PHY317 Hw # 1-Solutions

figueroa (mef2223) – Hw # 1 – Berk – (57445) 1

This print-out should have 45 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsLength is to meter as

1. density is to volume.

2. weight is to mass.

3. liter is to distance.

4. mass is to kilogram. correct

Explanation:The standard metric unit for length is the

meter, and the standard metric unit for massis the kilogram.

002 10.0 points

The prefix that means1

1000is

1. centi.

2. milli. correct

3. micro.

4. kilo.

Explanation:

003 10.0 pointsFind the speed of an object that covers 400 kmin 5 hr.

1. 800 km/hr

2. 80 km/hr correct

3. 40 km/hr

4. 2000 km/hr

Explanation:

speed =distance

time=

400 km

5 hr= 80 km/hr .

004 10.0 pointsVelocity is

1. the same as momentum.

2. speed in a specific direction. correct

3. the same as acceleration.

4. the same as speed.

Explanation:

005 10.0 pointsA man jogs at a speed of 0.94 m/s. His dogwaits 2 s and then takes off running at a speedof 3.1 m/s to catch the man.How far will they have each traveled when

the dog catches up with the man?

Correct answer: 2.69815 m.

Explanation:

Let : vman = 0.94 m/s ,

∆t = 2 s , and

vdog = 3.1 m/s .

The distance that separates the man andthe dog when the dog takes off is

d0 = vman∆t = (0.94 m/s) (2 s) = 1.88 m

and the dog catches up to the man when

xman = xdogd0 + vman t = vdog t

t =d0

vdog − vman

t =1.88 m

3.1 m/s− 0.94 m/s= 0.87037 s ,

so the distance is

d = vdog t = (3.1 m/s)(0.87037 s)

= 2.69815 m .


006 10.0 pointsThe time it takes for light, traveling at a speedof 3.0× 108 m/s, to cover 300 m is

1. 10−11 s

2. None of these

3. 106 s

4. 1011 s

5. 10−6 s correct

Explanation:The time to cover a distance d at a speed v

is given by

t =d

v=

300 m

3.0× 108 m/s= 10−6 s .

007 10.0 points

The graph shows the velocity v as a functionof time t for an object moving in a straightline.

t

v

0 tQ t

RtS

tP

Which graph shows the corresponding dis-placement x as a function of time t for thesame time interval?

1.

t

x

0 tQ t

RtS

tP

correct

2.

t

x

0 tQ t

RtS

tP

3. tx

0 tQ t

RtS

tP

4. tx

0 tQ t

RtS

tP

5.t

x

0 tQ t

RtS

tP

6.t

x

0 tQ t

RtS

tP

7.

t

x

0 tQ t

RtS

tP

8. None of these graphs is correct.

9.tx

0 tQ t

RtS

tP

Explanation:The displacement is the integral of the ve-

locity with respect to time:

~x =

∫

~v dt .

Because the velocity increases linearly fromzero at first, then remains constant, then de-creases linearly to zero, the displacement willincrease at first proportional to time squared,then increase linearly, and then increase pro-portional to negative time squared.From these facts, we can obtain the correct

answer.

t

x

0 tQ t

RtS

tP


008 10.0 pointsAn oceanic depth-sounding vessel surveys theocean bottom with ultrasonic waves thattravel 1530 m/s in seawater.How deep is the water directly below the

vessel if the time delay of the echo to theocean floor and back is 7 s?

Correct answer: 5355 m.

Explanation:

Let : v = 1530 m/s and

∆ t = 7 s .

The sound takes 3.5 s to reach the oceanfloor (and 3.5 s to return), so

d = v t = (1530 m/s) (3.5 s) = 5355 m .

009 10.0 pointsThe plot shows x(t) for a trainmoving along

a long, straight track.

x

t

Which statement is correct about the mo-tion?

1. The train moves at constant velocity.

2. The train continually speeds up.

3. The train at first speeds up, then slowsdown.

4. The train at first slows down, then speedsup.

5. The train in fact moves at constant speedalong a circular path described by the x(t)curve.

6. The train continually slows down andcomes to rest. correct

Explanation:The slope of x(t) continually decreases to

reach zero. Since vx(t) =d x

dtis by definition

the slope of x(t) at each t, the train continu-ally slows down until it stops.

010 10.0 pointsAn airplane starts from A and flies to B at aconstant speed. After reaching B it returnsto A at the same speed. There was no wind.Assuming there was a wind from A to B

of constant magnitude, when will the roundtrip take more time – when there is a wind orwhen there is no wind? Assume that the windspeed is less than that of the plane.

1. The same time in both cases.

2. Longer round trip time when there is nowind.

3. Longer round trip time when there is aconstant wind. correct

4. More information is needed.

Explanation:Assume the distance between A and B is d

and the speed of the plane in flight is v. Withno wind, the round trip time is

t1 =d

v+

d

v=

2 d

v.

When there is a constant wind speed of w, theround trip time is

t2 =d

v − w+

d

v + w=

2 v d

v2 − w2.

Compare the reciprocals of the times:

1

t2=

v2 − w2

2 v d=

v

2 d−

w2

2 v d<

v

2 d=

1

t1t1 < t2 .


011 10.0 pointsA particle moves in a straight line with veloc-ity

v(t) = 3t2 − 6t .

If it initially starts moving from 0 (wherex = 0), then its position x(t) is equal to

1. x(t) = t3 − 3t2 correct

2. x(t) = 6t

3. x(t) = t3 − 3t2 + 1

4. x(t) = 6t− 6

5. x(t) = 6t− 3t2

Explanation:

v(t) = 3t2 − 6t

d s(t)

dt= 3t2 − 6t

∫

d s(t)

dtdt =

∫

(

3t2 − 6t)

dt

s(t) = t3 − 3t2 + C

x(0) = 0, so 0 = C and

s(t) = t3 − 3t2 .

012 (part 1 of 2) 7.5 pointsA golfer takes two putts to get his ball intothe hole once he is on the green. The first puttdisplaces the ball 6.2 m east, and the second8.54 m south.What displacement would have been

needed to get the ball into the hole on thefirst putt?


Explanation:

Let : ∆x = 6.2 m and

∆y = −8.54 m .

6.20 m

8.54 mR

E

S θ

The east direction and the south directiondefine legs of a right triangle. The resultant isthe hypotenuse, so the displacement is

R =√

(∆x)2 + (∆y)2

=√

(6.2 m)2 + (−8.54 m)2

= 10.5533 m .

013 (part 2 of 2) 7.5 pointsWhat is the direction (in degrees S of E)?

Correct answer: 54.0205◦.

Explanation:

tan θ =∆y

∆x

θ = tan−1

(

∆y

∆x

)

= tan−1

(

−8.54 m

6.2 m

)

= −54.0205◦ ,

so the direction is 54.0205◦ of S of E.

014 (part 1 of 2) 7.5 pointsA particle has ~r(0) = (4 m) ̂ and ~v(0) =(2 m/s) ı̂.If its acceleration is constant and given by

~a = −(2 m/s2) (̂ı+ ̂), at what time t does theparticle first cross the x axis?

Correct answer: 2 s.

Explanation:

~r(t) = ~r(0) + ~v(0) t+1

2~a t2 , so


~r(t) = (4 m) ̂+ (2 m/s) ı̂ t

+(−1 m/s2) (̂ı+ ̂) t2

= [(2 m/s) t− (1 m/s2) t2] ı̂

+[4 m− (1 m/s2) t2] ̂

~r(t) will not have a y component when

4 m− (1 m/s2) t2 = 0

(1 m/s2) t2 = 4 m

t = 2 s .

015 (part 2 of 2) 7.5 pointsAt what time t is the particle moving parallelto the y axis; that is, in the ̂ direction?

Correct answer: 1 s.

Explanation:

~r(t) = ~r(0) + ~v(0) t+1

2~a t2

~v(t) =d~r

d t= ~v(0) + ~a t , so

~v(t) = (2 m/s) ı̂− (2 m/s2) (̂ı+ ̂) t

= [(2 m/s)− (2 m/s2) t] ı̂

−(2 m/s2) t ̂

~v(t) will not have an x component when

2 m/s− (2 m/s2) t = 0

t = 1 s .

At this time ~v(t) = −(2 m/s) ̂ parallel tothe y-axis.

016 10.0 pointsThe correct, general definitions of velocity

and acceleration, in terms of position vector

~r, are ~v =d~r

dt, and ~a =

d~v

dt. During a very

short time interval, the velocity of an objectchanges from ~vi to ~vf , as shown.

~vi~vf

What is the approximate direction of theacceleration during this time interval?

1.

2.

3.

4. correct

Explanation:

~a∆t = ~vf − ~vi .

~vi

~vf

~a∆t

017 (part 1 of 2) 7.5 pointsA particle has a constant acceleration ~a =(4 m/s2) ̂. At t = 2 sec, it has velocity~v = (̂ı+ ̂) (2 m/s).

What is its velocity at t = 0?

1. ı̂ (2 m/s) + ̂ (6 m/s)


2. ı̂ (4 m/s)− ̂ (4 m/s)

3. ı̂(4 m/s) + ̂ (4 m/s)

4. ~0

5. ı̂ (2 m/s)− ̂ (6 m/s) correct

Explanation:

~v(t) = ~v(0) + ~a t , so

~v(0) = ~v(t)− ~a t

= (̂ı+ ̂) (2 m/s)− ̂ (4 m/s2)(2 s)

= ı̂ (2 m/s) + ̂ (2 m/s− 8 m/s)

= ı̂ (2 m/s)− ̂ (6 m/s) .

018 (part 2 of 2) 7.5 pointsAfter 2 sec, its position is found to be ~r =(̂ı− ̂) (4 m).Where was it at t = 0?

1. ı̂ (2 m)− ̂ (6 m)

2. ı̂ (4 m) + ̂ (4 m)

3. ı̂ (2 m) + ̂ (6 m)

4. ı̂ (4 m)− ̂ (4 m)

5. ~0 correct

Explanation:

~r(t) = ~r(0) + ~v(0) t+1

2~a t2 , so

~r(0) = ~r(t)− ~v(0) t−1

2~a t2

= (̂ı− ̂) (4 m)

−[̂ı (2 m/s)− ̂ (6 m/s)](2 s)

−1

2(4 m/s2) ̂ (2 s)2

= ı̂(4 m− 4 m)

+̂(−4 m + 12 m− 8 m)

= ~0 .

019 10.0 pointsA ball rolling up a hill has vector velocities~v1 and ~v2 at times t1 and t2, respectively, asshown in the figure.

v1

initial

v2

final

Which vector diagram below most accu-rately depicts the direction of the ball’s aver-age acceleration over the interval?

1.

2.

correct

3.

4.

5.

6.

7.

8. Zero vector.

9.

Explanation:In fact the two forces exerting on the ball,

the gravitational force from the earth andthe force from the incline remain unchangedduring the interval, so the total accelerationshould be downward to the left and won’tchange either.


keywords:

020 10.0 pointsAt one instant the velocity of a particle is

b

and a very short time later its velocity has

changed tob

Qualitatively, what was the direction of theparticle’s acceleration during the interval be-tween the first and second instance?

1.

2. correct

3.

4.

5.

6.

7.

8.

Explanation:

From a =dv

dt, which for a finite time in-

terval looks like a =∆v

∆t, the acceleration is

in the direction of the vector ∆v :∆~v

~v1 ~v2∆~v = ~v2 − ~v1

~v2 = ~v1 +∆~v .

021 10.0 pointsWhich of the statements

A. A moving object does not accelerate if itsvelocity remains constant;

B. The difference between speed and velocityis that speed indicates an object’s direc-tion of motion;

C. A moving object is accelerating if its speedor direction varies.

is/are true?

1. C only

2. None is true.

3. A and C only correct

4. B and C only

5. B only

6. A only

7. A and B only

8. All are true.

Explanation:Velocity is speed in a particular direction.

022 10.0 points

A ball is thrown and follows the parabolicpath shown. Air friction is negligible. PointQ is the highest point on the path. Points Pand R are the same height above the ground.

Q

RP

How do the speeds of the ball at the threepoints compare?

1. ‖~vQ‖ < ‖~v

R‖ < ‖~v

P‖

2. ‖~vP‖ < ‖~v

Q‖ < ‖~v

R‖

3. ‖~vP‖ = ‖~v

R‖ = ‖~v

Q‖

4. ‖~vR‖ < ‖~v

Q‖ < ‖~v

P‖


5. ‖~vP‖ = ‖~v

R‖ < ‖~v

Q‖

6. ‖~vQ‖ < ‖~v

P‖ = ‖~v

R‖ correct

Explanation:The speed of the ball in the x-direction is

constant. Because of gravitational accelera-tion, the speed in the y-direction is zero atpoint Q. Since points P and R are located atthe same point above ground, by symmetrywe see that they have the same vertical speedcomponent (though they do not have the samevelocity). The answer is then “v

Q< v

P= v

R”.

023 10.0 points

A bowling ball accidentally falls out of thecargo bay of an airliner as it flies along in ahorizontal direction.

V

Z

Y

X

W

U

As observed by a person standing on theground and viewing the plane as in the fig-ure, which path would the bowling ball mostclosely follow after leaving the airplane?

1. V

2. Z

3. X correct

4. Y

5. W

6. U

Explanation:The horizontal direction of ball motion is

due to initial velocity it got from the air-liner, and the vertical is due to gravity.The resultant motion is on a parabolic curve

dotted curve X (example of projectile mo-

tion).Note: Dotted curves V and W are arcs of

circles. Dotted line Y is a straight line. Dot-ted curve U is a bezier curve. Dotted curve Zis a parabolic curve with axes inverted.

024 (part 1 of 3) 5.0 pointsNeglect: Air friction.Your teacher tosses a basketball. The ball

gets through the hoop (lucky shot).

b

ℓ

2.683m

3.048m

18m/s53◦

b

b

b

b

b

b

b

bb

bb b b b b

bb

b

b

b

Figure: Not drawn to scale.How long does it take the ball to reach its

maximum height?

Correct answer: 1.46688 s.

Explanation:

Let : α = 53◦ ,

v0 = 18 m/s ,

h1 = 2.683 m ,

h2 = 3.048 m , and

ytop = maximum height of

ball′s trajectory .

Note: The horizontal distance to the basket


ℓ is superfluous for Parts 1 and 2. This prob-lem has two distinct parts, vertical motion inParts 1 and 2, then horizontal motion in Part3.Basic Concepts:

v = v0 sinα− g t (1)

ytop − h1 =v20sin2 α

2 g(2)

ytop − h1 = v0 sinα t−1

2g t2 (3)

Solution: Using Eq. 1, the time t1 to reachthe maximum height, (i.e., vytop = 0, the ve-locity at the top) is

t1 =v0gsin(53◦)

=(18 m/s)

(9.8 m/s2)sin(53◦)

= 1.46688 s .

y1 =1

2g t21

=1

2(9.8 m/s2) (1.46688 s)2

= 10.5435 m , or

=v20sin2 α

2 g

=(18 m/s)2 sin2(53◦)

2 (9.8 m/s2)= 10.5435 m , so

ytop = y1 + h1

= (10.5435 m) + (2.683 m)

= 13.2265 m .

025 (part 2 of 3) 5.0 pointsHow long does it take the ball to reach thehoop?

Correct answer: 2.90815 s.

Explanation:The time t2 for the ball to decend from the

top into the basket is

t2 =

{

2 [ytop − h2]

g

}1/2

=

{

2 [(13.2265 m)− (3.048 m)]

(9.8 m/s2)

}1/2

= 1.44127 s .

Or symbolically, we have

t2 =

{

2 [ytop − h2]

g

}1/2

=

{

v20sin2 α

g2−

2 [h2 − h1]

g

}1/2

=

{

(18 m/s)2 sin2(53◦)

(9.8 m/s2)2

−2 [(3.048 m)− (2.683 m)]

(9.8 m/s2)

}1/2

= 1.44127 s .

The total time of the ball’s trajectory is

t = t1 + t2

= (1.46688 s) + (1.44127 s)

= 2.90815 s .

026 (part 3 of 3) 5.0 pointsWhat is the horizontal length ℓ of the shot?


Explanation:The horizontal length of the shot ℓ is

ℓ = v0 cosα t

= (18 m/s) cos(53◦) (2.90815 s)

= 31.503 m .

027 10.0 pointsA plane drops a hamper of medical suppliesfrom a height of 4610 m during a practice runover the ocean. The plane’s horizontal veloc-ity was 108 m/s at the instant the hamperwas dropped.What is the magnitude of the overall ve-

locity of the hamper at the instant it strikesthe surface of the ocean? The acceleration ofgravity is 9.8 m/s2 .


Correct answer: 319.406 m/s.

Explanation:This is a projectile motion problem. The

motion of the dropping hamper consists of twoparts: horizontally, it moves with the initialvelocity of the plane, i.e., vh = v = 108 m/s ;vertically, due to gravity, it moves as a freelyfalling body. Thus

v2 = 2 g h

vv =√

2 g h =√

2 (9.8 m/s2) (4610 m)

= 300.593 m/s

and the overall velocity at the instant thehamper strikes the surface of the ocean is

vf =√

v2v + v2h

=√

(300.593 m/s)2 + (108 m/s)2

= 319.406 m/s .

028 10.0 pointsA projectile is fired from a horizontal spring-loaded gun aimed directly (along the line ofsight) at a distant bull’s eye. Because of thepull of gravity during flight, the projectilemisses and hits a point at distance y beneaththe bull’s eye. To hit the bull’s eye, the gunshould be aimed along a line of sight abovethe bull’s eye, a vertical distance

1. of y, exactly.

2. slightly lower than y.

3. slightly higher than y. correct

Explanation:First, when the gun is tipped upward, the

horizontal component of velocity is less. Sothere’s a slightly longer time of flight andthe projectile falls a greater vertical distance.Second, depending on the mass of the projec-tile and strength of the spring, the emergingvelocity is less in the raised position becausethe spring not only pushes the projectile, butlifts it somewhat against the gravity.

See Paul Hewitt’s books for more examplesof “Figuring Physics”.

029 10.0 pointsA particle is moving in a circle at increasingspeed.If the radius of its circular path is 2 m, and

at a certain instant it is traveling at 2 m/s and

at that instantd v

dt= 2m/s2 ,what is the ratio

atar

of its tangential to its radial acceleration?

1. 4.0

2. 0.1

3. 0, since it is moving in a circle

4. 0.5

5. 0.25

6. 2.0

7. 1.0 correct

8. 16

9. 10

Explanation:

at =d v

dt= 2 m/s2 , so

ar =v2

r=

(2 m/s)2

2 m= 2 m/s2 = at

atar

= 1 .

030 10.0 pointsA particle moves along a circular path of ra-dius r with tangential speed v = b t2, whereb is a positive constant. If b = 1 m/s3, andr = 1m, at the time t = 1 sec what is the an-gle φ between the radian acceleration ar andthe full acceleration a = ar + at?


~aT

~ar

~aφ

1. 85.6◦

2. 63.4◦ correct

3. 0◦

4. 74.2◦

5. 10◦

6. 56.7◦

7. 45◦

8. 25.5◦

9. 90.0◦, by definition of ar.

Explanation:

In general, a = −r̂

(

v2

r

)

+ θ̂

(

d v

dt

)

.

v = b t2 sod v

dt= 2 b t = 2(1 m/s4)(1 s) = 2 m/s

2and

the angle φ between ar and a is

φ = tan−1

(

atar

)

= tan−1(2) = 63.4◦ .

031 10.0 points

A tennis ball is swinging at the end of astring, like a pendulum.

rest

max

Qualitatively, what is the direction of thetennis ball’s acceleration a when it is abouthalfway between its highest point (where itsspeed goes to zero) and its lowest point (whereits speed is a maximum?

1.

~arest

correct

2.

~a

rest

3.~a

rest

4.

~a

rest

Explanation:At the half-way position shown, the ball

has an acceleration at in the direction of v(the ball’s speed is increasing), and since it ismoving in a circle, it has an acceleration ar(the radial acceleration, also called centripetalacceleration) directed toward the center of thecircle. So the total acceleration a = ar + atis the resultant of these two perpendicularvectors.


032 10.0 pointsAn automobile moves at constant speeddown one hill and up another hill along thesmoothly curved surface as shown below.

car

Which of the following diagrams best rep-resents the directions of the velocity and theacceleration of the automobile at the instantthat it is at the lowest position as shown?

1.

v a

2.

v a

3.

v a

4.

v a

5.

v a

correct

Explanation:At the lowest position, the instantaneous

velocity of the automobile is directed horizon-tally to the right.Since the speed of the automobile is un-

changed, there is no tangential accelera-tion. Also since the path of the automobileis curved, there is centripetal acceleration,pointing upward.

033 (part 1 of 3) 5.0 pointsA river flows due east at 2.45 m/s. A boatcrosses the river from the south shore to thenorth shore by maintaining a constant veloc-ity of 10.1 m/s due north relative to the water.a) What is the magnitude of the velocity of

the boat as viewed by an observer on shore?


Explanation:

2.45 m/s10.1

m/s

vbeθ

Note: Figure is not drawn to scale.Basic Concepts:

~vbe = ~vbr + ~vre

The velocities are perpendicular, so

vbe =√

v2br + v2re

Given:Let east and north be positive:

vre = 2.45 m/s

vbr = 10.1 m/s

Solution:

vbe =√

(10.1 m/s)2 + (2.45 m/s)2

= 10.3929 m/s

034 (part 2 of 3) 5.0 points


b) How many degrees off course is the boatforced by the current?

Correct answer: 13.6351◦.

Explanation:Basic Concept:

tan θ =vrevbr

Solution:

θ = tan−1

(

vrevbr

)

= tan−1

(

2.45 m/s

10.1 m/s

)

= 13.6351◦

east of north.

035 (part 3 of 3) 5.0 pointsc) If the river is 217 m wide, how far down-stream is the boat when it reaches the northshore?


Explanation:Basic Concepts:

∆x = vre∆t

∆y = vbr∆t

Given:∆y = 217 m

Solution: The times are the same, so

∆t =∆y

vbr=

∆x

vre

so that

∆x =vre∆y

vbr

=(2.45 m/s)(217 m)

10.1 m/s= 52.6386 m

036 (part 1 of 3) 5.0 pointsRabid Texas basketball fans have been known

to rush out of their laboratory class at 5:00p.m. and immediately get on highway I-35to arrive in Waco, 100 miles away, at 7:00p.m. to see the Baylor-Longhorn basketballgame. They spend 4 hours watching and thencelebrating the outcome of the game, and thendrive 2 hours back to Austin.After they completed their trip, what is

their mean velocity for their trip? mph ismiles per hour

1. 100 mph

2. 0 mph correct

3. 50 mph

4. 25 mph

5. 12.5 mph

Explanation:The average velocity is

vav =final position− initial position

final time− initial time

=0 mi

8 h= 0 mph .

037 (part 2 of 3) 5.0 pointsAfter they completed their trip, what is theirmean speed for the trip?

1. 50 mph

2. 12.5 mph

3. 0 mph

4. 100 mph

5. 25 mph correct

Explanation:The average speed is

total distance

time=

200 mi

8 h= 25 mph .



What is their mean speed from when theyleave Austin and first arrive in Waco?

1. 50 mph South

2. 25 mph North

3. 50 mph North

4. 50 mph correct

5. 25 mph

6. 25 mph South

Explanation:Their mean speed is

total distance

time=

100 mi

2 h= 50 mph .

Speed is a scalar and does not have a direction.

039 10.0 pointsConsider the vectors ~A and ~B.

~A~B

Which sketch shows the vector ~A− ~B to thesame scale?

1.

2.

3. correct

4.

5.

6.

7.

8.

Explanation:

~A

−~B~A− ~B


Express the vector ~R

A

B

C

DP R

in terms of ~A, ~B, ~C, and ~D, the edges of aparallelogram.

1. ~R = ~A− ~D

2. ~R = ~D − ~A


3. ~R = ~B + ~A correct

4. ~R = ~A+ ~D

5. ~R = ~C + ~B

6. ~R = ~B − ~A

7. ~R = ~C + ~D

8. ~R = ~A− ~B

9. ~R = ~A− ~C

10. ~R = ~B + ~D

Explanation:Apply the parallelogram rule of addition:

join the tails of the two vectors ~A and ~B; theresultant vector is the diagonal of a parallel-ogram formed with ~A and ~B as two of itssides.


Express the vector ~P in terms of ~A, ~B, ~C, and~D ,

1. ~P = ~B + ~D

2. ~P = ~D − ~A

3. ~P = ~C + ~B

4. ~P = ~C + ~D

5. ~P = ~B + ~A

6. ~P = ~A− ~D

7. ~P = ~A+ ~D

8. ~P = ~B − ~A

9. ~P = ~C − ~A

10. ~P = ~A− ~B correct

Explanation:By the triangle method of addition

~B + ~P = ~A

~P = ~A− ~B .

042 10.0 pointsAman can throw a ball a maximum horizontaldistance of 107 m.The acceleration of gravity is 9.8 m/s2 .How far can he throw the same ball verti-

cally upward with the same initial speed?


Explanation:The range of a particle is given by the ex-

pression

R =v20sin 2 θ

g.

The maximum horizontal distance is obtainedwhen the ball is thrown at an angle θ = 45◦

and sin 2 θ = 1 . Solving for v0,

v0 =√

g R .

When the ball is thrown upward with thisspeed, the maximum height is obtained fromthe equation

v2f = v20 − 2 g h .

Let vf = 0, and solve for h

h =v20

2 g.

043 (part 1 of 3) 5.0 pointsA particle moves in the xy plane with constantacceleration. At time zero, the particle isat x = 1.5 m, y = 8.5 m, and has velocity~vo = (5 m/s) ı̂+(−8 m/s) ̂ . The accelerationis given by ~a = (6.5 m/s2) ı̂+ (5.5 m/s2) ̂ .What is the x component of velocity after

4.5 s?


Explanation:

Let : ax = 6.5 m/s2 ,

vxo = 5 m/s , and

t = 4.5 s .


After 4.5 s,

~vx = ~vxo + ~ax t

= (5 m/s) ı̂+ (6.5 m/s2) ı̂ (4.5 s)

= (34.25 m/s) ı̂ .

044 (part 2 of 3) 5.0 pointsWhat is the y component of velocity after4.5 s?


Explanation:

Let : ay = 5.5 m/s2 and

vyo = −8 m/s .

~vy = ~vyo + ~ay t

= (−8 m/s) ̂+ (5.5 m/s2) ̂ (4.5 s)

= (16.75 m/s) ̂ .

045 (part 3 of 3) 5.0 pointsWhat is the magnitude of the displacementfrom the origin (x = 0 m, y = 0 m) after4.5 s?


Explanation:

Let : do = (1.5 m, 8.5 m) ,

vo = (5 m/s,−8 m/s) , and

a = (6.5 m/s2, 5.5 m/s2) .

From the equation of motion,

~d = ~do + ~vo t+1

2a t2

=[

(1.5 m) ı̂+ (8.5 m) ̂]

+ [(5 m/s) ı̂+ (−8 m/s) ̂] (4.5 s)

+1

2

[

(6.5 m/s2) ı̂+ (5.5 m/s2) ̂]

(4.5 s)2

= (89.8125 m) ı̂+ (28.1875 m) ̂ , so

|~d| =√

d2x + d2y

=√

(89.8125 m)2 + (28.1875 m)2

= 94.1319 m .

PHY317 Hw # 1-Solutions

Documents

Transcript of PHY317 Hw # 1-Solutions