Chapter 3: Vectors and Two-Dimensional Motion

PHU 205 Mechanics for Life Sciences

Chapter 3:

Vectors and Two-Dimensional Motion

Prof. Liliana Braescu & Prof. Nouredine Zettili

Required text College Physics Raymond A. Serway and Chris Vuille 9th Edition, 2012 BROOKS/COLE CENGAGE Learning ISBN 10: 1-111-42745-3 ISBN 13: 978-1-111-42745-0

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Course Outline Chapter 3

Vectors and Two-Dimensional Motion 3.1 Vectors and Their Properties 3.2 Components of a Vector 3.3 Displacement, Velocity and Acceleration in Two

Dimensions 3.4 Motion in Two Dimensions 3.5 Relative Velocity

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Chapter 3 Vectors and Two-Dimensional Motion 3.1 Vectors and Their Properties 3.2 Components of a Vector 3.3 Displacement, Velocity and Acceleration in Two


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3.1 Vectors and Their Properties u A scalar quantity is defined by magnitude only,

✔e.g. mass, time, temperature u A vector quantity is defined by both magnitude and direction,

✔e.g. velocity, acceleration, force u Vector quantities are denoted by placing an arrow above the letter (or boldface letter): → vector → magnitude of vector u Equality of two vectors: For two vectors to be equal, they must have exactly the same magnitude and direction.

!v , !a,

!F

!A

!A = A

!A

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Adding (Subtracting) Vectors u Adding vectors: Only vector quantities with the

same physical units can be added/subtracted. u There are two graphical methods of adding/

subtracting vectors: Head-to-Tail or Triangle Method Tail-to-Tail or Parallelogram Method

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Properties of Addition/Subtraction u Properties of vector addition/subtraction: ✔Commutative: ✔Associative: ✔Additive Inverse

!A+!B =!B +!A

!A+

!B +!C( ) = !A+

!B( ) + !C

!A+ −

!A( ) = 0

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Multiplication, Division of Vectors with Scalars u Multiplication of a vector by a scalar: ✔ Vector has four times the magnitude and is oriented in the same direction. ✔ Vector has three times the magnitude and is oriented in the opposite direction. u Division of a vector by a scalar: ✔ Vector has half the magnitude and is oriented in the same direction. ✔ Vector has one third the magnitude and is oriented in the opposite direction.

4!A

−3!A

!A / 2

−!A / 3

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3.2 Components of a Vector (in Two-Dimensional Case) Vectors in plane (2D) vs. Vectors in space (3D)

(3, 2)

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3.2 Components of a Vector (in Two-Dimensional Case)

u is the x component of u is the y component of u When the magnitude and the direction of a vector are given, one can find the x and y components using the following relations: u One can also determine the magnitude and direction of a vector if given the x and y components:

Ax

!A Ay

cosxA A θ= sinyA A θ=

A= Ax( )2 + Ay( )2

θ = tan−1 Ay

Ax

⎛

⎝⎜⎞

⎠⎟, i.e. arctan

Ay

Ax

⎛

⎝⎜⎞

⎠⎟

!A

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3.2 Components of a Vector (in Two-Dimensional Case)

u Vectors can be added/subtracted using their components.

u To find the sum of three vectors, proceed in three steps:

1.  Find the x and y components of each individual vector. 2.  Calculate and 3.  Using these components, calculate the magnitude and direction of the resultant vector:

magnitude →

direction →

Rx = Ax + Bx +Cx Ry = Ay + By +Cy

R = Rx2 + Ry

2

1tan y

x

RR

θ − ⎛ ⎞= ⎜ ⎟

⎝ ⎠

!A Ax

, Ay( ), !B B

x, B

y( ), !C Cx,C

y( )

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Problems

u Problem 3.1 A golfer takes two putts to get his ball into the hole once he is on the green. The first putt displaces the ball 6.00 meters east, and the second, 5.40 meters south. Question: What displacement would have been needed to get the ball into the hole on the first putt?

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Problems u Problem 3.1 Solution

A total distance of = 8.07 m with an angle of arctan(5.40/6.00) = 42.0° south of east.

6.00( )2 + 5.40( )2

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Problems

u Problem 3.2 A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, then 6.00 blocks east.

Questions: (a) What is her resultant displacement? (b) What is the total distance she travels?

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Problems u Problem 3.2

Solution (a) Her resultant x (east-west) displacement is

-3.00 + 0 + 6.00 = 3.00 blocks, while her resultant y (north-south) displacement is

0 + 4.00 + 0 = 4.00 blocks. Her total displacement is thus 5.00 blocks at 53.1° north of east. (b) Her total distance is 3.00 + 4.00 + 6.00 = 13.00 blocks.

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Chapter 3 Vectors and Two-Dimensional Motion 3.1 Vectors and Their Properties 3.2 Components of a Vector 3.3 Displacement, Velocity and Acceleration in

Two Dimensions 3.4 Motion in Two Dimensions 3.5 Relative Velocity

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3.3 Displacement, Velocity and Acceleration in Two Dimensions

u In two dimensions, position is denoted by the vector u → initial position vector u → final position vector u Using this notation in the place of x, we can redefine two-dimensional displacement, velocity, and acceleration:

!r !ri

!rf

Δ!r = !rf −

!ri

!v = Δ!rΔt

=!rf −!ri

t f − ti

!a = Δ!vΔt

=!v f −!vi

t f − ti

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3.4 Motion in Two Dimensions

u Examples of projectile (two-dimensional) motion: motion of a fired bullet, a ballistic missile, etc. u For the purposes of this course, we need to make three assumptions when discussing projectile motion:

✔The acceleration due to gravity remains. constant over the range of the motion. ✔Air resistance can be neglected. ✔The effect of Earth’s rotation can be neglected.

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Projectile Motion u Since projectile motion takes place in two dimensions, the acceleration of the object must be described by both an x and y components: u Note that the object only experiences an acceleration along the y-axis because gravity is the only force acting upon it.

29.8 /ya g m s= − = −0xa =

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u When fired with an initial velocity at an angle θ, the velocity vector will have both an x and y component: uProjectile motion consists of two types of motion: ✔ free-fall motion along the y-axis (due to gravity), and ✔ uniform motion along the x-axis.

vox= !vo cosθ

voy

= !vo sinθ

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Projectile Motion

u Motion along x-axis: ax=0 u Motion along y-axis: ay=-g=-9.8 m/s2

vx = vo cosθ

x = vot cosθ

vy = vo sinθ − gt

y = 1

2(vy + vo sinθ )t

y = vot sinθ − 1

2gt2

vy2 − (vo sinθ )2 = −2gy

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Remarks:

u - constant from start of motion to end u at the highest point, and

vx = vo cosθ

vy = 0 vx = vo cosθ ≠ 0

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Remarks:

u Projectile motion is symmetrical; the time interval from the ground to the highest point equals the time interval from the highest point to the ground. u The speed at any given time is calculated using the x and y components of the speed: v = vx

2 + vy2 = vo

2 cos2θ + vy2

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Remarks:

u Maximum height reached by the projectile: u Maximum range (R) of the projectile: u Complimentary angles give the same range (provided the initial velocity is the same).

ymax =

voy

2

2g= vo

2 sin2θ2g

R = xmax =

2voxvoy

g= 2vo

2 sinθ cosθg

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Projectile Motion at Various Initial Angles

u Complementary values of the initial angle result in the same range, the heights will be different. u The maximum range occurs at a projection angle of 45o

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Problems

u Problem 3.3 A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 meters above a flat horizontal beach, as shown in the figure. Questions: ✔ How long after being released does the stone strike the beach below the cliff? ✔ With what speed and angle of impact does it land?

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Solution ✔ To begin, v0x = 18.0 m/s & v0y = 0.

We find the time of fall as y = voyt + (1/2)at2 or -50.0 m = (1/2)(-9.8)t2,

which gives t = 3.19 s. ✔ At impact, the horizontal component of velocity is

vx = v 0x = 18.0 m/s, and the vertical component is

vy = v0y + at = 0 + (-9.80 m/s2)(3.19 s) = - 31.3 m/s.

The resultant velocity is found from the Pythagorean Th. v = = 36.1 m/s, at an angle below the horizontal of arctan(31.3/18.0) = 60.1°.

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Problems

u Problem 3.4 A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 15 m/s. Question: If the brick is in flight for 3.0 seconds, how tall is the building?

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Solution We choose our origin at the initial position of the projectile. After 3 s, it is at ground level, y = -H. To find H, we use y = v0Yt + (1/2)at2.

-H = (15)(sin25°)(3) + (1/2)(-9.8)(3)2 = -25.1 m.

Thus, H = 25 m.

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Problems

u Problem 3.5 An artillery shell is fired with an initial speed of 1.70 x 103 m/s at an angle of 55.0° above the horizontal and returns to its original level before impact. Questions: Neglecting air resistance, find (a) the time it is in motion and (b) the horizontal distance traveled.

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Solution The components of the initial velocity are

v0 (hor) = 1700cos55.0° = 975 m/s and v0 (vert) = 1700sin 55.0° = 1393m/s.

(a) The time of flight is the time for the shell to return to its original level (y = 0), so we use

y = v0yt + (1/2)at2 or 0 = 1393t - 4.9 t2 and find t = 0 or t = 284 s. The time of flight is therefore 284 s or 4.73 min. (b) The horizontal distance traveled during this time (i.e., the range) is given by

x = v0t = 975(284) = 2.77 x 105 m = 277 km.

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3.5 Relative Velocity

u The pattern of subscripts can be useful in solving relative velocity problems u Assume the following notation:

✔E is an observer, stationary with respect to the earth ✔A and B are two moving cars

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3.5 Relative Velocity

u is the position of car A as measured by E u is the position of car B as measured by E u is the position of car A as measured by car B u The posi)on of car A rela)ve to car B is given by the vector subtrac)on equa)on

r!AE

r!BE

r!AB

r!AB = r!AE − r!BE

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Relative Velocity Equation

uThe rate of change of the displacements gives the relationship for the velocities Note: If there is something not explicitly noted as being relative to something else, it is probably relative to the earth

v!AB = v!AE − v!BE

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Relative Velocity, Example

u Need velocities: ✔Boat relative to river ✔River relative to the Earth ✔Boat with respect to the Earth (observer) u Equation:

=>

v!BE = v!RE + v!BR

v!BR = v!BE − v!RE

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Chapter 3 Ends

Quiz 3 is coming ….. before Chapter 4 For review, Solve the following problems: 4, 5 page 78; 11, 18, 19 page

79; 29 page 80; and 32, 36, 37 page 81. Textbook “College Physics” Raymond A. Serway and Chris Vuille 9th Edition, 2012

Test 1 is coming soon (Chapters 1,2,3)…

Chapter 3: Vectors and Two-Dimensional Motion

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Transcript of Chapter 3: Vectors and Two-Dimensional Motion