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Ch.3 Kinemtics In 2-D

Displacement vector is in the x-y plane (not in x or y direction)

t

r

tt

rrv

0

0

v = limitt

r

0t

Displacement 0rrr

2

Velocity Components In 2-D

Tangent to the path

(instantaneous velocity)

Average acceleration

Instantaneous Acceleration

t

v

tt

vva

0

0

a = limitt

v

0t

3

Motion Only In X Direction

x – motion equations

x

ax

vx

v0x

t

vx= v0x+axt (3.3a)

x=(1/2)(v0x+vx)t (3.4a)

x=v0xt+(1/2)axt2 (3.5a)

vx2=v0x

2+2axx (3.6a)

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Motion In Y Direction

y

ay

vy

v0y

t

vy= v0y+ayt (3.3b)

y=(1/2)(v0y+vy)t (3.4b)

y=v0yt+(1/2)ayt2 (3.5b)

vy2=v0y

2+2ayy (3.6b)

y component

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2 D Motion

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x Component Variable y Component

x Displacement y

axAcceleration ay

vxFinal velocity vy

v0xInitial velocity v0y

t Elapsed time t

vx= v0x+axt (3.3a) vy= v0y+ayt (3.3b)

x=(1/2)(v0x+vx)t (3.4a) y=(1/2)(v0y+vy)t (3.4b)

x=v0xt+(1/2)axt2 (3.5a) y=v0yt+(1/2)ayt2 (3.5b)

vx2=v0x

2+2axx (3.6a) vy2=v0y

2+2ayy (3.6b)

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Example 1

x ax vx v0x t

? +24m/s2 ? +22m/s 7s

y ay vy v0y t

? 12m/s2 ? 14 m/s 7 s

x-Direction Data y-Direction Data

mssmssmtatvx x 740)77)(/24(2

1)7)(/22(

2

1 2220

smssmsmtavv xxx /190)7()/24()/22( 20

8

mssmssmtatvy yy 390)77)(/12(2

1)7()/14(

2

1 2220

smssmsmtavv yyy /987)/12(/14 20

smvvv yx /210)98()190( 2222

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Projectile Motion

• 2-D motion under gravity

• What is the acceleration in X direction ?

• What is the acceleration in y direction ?

• 2-D motion under gravity

• What is the acceleration in X direction ? • What is the acceleration in y direction ?

ax= 0 ay = - 9.8 m/s2

Garden Horse – watering a planta Plane dropping a bomb

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Example 2: Falling Package

The time to hit the ground ?

y ay vy v0y t

-1050 -9.8 ? 0 ?

y = v0y t +(1/2) ay t2 = ½ (-9.8) t2

i.e., -1050 = ½ (-9.8) t2 st 6.14

8.9

)1050(2

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Velocity Of The Package When It Hits The Ground ?

vx = v0x + ax t = v0x = 115 m/s

vy = v0y + ay t = (-9.8 m/s2) 14.6 s = -143 m/s (Why Negative ?)

v2 = (115)2+ (-143)2 v = 184 m/s

vx

vy R

θ

y

x

v

vtan

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Shoot A Bullet Straight Up In A Moving Car

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Example 5: Height Of A Kickoff

y ay v0y vy t

H=?

-9.8 14 0 ?

v2 = v02 + 2ay

0 = (14)2 = 2*(-9.8)H

For y direction

v0x = v0 cosθ = 22 m/s cos 40.0o = 17m/s

v0y = v0 sinθ = 22 m/s sin 40.0o =14m/s

H = 14*14 /(2*9.8)= 10 m

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Time Of Flight ( Time In The Air)

Time until it hits the ground

y = v0yt + (½) ay t2

0 = 14 t + (½) (-9.8) t2

i.e., t(14 -4.9t) = 0 sect = 0 sec, or t= 14/4.9 = 2.9 sec (why two answers?)

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Time To Reach The Maximum Height

y ay v0y vy t

H=? -9.8 14 0 ?

vy = v0y +ay t

0 = 14 -9.8t

t = 14/9.8 = 1.45 sec

2t = 2.9 (time of flight) sec

Symmetry: time of flight = twice the time to reach the top --- Why?

RANGER = v0x t + (½ )ax t2

= 17 *2.9 = 49 m

vx

vy R

θy

x

v

vtan

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Same Height Same Speed

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Relative Velocity

vPG = vPT + vTG

= 2.0 m/s + 9.0 m/s

= 11.0 m/s

vPG = Velocity of Passenger relative to Ground

(vPG = - vGP)

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Check your understanding 4

Three cars A, B, C, are moving along a straight line, Relative velocities are given.

vAB vAC vCB

1. ? 40 m/s 30 m/s

2. ? 50 m/s -20 m/s

3. 60 m/s 20 m/s ?

4. -50 m/s ? 10 m/s

Three cars A, B, C, are moving along a straight line, Relative velocities are given.

vAB vAC vCB

1. 70 m/s 40 m/s 30 m/s

2. ? 50 m/s -20 m/s

3. 60 m/s 20 m/s ?

4. -50 m/s ? 10 m/s

Three cars A, B, C, are moving along a straight line, Relative velocities are given.

vAB vAC vCB

1. 70 m/s 40 m/s 30 m/s

2. 30 m/s 50 m/s -20 m/s

3. 60 m/s 20 m/s ?

4. -50 m/s ? 10 m/s

Three cars A, B, C, are moving along a straight line, Relative velocities are given.

vAB vAC vCB

1. 70 m/s 40 m/s 30 m/s

2. 30 m/s 50 m/s -20 m/s

3. 60 m/s 20 m/s 40 m/s

4. -50 m/s ? 10 m/s

Three cars A, B, C, are moving along a straight line, Relative velocities are given.

vAB vAC vCB

1. 70 m/s 40 m/s 30 m/s

2. 30 m/s 50 m/s -20 m/s

3. 60 m/s 20 m/s 40 m/s

4. -50 m/s -60 m/s 10 m/s

Three cars A, B, C, are moving along a straight line, Relative velocities are given.

vAB vAC vCB

1. ? 40 m/s 30 m/s

2. ? 50 m/s -20 m/s

3. 60 m/s 20 m/s ?

4. -50 m/s ? 10 m/s

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Crossing A RivervBS = vBW + vWS

R2 = (vBW)2 + (vWS)2

vWS

vBW

RvBS

BW

WS

v

vtan

BW

WS

v

v1tan

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10. Crossing A River Width of the river = 1800 m

vBS = vBW + vWS

= 4.0 m/s + 2.0 m/s

= 4.5 m/s

θ = tan-1(2) = 63º

0.2tan WS

BW

v

v

BWv

widthcrosstotime

s4500.4

1800

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11. Approaching Intersection

vAG = 25.0 m/s

vBG = 15.8 m/s

vAB = vAG + vGB

vAB = 25.0 m/s + 15.8 m/s

tanθ = 15.8/25

Θ = tan-1(15.8/25) = 32.4ºsmvAB /6.29)8.15()25( 22

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Raindrops On The Car

vRC = vRG + vGC

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Concepts: Circus Clowns

vox = 4.6 m/s

v0y = 10.0 m/s

No air resistance

Θ = tan-1(v0y/v0x)

= tan-1(10/4.6)

= 65º

220 )10()6.4( v

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Conceptual Question 2

REASONING AND SOLUTION An object thrown upward at an angle will follow the trajectory shown below. Its acceleration is that due to gravity, and, therefore, always points downward. The acceleration is denoted by ay in the figure. In general, the velocity of the object has two components, vx and vy. Since ax = 0, vx always equals its initial value. The y component of the velocity, vy, decreases as the object rises, drops to zero when the object is at its highest point, and then increases in magnitude as the object falls downward.

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–v0y

v0x

vf

ay

ay

v0y

ay

v0x

v0x

b.) In order for the velocity and acceleration to be parallel, the x component of the velocity would have to drop to zero. However, vx always remains equal to its initial value; therefore, the velocity and the acceleration can never be parallel.

a.) Since vy = 0 when the object is at its highest point, the velocity of the object points only in the x direction. As suggested in the figure below, the acceleration will be perpendicular to the velocity when the object is at its highest point and vy = 0.

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Conceptual Question 4

REASONING AND SOLUTION If a baseball were pitched on the moon, it would still fall downwards as it travels toward the batter. However the acceleration due to gravity on the moon is roughly 6 times less than that on earth. Thus, in the time it takes to reach the batter, the ball will not fall as far vertically on the moon as it does on earth. Therefore, the pitcher's mound on the moon would be at a lower height than it is on earth.

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Conceptual Questions 13

REASONING AND SOLUTION Since the plastic bottle moves with the current, the passenger is estimating the velocity of the boat relative to the water. Therefore, the passenger cannot conclude that the boat is moving at 5 m/s with respect to the shore.

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Conceptual Questions 16

REASONING AND SOLUTION The time required for any given swimmer to cross the river is equal to the width of the river divided by the magnitude of the component of the velocity that is parallel to the width of the river. All three swimmers can swim equally fast relative to the water; however, all three swim at different angles relative to the current. Since swimmer A heads straight across the width of the river, swimmer A will have the largest velocity component parallel to the width of the river; therefore, swimmer A crosses the river in the least time.

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Problem 4

REASONING AND SOLUTION The increase in altitude represents vy = 6.80 m/s. The movement of the shadow represents vx = 15.5 m/s. The magnitude of the glider's velocity is therefore

2 22 2 15.5 m/s 6.80 m/s 16.9 m/sx yv v v

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Problem 7

REASONING Trigonometry indicates that the x and y components of the dolphin’s velocity are related to the launch angle according to tan = vy /vx.

SOLUTION Using trigonometry, we find that the y component of the dolphin’s velocity is

smsmvvv xxy /4.535tan)/7.7(35tantan

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Problem 14

REASONING The vertical component of the ball’s velocity v0 changes as the ball approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal component does not change, assuming that air resistance can be neglected.

55o

15 m/s

vx=?

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SOLUTION Using trigonometry, we find that the horizontal component is

smsmvvx /6.855cos)/15(cos0

Hence, the horizontal component of the ball’s velocity when the opposing player fields the ball is the same as it was initially.

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Problem 16

REASONING AND SOLUTION Using vy = 0 and

voy = vo sin = (11 m/s) sin 65 = 1.0 101 m/s

and vy2 = voy

2 + 2ayy, we have

22 10

2

1.0 10 m/s5.1 m

2 2 9.80 m/s

y

y

vy

a

A 65O

v0=11m/s

v0y=0

B

h=?

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Problem 27

35

REASONING AND SOLUTION The time of flight of the motorcycle is given by

0 02

2 sin 2 33.5 m/s sin18.02.11 s

9.80 m/s

vt

g

The horizontal distance traveled by the motorcycle is then

x = vo cos o t = (33.5 m/s)(cos18.0°)(2.11 s) = 67.2 m

The daredevil can jump over (67.2 m)/(2.74 m/bus) = 24.5 buses. In even numbers, this means 24 buses

y=0, v0y=33.5sin18, t=?

y=v0t+(1/2)at2 2

00 8.92

1sin0 ttv

t=0, or

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Problem 40

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REASONING Using the data given in the problem, we can find the maximum flight time t of the ball using Equation 3.5b

1 20 2y yy v t a t

Once the flight time is known, we can use the definition of average velocity to find the minimum speed required to cover the distance x in that time.

SOLUTION Equation 3.5b is quadratic in t and can be solved for t using the quadratic formula. According to Equation 3.5b, the maximum flight time is (with upward taken as the positive direction)

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where the first root corresponds to the time required for the ball to reach a vertical displacement of as it travels upward, and the second root corresponds to the time required for the ball to have a vertical displacement of as the ball travels upward and then downward. The desired flight time t is 2.145 s.

y

yyy

y

yyy

a

yavv

a

yavv

t2

21

2

21

4 200

200

2

22

/8.9

1.2/8.920.50sin/0.150.50sin/0.15

sm

msmsmsm

sands 145.2200.0

2.10 my

2.10 my

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During the 2.145 s, the horizontal distance traveled by the ball is

Thus, the opponent must move in . The opponent must, therefore, move with a minimum average speed of

vmin

10.68 m

1.845 s5.79 m / s

mssmtvtvx x 68.20)145.2(0.50cos)/0.15()cos( 0

20.68 m –10.0 m 10.68 m2.145 s – 0.30 s 1.845 s

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Problem 41

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REASONING AND SOLUTION In the absence of air resistance, the bullet exhibits projectile motion. The x component of the motion has zero acceleration while the y component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the bullet is given by Equation 3.5a (with ):

0 0( cos )xx v t v t

with t equal to the time required for the bullet to reach the target. The time t can be found by considering the vertical motion. From Equation 3.3b

0y y yv v a t

0xa

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When the bullet reaches the target, . Assuming that up and to the right are the positive directions, we have

0y yv v

0 0 00

2 2 sin 2 sinand ( cos )y

y y y

v v vt x v

a a a

Using the fact that , we have 2sin cos sin 2

2 20 02 cos sin sin 2

y y

v vx

a a

Thus, we find that

32

2

1091.4)/427(

/8.94.91

sm

smm

43

2 0.281 or 2 180.000 0.281 179.719

and

0.141 and 89.860

Therefore

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Problem 48REASONING Since car A is moving faster, it will eventually catch up with car B. Each car is traveling at a constant velocity, so the time t it takes for A to catch up with B is equal to the displacement between the two cars (x = +186 m) divided by the velocity vAB of A relative to B. (If the relative velocity were zero, A would never catch up with B). We can find the velocity of A relative to B by using the subscripting technique developed in Section 3.4 of the text.

24.4 m/s 18.6 m/s

A B186 m

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vAB = velocity of car A relative to car B

vAG = velocity of car A relative to the Ground = +24.4 m/s

vBG = velocity of car B relative to the Ground = +18.6 m/s

We have chosen the positive direction for the displacement and velocities to be the direction in which the cars are moving. The velocities are related by

vAB = vAG + vGB

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SOLUTION The velocity of car A relative to car B is

vAB = vAG + vGB = +24.4 m/s + (18.6 m/s) = +5.8 m/s,

186 m32.1 s

+5.8 m/st

AB

x

v

where we have used the fact that vGB = vBG = 18.6 m/s. The time it takes for car A to catch car B is

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Problem 53REASONING Let represent the velocity of the hawk relative to the balloon and represent the velocity of the balloon relative to the ground. Then, as indicated by Equation 3.7, the velocity of the hawk relative to the ground is . Since the vectors and are at right angles to each other, the vector addition can be carried out using the Pythagorean theorem. (B: Balloon, H: Hawk, G: Ground)

SOLUTION Using the drawing at the right, we have from the Pythagorean theorem

HBv

BGv

HG HB BG v v v

HBv BGv

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v v vHG HB

2BG2

2 2 (2.0 m / s) (6.0 m / s) 6.3 m / s

vHB

EAST

vHG

vBG

NORTH

The angle is

eastofnorthsm

sm

v

v

BG

HB ,18/0.6

/0.2tantan 11

=2.0

=6.0

B: Balloon

H: Hawk

G: Ground

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Problem 56

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REASONING AND SOLUTION The velocity of the raindrops relative to the train is given by vRT = vRG + vGT

where vRG is the velocity of the raindrops relative to the ground and vGT is the velocity of the ground relative to the train

Since the train moves horizontally, and the rain falls vertically, the velocity vectors are related as shown in the figure at the right. Then

vRG

GTv

RTv25°

vGT = vRG tan = (5.0 m/s) (tan 25°) = 2.3 m/s

The train is moving at a speed of 2.3 m/s