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9/3/2009<br />

Understanding<br />

<strong>Kinematics</strong>/Newton’s <strong>Laws</strong><br />

Practice Questions<br />

An object that’s moving with constant speed travels once<br />

around a circular path. Which of the following is/are true<br />

concerning this motion?<br />

I <strong>The</strong> displacement is zero<br />

II <strong>The</strong> average speed is zero<br />

III <strong>The</strong> acceleration is zero<br />

(a) I only<br />

(b) I and II only<br />

(c) I and III only<br />

(d) III only<br />

(e) II and III only<br />

Understanding<br />

Understanding<br />

Which of the following is/are true?<br />

I. If an object’s acceleration is constant, then it must<br />

move in a straight line.<br />

II. If an object’s acceleration is zero, then its speed must<br />

remain constant.<br />

III. If an object’s speed remains constant, then its<br />

acceleration must be zero.<br />

(a) I and II only<br />

(b) I and III only<br />

(c) II only<br />

(d) III only<br />

(e) II and III only<br />

A baseball is thrown straight upward. What is the ball’s<br />

acceleration at its highest point?<br />

(a) 0<br />

(b) ½ g [down]<br />

(c) g [down]<br />

(d) ½ g [up]<br />

(e) g [up]<br />

Gravity is always turned on.<br />

1


9/3/2009<br />

Understanding<br />

Understanding<br />

Hint:<br />

How long would it take a car, starting from rest and<br />

accelerating uniformly in a straight line at 5 m/s 2 , to cover<br />

a distance of 200m?<br />

1<br />

s v0t at<br />

2<br />

(a) 9.0 s<br />

(b) 10.5 s<br />

(c) 12.0 s<br />

(d) 15.5 s<br />

(e) 20.0 s<br />

2<br />

m 1 m<br />

200m 0 t 5<br />

t<br />

2 <br />

s 2 s <br />

2 400m<br />

t <br />

m<br />

5<br />

2<br />

s<br />

t 8.9s<br />

2<br />

Hint:<br />

A rock is dropped off a cliff and strikes the ground with an<br />

impact of 30 m/s. How high was the cliff?<br />

We need both the<br />

initial and final velocity<br />

in the equation<br />

(a) 15.0 m<br />

(b) 20 m<br />

(c) 30 m<br />

(d) 46 m<br />

(e) 60 m<br />

v v 2a s<br />

2 2<br />

f i<br />

v v 2a s<br />

2 2<br />

f i<br />

<br />

v v<br />

s<br />

<br />

2a<br />

2 2<br />

f i<br />

<br />

<br />

<br />

2 2<br />

m m<br />

30 0<br />

<br />

s s<br />

<br />

<br />

m <br />

29.8<br />

2 <br />

s <br />

45.9m<br />

Understanding<br />

Understanding<br />

A stone is thrown horizontally with an initial speed of 10 m/s from a<br />

bridge. If air resistance could be ignored, how long would it take the<br />

stone to strike the water 80 m below the bridge?<br />

Hint:<br />

We need only think<br />

about the vertical<br />

components.<br />

(a) 1 s<br />

(b) 2 s<br />

(c) 4 s<br />

(d) 6 s<br />

(e) 8 s<br />

1<br />

s v0t at<br />

2<br />

m 1 m<br />

80m 0 t 9.8<br />

t<br />

2 <br />

s 2 s <br />

2 160m<br />

t <br />

m<br />

9.8<br />

2<br />

s<br />

t 4.0s<br />

2<br />

2<br />

A soccer ball, at rest on the ground, is kicked with an initial velocity of 10<br />

m/s at a launch angle of 30 0 . Calculate its total flight time, assuming air<br />

resistance is negligible?<br />

Hint:<br />

We need only think<br />

about the vertical time<br />

to travel up to its max<br />

height, then double it.<br />

(a) 0.5 s<br />

(b) 1 s<br />

(c) 1.7 s<br />

(d) 2 s<br />

(e) 4 s<br />

v v at<br />

f<br />

i<br />

vf<br />

vi<br />

t <br />

a<br />

m m <br />

0 10 sin 30<br />

s s<br />

<br />

<br />

m <br />

9.8<br />

2 <br />

s <br />

0.51s<br />

2


9/3/2009<br />

Understanding<br />

A stone is thrown horizontally with an initial speed of 30 m/s from a<br />

bridge. Find the stone’s total speed when it enters the water 4 seconds<br />

later?<br />

Hint:<br />

We need only think<br />

about the vertical time<br />

to travel to find vertical<br />

speed, then combine<br />

with horizontal speed<br />

(a) 30 m/ s<br />

(b) 40 m/s<br />

(c) 50 m/s<br />

(d) 60 m/s<br />

(e) 70 m/s<br />

v v at<br />

f<br />

i<br />

m m<br />

vyf<br />

0 9.8 2<br />

4s<br />

s s<br />

m<br />

39.2<br />

s<br />

v 2 2<br />

f<br />

vx vy<br />

m<br />

vxf<br />

30<br />

2 2<br />

s<br />

m m<br />

39.2 30<br />

2 2 <br />

s s <br />

m<br />

49.4<br />

2<br />

s<br />

Understanding<br />

Which of the following statements is true concerning the<br />

motion of an ideal projectile launched at an angle of 45 0<br />

to the horizontal?<br />

(a) <strong>The</strong> acceleration vector points opposite to the<br />

velocity vector on the way up and in the same<br />

direction as the velocity vector on the way down.<br />

(b) <strong>The</strong> speed at the top of the trajectory is zero.<br />

(c) <strong>The</strong> object’s total speed remains constant during the<br />

entire flight.<br />

(d) <strong>The</strong> horizontal speed decreases on the way down.<br />

(e) <strong>The</strong> vertical speed decreases on the way up and<br />

increases on the way down.<br />

Understanding<br />

This question concerns the motion of a car on a straight track; the car’s velocity<br />

as a function of time is plotted below<br />

20<br />

velocity (m/s)<br />

Understanding<br />

This question concerns the motion of a car on a straight track; the car’s velocity<br />

as a function of time is plotted below<br />

20<br />

velocity (m/s)<br />

10<br />

10<br />

time (s)<br />

1 2 3 4 5 6 7<br />

time (s)<br />

1 2 3 4 5 6 7<br />

−10<br />

−10<br />

−20<br />

(a) Describe what happened to the car at time t=1 s<br />

(b) How does the car’s average velocity between time t=0 sand t=1 s compare to its<br />

average velocity between times t=1 s and t=5 s?<br />

(c) What is the displacement of the car from time t=0 to time t=7s?<br />

(d) Plot the car’s acceleration during this interval as a function of time.<br />

(e) Plot the object’s position during this interval as a function of time. Assuming that<br />

the car begins at s=0.<br />

−20<br />

(a) Describe what happened to the car at time t=1 s<br />

<strong>The</strong> car’s velocity remains at 20 m/s, but the acceleration changes from<br />

positive to negative at this time (the foot leaves the accelerator)<br />

3


9/3/2009<br />

Understanding<br />

This question concerns the motion of a car on a straight track; the car’s velocity<br />

as a function of time is plotted below<br />

20<br />

velocity (m/s)<br />

Understanding<br />

This question concerns the motion of a car on a straight track; the car’s velocity<br />

as a function of time is plotted below<br />

20<br />

velocity (m/s)<br />

10<br />

10<br />

time (s)<br />

1 2 3 4 5 6 7<br />

time (s)<br />

1 2 3 4 5 6 7<br />

−10<br />

−10<br />

−20<br />

(b) How does the car’s average velocity between time t=0 sand t=1 s compare to its<br />

average velocity between times t=1 s and t=5 s?<br />

Between t=0 and t=1: the average velocity is ½(0 m/s + 20 m/s)=10 m/s<br />

Between t=1 and t=5: the average velocity is ½(20m/s + 0m/s)=10 m/s<br />

−20<br />

(c) What is the displacement of the car from time t=0 to time t=7s?<br />

Displacement is the net area between the graph as the time axis<br />

Area=1/2(5*20) – ½(2*10) =40m<br />

Understanding<br />

This question concerns the motion of a car on a straight track; the car’s velocity<br />

as a function of time is plotted below<br />

20<br />

velocity 20 velocity (m/s) (m/s)<br />

Understanding<br />

This question concerns the motion of a car on a straight track; the car’s velocity<br />

as a function of time is plotted below<br />

50<br />

velocity (m/s) 20 velocity (m/s)<br />

10<br />

10<br />

40<br />

10<br />

time (s)<br />

1 2 4 5 6 7<br />

3 time (s)<br />

1 2 3 4 5 6 7<br />

30<br />

time (s)<br />

1 2 3 4 5 6 7<br />

−10<br />

20<br />

−10<br />

−10<br />

−20<br />

10<br />

−20<br />

−20<br />

(d) Plot the car’s acceleration during this interval as a function of time.<br />

time (s)<br />

1 2 3 4 5 6 7<br />

(e) Plot the object’s position during this interval as a function of time. Assuming that<br />

the car begins at s=0.<br />

4


9/3/2009<br />

Understanding<br />

A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0<br />

toward a castle wall 220m away. <strong>The</strong> height of the wall is 30 m. Assume that<br />

effects due to the air are negligible.<br />

(a) Show that the cannonball will strike the wall.<br />

(b) How long will it take for the cannonball to strike the wall?<br />

(c) At what height above the base of the wall will the cannonball strike?<br />

Understanding<br />

A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0<br />

toward a castle wall 220m away. <strong>The</strong> height of the wall is 30 m. Assume that<br />

effects due to the air are negligible.<br />

(a) Show that the cannonball will strike the wall.<br />

1 2<br />

x v0<br />

xt<br />

y v0<br />

yt gt<br />

2<br />

x<br />

2<br />

t <br />

x 1 x <br />

v<br />

v0<br />

sin<br />

<br />

2<br />

0x<br />

g<br />

v0cos<br />

<br />

<br />

2 v0cos<br />

<br />

<br />

<br />

x<br />

2<br />

<br />

gx<br />

v cos x tan <br />

<br />

2 2<br />

0 <br />

<br />

2v0<br />

cos <br />

<br />

m <br />

2<br />

9.8 2 220m<br />

s<br />

220m tan 40<br />

<br />

<br />

2<br />

m 2<br />

250 cos 40<br />

s <br />

23m<br />

v 2<br />

sin 0 2 <br />

R <br />

g<br />

m <br />

50 sin 240<br />

s<br />

R <br />

<br />

m <br />

9.8<br />

2 <br />

s <br />

251m<br />

We need now only show that the<br />

height of the ball is below 30 m<br />

when the horizontal displacement<br />

is 220m<br />

This is less than 30 m, so contact<br />

Understanding<br />

A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0<br />

toward a castle wall 220m away. <strong>The</strong> height of the wall is 30 m. Assume that<br />

effects due to the air are negligible.<br />

b) How long will it take for the cannonball to strike the wall?<br />

x<br />

t <br />

v<br />

0x<br />

x<br />

<br />

v cos 0<br />

<br />

<br />

220m<br />

m <br />

50 cos40<br />

s <br />

5.7s<br />

<br />

Understanding<br />

A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0<br />

toward a castle wall 220m away. <strong>The</strong> height of the wall is 30 m. Assume that<br />

effects due to the air are negligible.<br />

c) At what height above the base of the wall will the cannonball strike?<br />

From (a) we have 23 m<br />

5


9/3/2009<br />

Understanding<br />

A physics student is driving home after class. <strong>The</strong> car is travelling at 14.7 m/s<br />

when it approaches an intersection. <strong>The</strong> student estimates that he is 20.0 m<br />

from the entrance to the intersection (10.0 m wide) when the traffic light changes<br />

from green to yellow. <strong>The</strong> light will change from yellow to red in 3.00 seconds.<br />

<strong>The</strong> maximum safe deceleration of the car is 4.00 m/s 2 while the maximum<br />

acceleration of the car is 2.00 m/s 2 . Should the physics student a) decelerate<br />

and stop or b) accelerate and travel through the intersection? Note: there is a<br />

police car directly behind the student.<br />

Understanding<br />

A physics student is driving home after class. <strong>The</strong> car is travelling at 14.7 m/s when it<br />

approaches an intersection. <strong>The</strong> student estimates that he is 20.0 m from the entrance to<br />

the intersection (10.0 m wide) when the traffic light changes from green to yellow. <strong>The</strong> light<br />

will change from yellow to red in 3.00 seconds. <strong>The</strong> maximum safe deceleration of the car<br />

is 4.00 m/s 2 while the maximum acceleration of the car is 2.00 m/s 2 . Should the physics<br />

student a) decelerate and stop or b) accelerate and travel through the intersection? Note:<br />

there is a police car directly behind the student.<br />

Let’s try braking first and see how far the car travels before stopping.<br />

Data Inventory<br />

m<br />

a 4.0<br />

2<br />

s<br />

x 0m<br />

i<br />

m<br />

vi<br />

14.7<br />

s<br />

m<br />

v<br />

f<br />

0<br />

s<br />

t 3.0 s<br />

x ?<br />

f<br />

1 2<br />

We shall use: x<br />

f<br />

xi vit at<br />

2<br />

m 1 m<br />

x<br />

f<br />

0m 14.7 3.0s 4 3.0s<br />

2 <br />

s 2 s <br />

0m 44.1m 18.0m<br />

26.1m<br />

2<br />

<strong>The</strong>refore the student will travel 6.1 m into the intersection<br />

(not good with police car behind him)<br />

Understanding<br />

A physics student is driving home after class. <strong>The</strong> car is travelling at 14.7 m/s when it<br />

approaches an intersection. <strong>The</strong> student estimates that he is 20.0 m from the entrance to<br />

the intersection (10.0 m wide) when the traffic light changes from green to yellow. <strong>The</strong> light<br />

will change from yellow to red in 3.00 seconds. <strong>The</strong> maximum safe deceleration of the car<br />

is 4.00 m/s 2 while the maximum acceleration of the car is 2.00 m/s 2 . Should the physics<br />

student a) decelerate and stop or b) accelerate and travel through the intersection? Note:<br />

there is a police car directly behind the student.<br />

Now let’s try accelerating and see how far the car travels before the light turns red.<br />

Data Inventory<br />

m<br />

a 2.0<br />

2<br />

s<br />

x 0m<br />

i<br />

m<br />

vi<br />

14.7<br />

s<br />

v ?<br />

f<br />

t 3.0 s<br />

x ?<br />

f<br />

1 2<br />

We shall use: x<br />

f<br />

xi vit at<br />

2<br />

m 1 m<br />

x<br />

f<br />

0m 14.7 3.0s<br />

2.0 3.0s<br />

2 <br />

s 2 s <br />

0m 44.1m 9.0m<br />

53.1m<br />

2<br />

<strong>The</strong>refore the student will travel about 23 m past the<br />

intersection before the light turns red.<br />

Understanding<br />

A car traveling at a constant speed of 30 m/s passes a highway patrol police car<br />

which is at rest. <strong>The</strong> police officer accelerates at a constant rate of 3.0 m/s 2 and<br />

maintains this rate of acceleration until he pulls next to the speeding car.<br />

Assume that the police car starts to move at the moment the speeder passes<br />

the car. Determine:<br />

a) <strong>The</strong> time required for the police officer to catch the speeder?<br />

b) <strong>The</strong> distance travelled during the chase?<br />

6


9/3/2009<br />

Understanding<br />

A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is<br />

at rest. <strong>The</strong> police officer accelerates at a constant rate of 3.0 m/s 2 and maintains this rate<br />

of acceleration until he pulls next to the speeding car. Assume that the police car starts to<br />

move at the moment the speeder passes the car. Determine:<br />

a) <strong>The</strong> time required for the police officer to catch the speeder?<br />

Plan of attack: If we can find a position function for both the motorist and the<br />

police car in terms of time, then we can set both functions equal to each other<br />

(same position) and solve for time<br />

Data Inventory motorist<br />

m<br />

a 0<br />

2<br />

s<br />

x 0m<br />

i<br />

m<br />

vi<br />

30<br />

s<br />

m<br />

v<br />

f<br />

30<br />

s<br />

t ?<br />

x ?<br />

f<br />

x x vt<br />

f<br />

i<br />

m <br />

x<br />

f<br />

0m 30 t<br />

s <br />

m <br />

xf<br />

30 t<br />

s <br />

Data Inventory Police<br />

m<br />

a 3.0<br />

2<br />

s<br />

x 0m<br />

i<br />

m<br />

vi<br />

0<br />

s<br />

v ?<br />

f<br />

t ?<br />

x ?<br />

f<br />

1<br />

x<br />

f<br />

xi vit at<br />

2<br />

m 1 m<br />

x<br />

f<br />

0m 0 t 3.0<br />

t<br />

2 <br />

s 2 s <br />

m 2<br />

xf<br />

1.5 t<br />

2 <br />

s <br />

2<br />

2<br />

Let’s set then equal and solve<br />

m m<br />

2<br />

30 t<br />

1.5<br />

t<br />

2 <br />

s s <br />

m m<br />

0 30 t<br />

1.5 t<br />

s s <br />

2<br />

2 <br />

m m<br />

<br />

0 t<br />

30 1.5 t<br />

2 <br />

s s <br />

<br />

Understanding<br />

<strong>The</strong>refore the police car catches up with<br />

the speeder at 20s (the 0s is when the<br />

car initially passes the police car)<br />

<strong>The</strong>refore<br />

t 0<br />

or<br />

m m<br />

30 1.5 0<br />

2 t <br />

s s <br />

m m<br />

1.5 t 30<br />

2 <br />

s s<br />

m<br />

30<br />

t s<br />

m<br />

1.5<br />

2<br />

s<br />

20s<br />

Understanding<br />

Understanding<br />

A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is<br />

at rest. <strong>The</strong> police officer accelerates at a constant rate of 3.0 m/s 2 and maintains this rate<br />

of acceleration until he pulls next to the speeding car. Assume that the police car starts to<br />

move at the moment the speeder passes the car. Determine:<br />

b) <strong>The</strong> distance travelled during the chase?<br />

A stone is thrown vertically upward from the edge of a building 19.6 m high with<br />

an initial velocity of 14.7 m/s. <strong>The</strong> stone just misses the building on the way<br />

down and strikes the street below. Determine:<br />

a) <strong>The</strong> time of flight?<br />

b) <strong>The</strong> velocity of the stone just before it strikes the ground ?<br />

Since we know the time (20s) from part a), we need only plug it into the<br />

distance formula from either car. [let’s use the speeder ]<br />

x<br />

f<br />

vt<br />

m <br />

30 20s<br />

s <br />

600m<br />

<br />

7


9/3/2009<br />

Understanding<br />

Understanding<br />

A stone is thrown vertically upward from the edge of a building 19.6 m high with<br />

an initial velocity of 14.7 m/s. <strong>The</strong> stone just misses the building on the way<br />

down and strikes the street below. Determine:<br />

a) <strong>The</strong> time of flight?<br />

Given:<br />

Required:<br />

m<br />

vi<br />

14.7<br />

s<br />

m<br />

a 9.80<br />

2<br />

s<br />

x 19.6m<br />

i<br />

x 0m<br />

f<br />

t<br />

Relationship:<br />

Work:<br />

1<br />

x<br />

f<br />

xi vit at<br />

2<br />

2<br />

1 2<br />

x<br />

f<br />

xi vit at<br />

2<br />

1<br />

0 <br />

2<br />

2<br />

at vit xi x<br />

f<br />

2 1<br />

<br />

vi vi 4<br />

a 2<br />

xi x<br />

f<br />

<br />

t <br />

1<br />

<br />

2<br />

a <br />

2 <br />

2<br />

v vi 2a xi x<br />

f<br />

<br />

a<br />

<br />

2<br />

m m m <br />

14.7 14.7 2 9.8 2 19.6m<br />

0m<br />

s s s <br />

<br />

m<br />

9.8<br />

2<br />

s<br />

<br />

<br />

<strong>The</strong>refore<br />

t 1s<br />

t 4s<br />

Flight time is 4 s<br />

A stone is thrown vertically upward from the edge of a building 19.6 m high with<br />

an initial velocity of 14.7 m/s. <strong>The</strong> stone just misses the building on the way<br />

down and strikes the street below. Determine:<br />

b) <strong>The</strong> velocity of the stone just before it strikes the ground ?<br />

m<br />

v 14.7<br />

Given: i<br />

<br />

s<br />

Work:<br />

m<br />

a 9.80<br />

2<br />

s<br />

v vi<br />

at<br />

f<br />

xi<br />

19.6m<br />

m m<br />

14.7 9.<br />

80 2 4.<br />

0s<br />

xf<br />

0m<br />

s s <br />

t 4s<br />

Required: v<br />

f<br />

Relationship:<br />

m<br />

245<br />

.<br />

s<br />

<strong>The</strong> negative value in the velocity indicates that<br />

the stone is travelling downward.<br />

v<br />

f<br />

vi<br />

at<br />

`<br />

8

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