formula for centripetal acceleration, Equation 3.30. We need the radius of the track which is half the diameter: / 2 (45 m)/2 = 22.5 m.R D= = Solve:

2 2

2(15 m/s) 10 m/s22.5 m

var

= = =

We now convert this acceleration to units of g:

2 2210 m/s 10 m/s 1.0

9.8 m/sg g= =

Assess: The greyhounds are accelerating at around the acceleration of free fall!

P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled by the speck in one period. Then the acceleration is given by Equation 3.30. Solve: (a) The disk’s frequency can be converted as follows:

rev rev 1 min rev10,000 10,000 167 min min 60 sec sec

= =

The period is the inverse of the frequency:

1 1 6.00 ms167 rev/s

Tf

= = =

(b) The speed of the speck equals the circumference of its orbit divided by the period:

2 2 (6.0 cm) 1000 ms 1 m 62.8 m/s,6.00 ms 1 s 100 cm

rvTπ π = = =

which rounds to 63 m/s.

(c) From Equation 3.30, the acceleration of the speck is given by 2 /v r :

2 2

2(62.8 m/s) 100 cm 65,700 m/s ,6.0 cm 1 m

var

= = =


Which rounds to 266,000 m/s . In units of g, this is as follows:

2 22

165,700 m/s 65,700 m/s 6,7009.8 m/s

g g = =

Assess: The speed and acceleration of the edge of a CD are remarkable. The speed, 63 m/s, is about 140 mi/hr. As you will learn in chapter 4, very large forces are necessary to create large accelerations like 6,700 .g

P3.38. Prepare: The acceleration the rider experiences in the centrifuge is a centripetal acceleration. Solve: Since a = v2/r, we have

2 2(98 m/s )(12 m) 34 m/sv ar v= = ⇒ =

Assess: 34 m/s ≈ 70 mph is a large yet understandable speed.

P3.39. Prepare: Examine the formula carefully. 2

= var

Before plugging in numbers, notice that if the speed is held constant (as in part (a)), then a and r are inversely proportional to each other: doubling one halves the other. And if r is held constant (as in part (b)), then there is a square relationship between a and v: doubling v quadruples a. Solve: It is convenient to use ratios to solve this problem, because we never have to know any specific values for r or v. We’ll use unprimed variables for the original case (a = 8.0 m/s2), and primed variables for the new cases. (a) With the speed held constant, =v v′ but = 2 .r r′

2 2

2 22 1

2

v vr rv vr r

aa

′′′

= = =

So 2 21 12 2 (8.0 m/s ) 4.0 m/s .a a′ = = =

(b) With the radius held constant, =r r′′ but = 2 .v v′′ 22

2 2

(2 )22 4

vvr rv vr r

aa

′′′′′′

= = = =

So 2 24 4(8.0 m/s ) 32 m/s .a a′′ = = = Assess: Please familiarize yourself with this ratio technique and look for opportunities to use it. Again, the advantage is not needing to know any specific values of r or v—not only not having to know them, but realizing that the result is independent of them. The daily life lesson is that driving around a curve with a larger radius produces gentler acceleration. Going around a given curve faster, however, requires a much larger acceleration (produced by the friction between the tires and the road), and the relationship is squared. If your tires are bald or the road slippery there won’t be enough friction to keep you on the road if you go too fast. And remember that it is a squared relationship, so going around a curve twice as fast requires four times as much friction.

P3.40. Prepare: From Equation 3.30, centripetal acceleration is directly proportional to velocity squared. Solve: Since a is proportional to 2v , if v is doubled, a is quadrupled to 212 m/s . One way to see this is to look at the ratio of the velocity after the doubling to the velocity before the doubling and also the acceleration after the doubling to the acceleration before the doubling:

22 22 2 2 2

2 21 1 1 1

//

a v r v va v r v v

= = =

This tells us that if we increase v by any factor, a will be increased by the square of that factor. Assess: This shows how important it is to avoid making turns too fast. Centripetal acceleration depends very sensitively on speed. As you will learn in chapter 5, the centripetal acceleration of a car is provided by friction and if the friction cannot provide the needed acceleration, 212 m/s in this case, the car will run off the road.

3-26 Chapter 3

P3.41. Prepare: The magnitude of centripetal acceleration is given in Equation 3.30. Solve: The centripetal acceleration is given as 1.5 times the acceleration of gravity, so

2 2(1.5)(9.80 m/s ) 15 m/sa = =

Using Equation 3.30, the radius of the turn is given by 2 2

2

(20 m/s) 27 m15 m/s

vra

= = =

Assess: This seems reasonable.

P3.42. Prepare: The vectors , ,A B

and C A B= +

are shown. Because x yA A A= +

and ,x yB B B= +

so the

components of the resultant vector are x x xC A B= + and .y y yC A B= +

Solve: (a) With Ax = 5, Ay = 2, Bx = –3, and By = –5, we have Cx = 2 and Cy = –3. (b) Vectors ,A

,B

and C

are shown with their tails together.

(c) Since 2,xC = and 3,yC = − the magnitude and direction of C

are 2 2(2) ( 3) 3.6.C = + − =

1 1| | 3tan tan 562

y

x

CC

θ − − = = = °

Assess: The vector C

is to the right and down, thus implying a negative y-component and positive x-component, as obtained above. Also 45θ > ° since | | | | .y xC C>


and D A B= −

are shown. Because x yA A A= +

and ,x yB B B= +

so

the components of the resultant vector are x x xD A B= − and .y y yD A B= −


Solve: (a) With Ax = 5, Ay = 2, Bx = –3, and By = –5, we have Dx = 8 and Dy = 7. (b) Vectors ,A

B

and D

are shown in the above figure.

(c) Since 8xD = and 7,yD = the magnitude and direction of D

are

2 2 1 1 7(8) (7) 11 tan tan 418

y

x

DD

Dθ − − = + = = = = °

Assess: Since | | | |,y xD D< the angle θ is less than 45°, as it should be.

P3.44. Prepare: Because ,x yA A A= +

and ,x yB B B= +

so the components of the resultant vector are Ex = 2Ax +

3Bx and Ey = 2Ay + 3By. The vectors , ,A B

and 2 3E A B= +

are shown.

Solve: (a) With Ax = 5, Ay = 2, Bx = –3, and By = –5, we have Ex = 1 and Ey = –11. (b) Vectors ,A

,B

and E

are shown in the above figure.

(c) From the E

vector, 1xE = and 11.yE = − Therefore, the magnitude and direction of E

are

2 2(1) ( 11) 11E = + − = 1 1 1tan tan 5.2| | 11

x

y

EE

φ − − = = = °

Assess: Note that φ is the angle made with the y-axis, and that is why 1tan ( /| | )x yE Eφ −= rather than 1tan (| |/ )y xE E− which would be the case if φ were the angle made with the x-axis.

P3.45. Prepare: Refer to Figure P3.45 in your textbook. Because ,x yA A A= +

,x yB B B= +

and ,x yC C C= +

so the components of the resultant vector are x x x xD A B C= + + and .y y y yD A B C= + + Dx and Dy are given and we

will read the components of A

and C

off Figure P3.45.

Solve: (a) Ax = 4, Cx = 0, and Dx = 2, so Bx = Ax – Cx + Dx = –2. Similarly, Ay = 0, Cy = –2, and Dy = 0, so By = –Ay – Cy + Dy = 2.

3-28 Chapter 3

(b) With the components in (a), 2 2( 2) (2) 2.8B = − + =

1 1 2tan tan 45| | 2

y

x

BB

θ − −= = = °

Since B

has a negative x-component and a positive y-component, the angle θ made by B

is with the –x-axis and it is above the –x-axis. Assess: Since | | | | ,y xB B= θ = 45° as is obtained above.


and C

are shown. We will first calculate the x- and y-components of each

vector and then obtain the magnitude and the direction of the vector .D

Solve: (a) The vectors , ,A B

and C

are drawn above.

(b) The components of the vectors , ,A B

and C

are (3 m) cos 20 2.82 mxA = ° = and (3 m) sin 20yA = − ° =

1.03 m;− 0 mxB = and 2 m;yB = (5 m) cos 70 1.71 m and (5 m) sin 70 4.70 m.x yC C= − ° = − = − ° = −

(c) We have ,x yD A B C D D= + + = +

which means Dx = 1.11 and 3.73yD = − .

2 2 1 13.73(1.11 m) (3.73 m) 3.9 m tan tan 3.36 731.11

D θ − −= + = = = = °

The direction of D

is south of east, 73° below the positive x-axis.

P3.47. Prepare: We need to draw a right triangle which represents the staircase. The hypotenuse is given by multiplying speed by distance and then the other two sides are given from trigonometric ratios.

Solve: The distance traveled, D, is speed multiplied by time: (3.5 m/s)(2.0 s)=7.0 m.D = This is the hypotenuse

of the staircase. The height gained is opposite the 38 angle and so is obtained using the sine function:

(7.0 m) sin 38 4.3 my∆ = =

The horizontal distance travelled is adjacent to the 38 angle, so we need to use the cosine function.


(7.0 m/s) cos 38 5.5 mx∆ = =

Assess: It makes sense that the horizontal distance traveled is greater than the height gained since the angle of the stairs, 38 ,° is less than 45 .° At that angle, the vertical and horizontal distances would have been equal.

P3.48. Prepare: The minute hand of the watch is shown in the figure.

Solve: (a) We have 8:00 (2.0 cm, north)S =

and 8:20 (2.0 cm cos 30 , east) (2.0 cm sin 30 , south).S = ° + °

The displacement vector is

8:20 8:00 (1.74 cm, east) (3.00 cm, south)r S S∆ = − = +

(b) We have 8:00 (2.0 cm, north)S =

and 9:00 (2.0 cm, north).S =

The displacement vector is 9:00 8:00r S S∆ = −

= 0. Assess: The displacement vector in part (a) has positive x-component (toward east) and negative y-component (toward south). The vector thus is to the right and points down, in the IV quadrant. This is what the vector drawn from the tip of the 8:00 A.M. arm to the tip of the 8:20 A.M. arm will be.

P3.49. Prepare: In the coordinate system shown below, the mouse starts from the origin, so his initial position vector is zero. His net displacement is then just the final position vector, which is just the sum of the three vectors , , and .A B C

Solve: We are given (5 m, east)A =

and (1 m, down).C =

Using trigonometry,

(3 cos 45 m, east) (3 sin 45 m, south).B = ° + °

The total displacement is (7.12 m, east) (2.12 m, south) (1 m, down).r A B C= + + = + +

The magnitude of r is

2 2 2(7.12) (2.12) (1) m 7.5 mr = + + =

Assess: A displacement of 7.5 m is a reasonable displacement.

3-30 Chapter 3

P3.50. Prepare: The ideas of vector addition and subtraction can be used here to find the pilot’s third displacement. Solve: See the following diagram. We place the origin of coordinates at the origin of the plane’s trip. Note that northeast means exactly 45° north of east.

The three legs of his trip are labeled 1,D

2 ,D

and 3.D

His net displacement is labeled by .R

His trip and net displacement can be described by the vector equation

1 2 3R D D D= + +

We need to calculate 3.D

Solving the equation for 3D

we find

3 1 2D R D D= − −

We can use the technique of algebraic addition to solve this equation. The components of each of the vectors on the right hand side of the equation are

1 1

1 1

2

2

0.00 km70.0 km

cos ( ) (26.0 km) cos (45 ) 18.4 kmsin ( ) (26.0 km) sin (45 ) 18.4 km

0.00 km45.0 km

x

y

x

y

x

y

RR

D DD DDD

θθ

==

= = ° == = ° =

==

We can calculate the components of 3D

by subtracting the components of 1D

and 2D

from the components of .R

3 1 2

3 1 2

0.00 km 18.4 km 0.00 km 18.4 km70.0 km 18.4 km 45.0 km 6.62 km

x x x x

y y y y

D R D DD R D D

= − − = − − = −= − − = − − =

The vector is drawn as shown.


The magnitude of 3D

is ( ) ( )223 3 3 19.5 km.x yD D D= + = From the previous diagram, the direction of 3D

is

31 1

3

6.62 kmtan tan 20 north of west| | 18.4 km

y

x

DD

θ − − = = = °

Assess: This is reasonable. The pilot must travel west to arrive directly north of his starting point, and must have continued north for a few kilometers to be so far north of his starting point.

P3.51. Prepare: We draw a picture of the plane’s path. The distance traveled by the plane is given by speed multiplied by time and forms the hypotenuse. The shortest distance to the equator is 100 km and this is the length of the side adjacent to the 30 angle. We can relate these two sides of a triangle using trigonometry.

Solve: If D is the distance traveled by the airplane, then we can write cos30 (100 km)/D= and solve for the distance traveled: (100 km)/cos 30 115 kmD = =

The time of flight is equal to the distance traveled divided by speed:

(115 km)/(150 km/hr) 0.767 hrt∆ = =

Finally, we convert this time to minutes:

60 min0.767 hr 0.767 hr 46 min1 hr

= =

It takes the pilot 46 min to reach the equator.

3-32 Chapter 3

Assess: The pilot’s time to reach the equator is greater than it would be if he were flying directly toward the equator. If the flight were direct, the distance would be 100 km instead of 115 km and the time of flight would be 40 min instead of 46 min.

P3.52. Prepare: We will need to add the individual displacements as vectors to calculate the bacterium’s net displacement. Solve: (a) The components of the displacement vectors can be read directly from Figure P3.52.

50 m 10 m

0 m 10 m

40 m 10 m

50 m 50 m

ABx ABy

BCx BCy

CDx CDy

DEx DEy

D D

D D

D D

D D

µ µ

µ µ

µ µ

µ µ

= + = +

= = +

= + = +

= − = −

Note both of the components of 4D

are negative. Note also that the components are read from the tail to the tip of the vector relative to the tail. The bacterium is moving with constant speed 20 m/sv µ= along each portion of the path. We will calculate the angles in the following figure in order to calculate components of the velocity. During each segment of the motion, the bacterium is headed exactly in the direction of its displacement.

For ,AB

D

the angleAB

θ indicated is 1tan (10 m/50 m) = 11AB

θ µ µ−= ° so the components of the velocity

vectorAB

v are

cos ( ) (20 m/s) cos (11 ) 19.6 m/s

sin ( ) (2 m/s) sin (11 ) 3.82 m/sABx AB

ABy AB

v v

v v

θ µ µ

θ µ µ

= = ° =

= = ° =

The displacementBC

D

has no x-component, so the components of velocity vectorCD

v are

0.00 m/s

20.0 m/sBCx

BCy

v

v

µ

µ

=

=

For ,CD

D

the angle CD

θ indicated is 1tan (10 m/40 m) = 14CD


vectorCD

v are

cos ( ) (20 m/s) cos (14 ) 19.4 m/s

sin ( ) (20 m/s) sin (14 ) 4.84 m/sCDx CD

CDy CD

v v

v v

θ µ µ

θ µ µ

= = ° =

= = ° =

For ,DE

D

the angle DE

θ indicated is 1tan (50 m/50 m)= 45DE


vectorDE

v are


cos ( ) (20 m/s) cos (45 ) 14.1 m/s

sin ( ) (20 m/s) sin (45 ) 14.1 m/sDEx DE

DEy DE

v v

v v

θ µ µ

θ µ µ

= − = − ° = −

= − = − ° = −

Additional significant figures have been kept in reporting the components. Each should be reported to two significant figures. (b) We need to calculate the length of each displacement vector using the components obtained in part (a).

( ) ( )

( ) ( )( ) ( )

222 2 2

222 2 2

22 2 2

(50 m) (10 m) 51.0 m

10 m

(40 m) (10 m) 41.2 m

( 50 m) ( 50 m) 70.7 m

AB ABx ABy

BC

CD CDx CDy

DE DEx DEy

D D D

D

D D D

D D D

µ µ µ

µ

µ µ µ

µ µ µ

= + = + =

=

= + = + =

= + = − + − =

Additional significant figures have been reported in these results for use later. Each of these distances should be reported to two significant figures. The total distance traveled is the sum of these distances, 51.0 m 10 m 41.2 m 70.7 m=172.9 m,µ µ µ µ µ+ + + which should be reported as170 mµ to two significant figures. We will use the more precise result in part (c). The magnitude of the net displacement can be found from the components of the vector sum of the four vectors in the diagram. The components of the resultant D

are the sum of the components of the displacements found in part (a).

50 m 0.0 m 40 m 50 m 40 m

10 m 10 m 10 m 50 m 20 mx ABx BCx CDx DEx

y ABy BCy CDy DEy

D D D D D

D D D D D

µ µ µ µ µ

µ µ µ µ µ

= + + + = + + + − =

= + + + = + + + − = −

The components of the resultant could have also been read directly from the diagram. The magnitude of the resultant is

2 2 2 2( ) ( ) (40 m) ( 20 m) 44.7 mx yD D D µ µ µ= + = + − =

Where additional figures have been kept for use in part (c). This should be reported to two significant figures as 45 m.D µ= (c) Since the bacterium’s speed is constant, its average speed is 20 m.µ The bacterium’s average velocity would be the magnitude of its net displacement divided by the duration of its travel. The total time it took to travel is given in terms of the total distance it traveled divided by its speed.

172.9 m 8.6 s20 m/s

t µµ

∆ = =

Its net displacement was calculated as 44.7 mµ so its average velocity is

44.7 m 5.2 m/s8.6 s

Dvt

µ µ= = =∆

From Equation 3.1, the average velocity is in same direction as the net displacement. Refer to the diagram above. The angle in the diagram is given by

1 1| | 20 mtan tan 2740 m

y

x

DD

µθµ

− − = = = °

Assess: The answers here make sense. In part (c), the average velocity is much less than the speed of the bacterium because while the bacterium has traveled a large distance, it’s displacement from the origin is small. Note the direction of the velocity vector is the same as the direction of the net displacement.

P3.53. Prepare: The skier’s motion on the horizontal, frictionless snow is not of any interest to us. The skier’s speed increases down the incline due to acceleration parallel to the incline, which is equal to g sin10°. A visual overview of the skier’s motion that includes a pictorial representation, a motion representation, and a list of values is shown.

3-34 Chapter 3

Solve: Using the following constant-acceleration kinematic equations, 2 2f i f i

2 2 2f f

f i f i2

f f

2 ( )

(15 m/s) (3.0 m/s) 2(9.8 m/s ) sin 10 ( 0 m) 64 m( )

(15 m/s) (3.0 m/s) (9.8 m/s )(sin 10 ) 7.1 s

x

x

v v a x x

x xv v a t t

t t

= + −

⇒ = + ° − ⇒ == + −

⇒ = + ° ⇒ =

Assess: A time of 7.1 s to cover 64 m is a reasonable value.

P3.54. Prepare: A visual overview of the skier’s motion that includes a pictorial representation, a motion diagram, and a list of values is shown. We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the motion because the speed does not change. So, the problem “starts” at the bottom of the uphill ramp and “ends” at the bottom of the downhill ramp. The final speed from the uphill roll (first problem) becomes the initial speed of the downhill roll (second problem). Because the axes point in different directions, we can avoid possible confusion by calling the downhill axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill velocities as v. Note that the height information, h = 1 m, has to be transformed into information about positions along the two axes.

Solve: (a) The uphill roll has 20 sin 6.93 m/s .a g α= − = − The speed at the top is found from

2 2 2 2 21 0 0 1 0 1 0 0 12 ( ) 2 (5 m/s) 2( 6.93 m/s )(1.414 m) 2.32 m/s 2.3 m/sv v a x x v v a x= + − ⇒ = + = + − = ≈

(b) The downward roll starts with velocity 1 1 2.32 m/su v= = and 21 sin 4.90 m/s .a g β= + = Then,

2 2 2 22 1 1 2 1 22 ( ) (2.32 m/s) 2(4.90 m/s )(2.00 m 0 m) 5.0 m/su u a z z u= + − = + − ⇒ =

(c) The final speed is equal to the initial speed, so the percentage change is zero! This result may seem surprising, but can be more easily understood after we introduce the concept of energy.

P3.55. Prepare: A visual overview of the puck’s motion that includes a pictorial representation, a motion diagram, and a list of values is shown as follows.


Solve: The acceleration, being the same along the incline, can be found as

2 2 2 2 21 0 1 02 ( ) (4.0 m/s) (5.0 m/s) 2 (3.0 m 0 m) 1.5 m/sv v a x x a a= + − ⇒ = + − ⇒ = −

We can also find the total time the puck takes to come to a halt as 2

2 0 2 0 2 2( ) 0 m/s (5.0 m/s) ( 1.5 m/s ) 3.3 sv v a t t t t= + − ⇒ = + − ⇒ =

Using the above obtained values of a and t2, we can find x2 as follows:

2 2 22 0 0 2 0 2 0

1 1( ) ( ) 0 m (5.0 m/s)(3.3 s) ( 1.5 m/s )(3.3 s) 8.3 m2 2

x x v t t a t t= + − + − = + + − =

That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position x0 and the puck stops 0.2 m or 20 cm before the ramp ends, you are not a winner.

P3.56. Prepare: Assume motion along the x-direction. Let xf – xi be the displacement from your gate to the baggage claim. We will use the technique of Equation 3.22: yg ym mg( ) ( ) ( ) .x x xv v v= +

Solve: In the first case, when the moving sidewalk is broken, we can find your velocity

f iY

( )50 s

x xv −=

In the second case, when you stand on the moving sidewalk, the velocity of the sidewalk relative to the ground is

f isg 75 s

x xv −=

In the third case, when you walk while riding, we can use the equation

yg ys sg( ) ( ) ( )v v v= +

That is, your velocity relative to the ground, when you are walking on the moving sidewalk, is equal to your velocity relative to the moving sidewalk (which is vY) plus the sidewalk’s velocity relative to the ground. Thus,

1 0 1 0 1 0

50 s 75 sx x x x x x

t− − −

= + 30 st⇒ =

Assess: A time smaller than 50 s was expected.

P3.57. Prepare: Both ships have a common origin at t = 0 s. Solve: (a) The velocity vectors of the two ships are as follows:

A

B

(20 mph cos 30 , north) (20 mph sin 30 , west) (17.32 mph, north) (10.0 mph, west)(25 mph cos 20 , north) (25 mph sin 20 , east) (23.49 mph, north) (8.55 mph, east)

vv

= ° + ° = += ° + ° = +

Since ,r v t= ∆

A A (2 hr)r v= =

(34.64 mph, north) (20.0 mph, west)+

B B(2 hr) (46.98 miles, north) (17.10 miles, east)r v= = +

So, the displacement between the two ships is

A B (12.34 miles, south) (37.10 miles, west) 39.1 milesR r r R= − = + ⇒ =

3-36 Chapter 3

(b) The speed of A as seen by B is: A B (6.17 mph, south) (18.55 mph, west) 19.5 mph.V v v V= − = + ⇒ =

Assess: The value of the speed is reasonable.

P3.58. Prepare: The boat’s motion is shown below.

Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of 2.0 m/s, it will take her 50 s to row across. But while she’s doing so, the current sweeps her boat sideways a distance 1 m/s × 50 s = 50 m. Mary’s net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the river’s current. She lands 50 m east of the point that was directly across the river from her when she started. (b) Mary’s net displacement is shown on the figure.

P3.59. Prepare: A visual overview of the ducks’ motion is shown below. The resulting velocity is given by fly wind ,v v v= +

where wind 6 m/s, east)v = and fly fly( sin , west)v v θ= +

fly( cos , south).v θ

Solve: Substituting the known values we get (8 m/s sin , west) (8 m/s cos , south) (6 m/s, east).v θ θ= + + That

is, v = ( 8 m/s sin , east) (8 m/s cos , south) (6 m/s, east).θ θ− + + We need to have vx = 0. This means 0 = –8 m/s sin θ + 6 m/s, so 6

8sinθ = and 48.6 .θ = ° Thus the ducks should head 48.6° west of south.

P3.60. Prepare: The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he needs to point at angle θ west of north. The direction the kayaker should paddle can be obtained from the technique of Equation 3.22: kg kw wg ,v v v= +

where kg kg( , north)v v= and wg wg( , east)v v=

with wg 2.0 m/s.v = Also,

kwv =

kw kw[( cos ,north) ( sin ,west)]v vθ θ+ with vkw = 3.0 m/s. Solve: (a) kg kw wg ,v v v= +

kg( , north)v = [(3.0 m/s cos , north) (3.0 m/s sin , west)]θ θ⋅ + ⋅ + (2.0 m/s, east) = [( 3.0 sin 2.0) m/s, east] (3.0 cos m/s, north)θ θ− + +

In order to go straight north in the earth frame, the kayaker’s velocity along the east must be zero. This will be true if

12.0 2.0sin sin 41.8 423.0 3.0

θ θ − = ⇒ = = ° = °

Thus he must paddle in a direction 42° west of north. (b) His northward speed is vkg = 3.0 cos (41.8°) m/s = 2.236 m/s. The time to cross is


100 m 45 s2.236 m/s

t = =

P3.61. Prepare: A visual overview of the plane’s motion is shown in the following figure. The direction the pilot must head the plane can be obtained from pg pa ag ,v v v= +

where pa pa( sin , south)v v θ= +

pa( cos , east),v θ

pa 200 mph,v = agv = ag ag( sin 30 , north) ( cos 30 , east),v v° + ° and pg pg( , east).v v=

Solve: (a) Writing the equation pg pa agv v v= + in the form of components

pg( , east)v = [ (200 mph sin , south) (200 mph cos , east)θ θ+ ] + [ (50mph sin 30 , north) (50mph cos 30 , east)° + ° ]

Because pg( )v should have no component along north,

50 sin 30 200 sin 0θ° − = ⇒ θ = 7.2°

(b) The pilot must head 7.18° south of east. Substituting this value of θ in the above velocity equation gives

pg( , east)v = (200 mph cos7.18 , east)° + (50 mph cos 30 , east)° (240 mph, east).= At a speed of 240 mph, the trip takes 600 mi/240 mpht = = 2.5 hours.

P3.62. Prepare: We need to relate the velocity of the ship with respect to the water to the velocity of the ship with respect to the land using the relative velocity formula, Equation 3.22. The velocity with respect to the land is easy to find from the distance traveled by the ship and the time of travel, but the problem asks for the velocity with respect to the water.

Solve: (a) The ship’s velocity relative to the land is SL ( , 0 m/s)v v= −

and the velocity of the water with

respect to the land is WL (0,3.6 m/s).v = The velocity of the ship with respect to the water is given by:

SW ( (10 m/s) cos , (10 m/s) sin ).v θ θ= − − From Equation 3.22, we know that SL SW WLv v v= +

which may written as follows:

( ,0 m/s) ( (10 m/s) cos , (10 m/s) sin ) (0,3.6 m/s)v θ θ− = − − +

This equation contains two equations. For the y components of the left and right hand sides to be the same, we need to have 0 m/s (10 m/s)sin 3.6 m/sθ= − + which can be solved to yield the following:

1sin (3.6 /10) 21.1θ −= = °

3-38 Chapter 3

which rounds to 21 .° The ship should sail 21° south of west. (b) Now for the x components of the left and right hand sides to be equal, we need (10 m/s)cos ,v θ− = − which can be solved as

(10 m/s) cos (10 m/s) cos 21.1 9.3 m/sv θ= = °=

The ship’s speed with respect to land is 9.3 m/s. Assess: A fairly low angle like 21° is reasonable because the ship’s speed relative to the water is much larger

than the speed of the current. In the limiting case of no current, the angle would be 0 . In other words, the ship would not have to overcome a current and would sail directly west. Also it is reasonable that the speed of the ship relative to land, 9.3 m/s, is less than its speed relative to the water because the ship is fighting the current and the current makes the ship progress less rapidly.

P3.63. Prepare: We will apply the velocity and acceleration concepts for projectile motion as shown in Figure 3.32. A visual overview is shown in the following figure.

Solve: (a) We know the velocity 1 1 1( ) ( )x yv v v= + with 1( ) 2.0 m/sxv = and 1( ) 2.0 m/syv = at t = 1 s. The ball is at

its highest point at t = 2 s, so vy = 0 m/s. The horizontal velocity is constant in projectile motion, so vx = 2.0 m/s at all times. Thus, 2 2 2( ) ( ) ,x yv v v= +

with 2( ) 2.0 m/sxv = and 2( ) 0 m/syv = at t = 2 s. We can see that the

- componenty of velocity changed by ∆vy = –2.0 m/s between t = 1 s and t = 2 s. Because ay is constant, vy changes by –2.0 m/s in any 1-s interval. At t = 3 s, vy is 2.0 m/s less than its value of 0 at t = 2 s. At t = 0 s, vy must have been 2.0 m/s more than its value of 2.0 m/s at t = 1 s. Consequently, at t = 0 s,

0 0 0( ) ( ) ,x yv v v= + with 0( ) 2.0 m/sxv = and 0( ) 4.0 m/syv =

At t = 1 s,

1 1 1( ) ( ) ,x yv v v= + with 1( ) 2.0 m/sxv = and 1( ) 2.0 m/syv =

At t = 2 s,

2 2 2( ) ( ) ,x yv v v= + with 2( ) 2.0 m/sxv = and 2( ) 0 m/syv =

At t = 3 s,

3 3 3( ) ( ) ,x yv v v= + with 3( ) 2.0 m/sxv = and 3( ) 2.0 m/syv = −

(b) Because vy is changing at the rate –2.0 m/s per s, the y-component of acceleration is ay = –2.0 m/s2. But ya g= − for projectile motion, so the value of g on Exidor is g = 2.0 m/s2.

(c) From part (a) the components of 0v are 0 0( ) 2.0 m/s and ( ) 4.0 m/s.x yv v= = This means


01 1

0

( ) 4.0 m/stan tan 63( ) 2.0 m/s

y

x

vv

θ − − = = = °

above +x

Assess: The y-component of the velocity vector decreases from 2.0 m/s at t = 1 s to 0 m/s at t = 2 s. This gives an acceleration of −2 m/s2. All the other values obtained above are also reasonable.

P3.64. Prepare: We will apply the constant-acceleration kinematic equations to the horizontal and vertical motions of the ball as described by Equations 3.25. A visual overview is shown in the next figure. The effect of air resistance on the ball is ignored.

Solve: Using 21

f i i f i f i2( ) ( ) ( ) ,x xx x v t t a t t= + − + −

f50 m 0 m (25 m/s)( 0 s) 0t= + − + m f 2.0 st⇒ =

Now, using 21f i i f i f i2( ) ( ) ( ) ,y yy y v t t a t t= + − + −

2 21i 20 m 0 m ( 9.8 m/s )(2.0 s 0 s)y= + + − − i 19.6 my⇒ =

Assess: A magnitude of 20 m for the height is reasonable.

P3.65. Prepare: This problem is somewhat similar to Problem 3.27 with all of the initial velocity in the horizontal direction. We will use the vertical equation for constant acceleration to determine the time of flight and then see how far Captain Brady can go in that time. Of interest is the fact that we will do this two-step problem completely with variables in part (a) and only plug in numbers in part (b). We could do part (b) in feet (using g = 32 ft/s2), but to compare with the world record 100 m dash, let’s convert to meters. L = 22 ft = 6.71 m and h = 20 ft = 6.10 m. Solve: (a) Given that (vy) = 0.0 ft/s we can use Equation 2.14.

21= ( )2 yy a t∆ ∆

With up as the positive direction, y∆ is negative and ay = –g; those signs cancel leaving

21= ( )2

h g t∆

Solve for .t∆

2= htg

∆

Now use that expression for t∆ in the equation for constant horizontal velocity.

2= = =x xhL x v t v

g∆ ∆

Finally solve for v = vx in terms of L and h.

2= =

2hg

L gv Lh

3-40 Chapter 3

(b) Now plug in the numbers we are given for L and .h

29.8 m/s= = (6.71 m) = 6.0 m/s2 2(6.10 m)gv Lh

Compare this result (v = 6.0 m/s) with the world-class sprinter (v = 10 m/s); a fit person could make this leap. Assess: The results are reasonable, and not obviously wrong. 6.0 m/s ≈ 13 mph, and that would be a fast run, but certainly possible. By solving the problem first algebraically before plugging in any numbers, we are able to substitute other numbers as well, if we desire, without re-solving the whole problem.

P3.66. Prepare: We convert the 83 yards to meters and then use the formula for range at a launch angle of 45 .°

36 in 1 m83 yd 83 yd 75.9 m1 yd 39.37 in

= =

Solve: The record range is 83 yards or 75.9 m. We use an extra significant figure since this datum will be used to

get another result. Based on the formula for range, we can say: / sin (2 )v gR θ= . Using 45° for the angle gives

the following:

( )29.8 m/s (75.9 m) 27 m/sv = =

The launch speed of this record pass was 27 m/s.

Assess: The launch speed is about 60 mi/hr. This is impressive for a football. We would expect the launch speed to be less than that for a baseball because of a football’s greater mass.

P3.67. Prepare: We will use the initial information (that the marble goes 6.0 m straight up) to find the speed the marble leaves the gun. We also need to know how long it takes something to fall 1.5 m from rest in free fall so we can then use that in the horizontal equation. Assume that there is no air resistance (ay = –g) and that the marble leaves the gun with the same speed (muzzle speed) each time it is fired. Solve: To determine the muzzle speed in the straight-up case, use Equation 2.13.

2 2f i( ) = ( ) 2y y yv v a y+ ∆

where at the top of the trajectory f( ) 0.0 m/syv = and 6.0 m.y∆ =

2i i( ) = 2 ( ) = 2 = 10.8 m/sy yv g y v g y∆ ⇒ ∆

We also use Equation 2.14 to find the time for an object to fall 1.5 m from rest: y∆ = –15 m now instead of the 6.0 m used previously.

21= ( )2 yy a t∆ ∆

2

2 2( 1.5 m)= 0.553 s9.8 m/s

ytg∆ −

∆ = =− −


At last we combine this information into the equation for constant horizontal velocity.

= = (10.8 m/s)(0.553 s) = 6.0 mxx v t∆ ∆

Assess: Is it a coincidence that the marble has a horizontal range of 6.0 m when it can reach a height of 6.0 m when fired straight up, or will those numbers always be the same? Well, the 6.0 m horizontal range depends on the height (1.5 m) from which you fire it, so if that were different the range would be different. This leads us to conclude that it is a coincidence. You can go back, though, and do the problem algebraically (with no numbers) and find that g cancels and that the horizontal range is 2 times the square root of the product of the vertical height it can reach and the height from which you fire it horizontally.

P3.68. Prepare: We will apply the constant-acceleration kinematic equations to the horizontal and vertical motions of the shot as described by Equations 3.25. A visual overview is shown.

Solve: (a) Using 21f i i f i f i2( ) ( ) ( ) ,y yy y v t t a t t= + − + −

2 21i f f2

2 2f f f

0 m 1.8 m sin 40 ( 0 s) ( 9.8 m/s )( 0 s)

1.8 m (7.713 m/s) (4.9 m/s ) 0.206 s and 1.780 s

v t t

t t t

= + ° − + − −

= + − ⇒ = −

The negative value of tf is unphysical for the current situation. Using tf = 1.780 s and f i i f i( ) ( ),xx x v t t= + − we get

f i0 ( cos40 m/s)(1.780 s 0 s) (12 m/s)cos40 (1.78 s)x v= + ° − = ° = 16.36 m = 16.4 m

(b) We can repeat the calculation for each angle. A general result for the flight time at angle q is

( )2f 12 sin 144 sin 35.28 /9.8 st θ θ= + +

and the distance traveled is xf = 12 cos θ × tf. We can put the results in a table.

θ tf xf 40.0° 1.780 s 16.36 m 42.5° 1.853 s 16.39 m 45.0° 1.923 s 16.31 m 47.5° 1.990 s 16.13 m

Maximum distance is achieved at θ ≈ 42.5°. Assess: The well-known “fact” that maximum distance is achieved at 45° is true only when the projectile is launched and lands at the same height. That isn’t true here. The extra 0.03 m = 3 cm obtained by increasing the angle from 40.0° to 42.5° could easily mean the difference between first and second place in a world-class meet.

P3.69. Prepare: We will apply the constant-acceleration kinematics equations to the horizontal and vertical motions of the tennis ball as described by Equations 3.25. A visual overview is shown as follows. To find whether the ball clears the net, we will determine the vertical fall of the ball as it travels to the net.

3-42 Chapter 3

Solve: The initial velocity is

i i( ) cos 5 (20 m/s) cos 5 19.92 m/sxv v= ° = ° =

i i( ) sin 5 (20 m/s) sin 5 1.743 m/syv v= ° = ° =

The time it takes for the ball to reach the net is

f i i f i( ) ( )xx x v t t= + − f f7.0 m 0 m (19.92 m/s)( 0 s) 0.351st t⇒ = + − ⇒ =

The vertical position at tf = 0.351 s is 21

f i i f i f i2

2 212

( ) ( ) ( )

(2.0 m) (1.743 m/s)(0.351 s 0 s) ( 9.8 m/s )(0.351s 0 s) 2.0 my yy y v t t a t t= + − + −

= + − + − − =

Thus the ball clears the net by 1.0 m. Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial vertical velocity and the ball clears by a larger distance. The above result is reasonable.

P3.70. Prepare: We can use the equation for vertical motion at constant acceleration to find the time of fall and then use the time to find the final velocity.

Solve: Since the water is launched horizontally, its time of flight and vertical displacement are related by the

equation: 21 .2

y g t∆ = − ∆ Solving for the time, we have

22 / 2(53 m) / 9.8 m/s 3.29 st y g∆ = ∆ = =

The horizontal component of the velocity, xv , is constant, but the vertical component is given by the equation:

f i( ) ( ) .y y yv v a t= + ∆ At the moment the water strikes the pool, the vertical component is

( )2y f( ) 0 m/s 9.8 m/s (3.29 s) 32.2 m/sv = − = −

At the moment of impact the velocity of the water is: (9.0 m/s, 32.2 m/s)− . The angle that the water makes with the vertical is given by

1tan ((9.0 m/s)/(32.2 m/s)) 16φ −= =

The water is falling at an angle of 16 with the vertical.


Assess: Even though the water is launched at a fairly high speed ( 9.0 m/s is about 20 mi/hr ), it is close to the vertical when it lands because it spends such a long time in the air during which time the absolute value of yv

increases steadily.

P3.71. Prepare: Projectiles launched at higher angles stay in the air longer. Since objects launched at 30 and at 60° have the same range, we can conclude that their horizontal component of velocity multiplied by time

of flight is the same. This fact can be written as an equation: i i30 60cos 30 cos 60v t v t°∆ = °∆

.

Solve: The above equation can be solved to obtain the following:

( )60 30cos 30 / cos 60 3(2.0 s) 3.5 st t∆ = ° ° ∆ = =

The projectile launched at 60° spends 3.5 s in the air.

Assess: The projectile launched at a higher angle stays in the air much longer because it has a higher vertical component of velocity. But the two projectiles have the same range because their launch angles are complementary.

P3.72. Prepare: We will apply the constant-acceleration kinematic equations to the horizontal and vertical motions of the food package as described by Equations 3.25. A visual overview is shown.

Solve: For the horizontal motion,

21f i i f i f i2( ) ( ) ( )x xx x v t t a t t= + − + − f0 m (150 m/s)( 0 s) 0 mt= + − + f(150 m/s)t=

We will determine tf from the vertical y-motion as follows: 21

f i i f i f i2( ) ( ) ( )y yy y v t t a t t= + − + −

2 21f20 m 100 m 0 m ( 9.8 m/s )t⇒ = + + − ⇒ f 2

200 m 4.518 s9.8 m/s

t = =

From the above x-equation, the displacement is f (150 m/s)(4.518 s) 680 m.x = = Assess: The horizontal distance of 680 m covered by a freely falling object from a height of 100 m and with an initial horizontal velocity of 150 m/s (≈ 335 mph) is reasonable.

3-44 Chapter 3

P3.73. Prepare: First draw a picture.

In part (a) use tilted axes so the x-axis runs down the slide. The acceleration will be ax = g sin .θ Part (b) is a familiar two-step projectile motion problem where we use the vertical direction to determine the time of flight and then plug it into then equation for constant horizontal velocity. Use axes that are not tilted for part (b). Solve: (a) We use Equation 2.13 with i( ) = 0.0 m/s.xv

2f( ) = 2x xv a x∆

2 of( ) = 2( sin ) = 2(9.8 m/s )(sin 40 )(3.0 m) = 6.1 m/sxv g xθ ∆

(b) We use Equation 2.14 to find the time for an object to fall 0.4 m from rest: = 0.4m.y∆ −

21= ( )2 yy a t∆ ∆

2

2 2( 0.4 m) 0.286 s9.8 m/s

ytg∆ −

∆ = = =− −

At last we combine this information into the equation for constant horizontal velocity.

= = (6.15 m/s)(0.286 s) = 1.8 mxx v t∆ ∆

Assess: We reported the speed at the bottom of the slide to two significant figures, but kept track of a third to use as a guard digit because this result is also an intermediate result for the final answer. We also kept a third significant figure on the t∆ as a guard digit. The result of landing 1.8 m from the end of the frictionless slide seems just a bit large because this slide was frictionless and real slides aren’t, but it doesn’t seem to be too far out of expectation, so our result is probably correct.

P3.74. Prepare: Equations 2.8, 2.13, and 2.16 can be used to calculate the accelerations. Equation 3.30 can be used to calculate the minimum turning radius. We will convert to SI units. Solve: (a) Converting 60 mph to SI units, we have

0.447 m/s60 mph 60 mph 26.8 m/s1 mph

= =

An additional significant figure has been kept for use in part (b). This should be reported as 27 m/s to two significant figures. For reaching “60 mph in 5 seconds flat”, the acceleration can be calculated with Equation 2.8.

227 m/s 5.4 m/s5 s

xx

vat

∆= = =

∆

The car can corner at 0.85g. The acceleration of the car during cornering is then 2 2(0.85)(9.80 m/s ) 8.33 m/sa = =


An additional significant figure has been kept for use in part (b). This should be reported to two significant figures as 28.3 m/s .a = During a stop, we can calculate the acceleration using Equation 2.13. Converting to SI units

0.447 m/s70 mph 70 mph 31 m/s1 mph

= =

Converting 168 ft to SI units

( ) 0.305 m168 ft 168 ft 51 m1ft

= =

The final velocity of the car is 0 m/s. We can calculate the acceleration using Equation 2.13. Solving for acceleration in that equation we have:

2 22i( ) (31 m/s) 9.4 m/s

2 2(51 m)x

xva

x− −

= = = −∆

The magnitude of the car’s acceleration is the largest when it is stopping. (b) In part (a) we found the car’s maximum acceleration while cornering. During cornering, assuming that the car is moving on a portion of a circle, the magnitude of its acceleration is given in Equation 3.30: 2 / .a v r= Solving that equation for the radius and using the previous conversion of 60 mph,

2 2

2

(26.8 m/s) 86 m(8.33 m/s )

vra

= = =

Assess: These results make sense. A turning radius of 86 m is reasonable for a sports car traveling at 60 mph.

P3.75. Prepare: We need to convert the radius of the Mini Cooper’s turn to meters and convert the final speed of the Mustang to meters per second. The radius of the Mini Cooper’s turn is the following:

1 m17 ft = 17 ft 5.18 m3.28 ft

=

And the final speed of the Mustang is as follows:

mi mi 1609 m 1 hr60 60 26.8 m/shr hr 1 mi 3600 s

= =

The acceleration of the Mustang is given by /a v t= ∆ ∆

2(26.8 m/s)/(5.6 s) 4.79 m/sa = =

Solve: To match the Mustang’s acceleration, the Mini Cooper must have a centripetal acceleration of 24.79 m/s . Given the formula for centripetal acceleration, 2 / ,a v r= we can solve for the necessary radius as

follows:

( )2 m 1 mi 3600 s4.79 m/s (5.18 m) 4.98 m/s 4.98 11 mphs 1609 m 1 hr

v ar = = = = =

The Mini Cooper must travel at 11 mph to have the same acceleration as the Mustang.

Assess: Even at a fairly low speed, 11 mph, the acceleration is high. This is because the radius of the turn is so small—17 ft.

3-46 Chapter 3

P3.76. Prepare: As with any circular motion problem, at constant speed, the acceleration of an object is just the centripetal acceleration given by Equation 3.30.

Solve: (a) The acceleration the riders experience is given by 2 /a v r=

2 2

2 22

(30 m/s) 136 m/s 36 m/s 3.725 m 9.8 m/s

v g gr

= = = =

They accelerate at 236 m/s or 3.7 .g

(b) We can use the relative velocity formula to find the speed at which the riders pass one another. If riders on the first swing have a velocity of 30 m/s relative to the ground and riders on the second swing have a velocity of

30 m/s− relative to the ground, then we can say the following:

12 1G G2 1G 2G 30 m/s ( 30 m/s) 60 m/sv v v v v= + = − = − − =

So the velocity of riders on the first swing relative to riders on the second swing is 60 m/s.

Assess: An acceleration of 3.7g is very large but this is not surprising considering the large speed of the

riders 30 m/s– comes to about 67 mi/hr. When riders on the first swing see riders on the second swing going twice as fast as the swing is moving relative to the ground, it makes them feel like they are moving even faster than they actually are and makes the ride more impressive.

P3.77. Prepare: We will use Equation 3.30 to relate the acceleration to the speed. But first we need to convert the speed of the car to m/s.

mi mi 1609m 1 hr40 40 17.9 m/shr hr 1 mi 3600 s

= =

Solve: (a) Your acceleration is given from the equation 2 /a v r=

2

2(17.9 m/s) 2.91 m/s110 m

=

which converts as follows:

( )2 22

12.91 m/s 2.91 m/s 0.309.8 m/s

g g = =

The acceleration is 22.9m/s or 0.30 .g

(b) The formula for centripetal acceleration, 2 /a v r= can be solved for v as follows: .v ar= In this form we

see that if the acceleration is doubled, then the velocity is multiplied by 2. So we multiply the 40 mph speed

limit by 2 : (40 mph) 2 57 mph.= At 57 mph the acceleration would be twice the acceleration at 40 mph. Assess: As noted in the solution to Problem 40, a small change in velocity can produce a large change in centripetal acceleration. Here, with an increase in speed of less than 50%, the acceleration doubles and the friction needed for the turn also doubles.

P3.78. Prepare: Assume that water slides are frictionless. Use tilted axes so the x -axis runs along the slide. This is much like Problem 3.73. Solve: We use Equation 2.13 with i( ) 0.0 m/s.xv =

2f( ) = 2x xv a x∆


2 of( ) = 2( sin ) = 2(9.8 m/s )(sin 45 )(6.0 m) = 9.1 m/sxv g xθ ∆

So the answer is C. Assess: This speed seems fast, but not out of the question for a frictionless slide this long. And it isn’t as fast as an object would be going in free fall after falling from rest for 6.0 m.

P3.79. Prepare: Equation 3.30 governs circular motion. Solve: From Equation 3.30 we can see the centripetal acceleration of a car has a quadratic relationship to velocity and an inverse relationship to the radius of the circular motion. If the velocity of the object is decreased, the acceleration decreases. If the radius of the motion is increased the acceleration of the object decreases. We can either decrease the velocity or increase the radius. The choices C and D act to increase the velocity, so the correct answer is B. Assess: Centripetal acceleration decreases with increasing radius.

P3.80. Prepare: Do not use tilted axes for this problem. Solve: We use Equation 2.13 with i( ) 0.0 m/s.yv =

2f( ) = 2y yv a y∆

2f( ) = 2( ) = 2( 9.8 m/s )( 0.60 m) = 3.4 m/sxv g y− ∆ − −

So the answer is B. Assess: 3.4 m/s seems like a reasonable vertical speed to hit the water. The answer to this question is independent of the length of the slide since the vertical speed is zero as the rider leaves the slide horizontally.

P3.81. Prepare: The riders are in free fall during this part of the motion. We can use Equations 3.25. Solve: The riders have no initial velocity in the vertical direction after they leave the slide. Their initial velocity in the horizontal direction will be determined by the first and second sections and is unaffected by changing the height of the slide exit above the water, assuming that the first two sections are unchanged. The time it takes the riders to hit the water after leaving the slide can be calculated with Equation 2.12.

2f i i

1( ) ( )2yy y v t g t= + ∆ − ∆

Taking the origin as the exit of the slide and i( ) 0 m/syv = this equation becomes

f2ytg

∆ = −

The initial horizontal velocity of the rider is i( ) .xv The horizontal displacement of the rider once leaving the ramp will be i( ) .xv t∆ To double the distance the rider travels before hitting the water, we only need to double the time the rider takes to hit the water. From the previous equation, to double the time, we must quadruple the vertical distance. So the answer is (4)(0.6 m) = 2.4 m. The correct choice is C. Assess: This answer is reasonable, since the distance traveled vertically in free fall has a quadratic relationship to time.

P3.82. Prepare: We are simply asked to compare three accelerations, the middle of which is the centripetal acceleration 2/a v r= during the second section. Solve: The acceleration of the rider in the air is certainly greater than the acceleration of the rider coming down the slide because the magnitude is g in the air but sing θ on the slide (and sin < 1θ for < 90θ ° ). So we know the answer isn’t “during the first section of the motion” (nor is it the same in all sections). Now lets compute the centripetal acceleration during the second section of the motion. We assume the speed during the circular portion of the motion is constant and equal to the value reached at the bottom of the slide (see Problem P3.78).

2 22(9.1 m/s) 55 m/s

1.5 mvar

= = =

3-48 Chapter 3

This is much bigger than 2= 9.8m/s ,g so section two wins. The answer is B. Assess: The fast speed and tight curvature make for a large centripetal acceleration—5.6 g in this case. The problem only asked about the magnitude of the acceleration, but a quick look at the direction is interesting. In section one it is directed down the slide, in section two it is toward the center of curvature, and in section three it is downward (as it is for all objects in free fall).

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