9/3/2009
Understanding
Kinematics/Newton’s Laws
An object that’s moving with constant speed travels once
around a circular path. Which of the following is/are true
concerning this motion?
I The displacement is zero
II The average speed is zero
III The acceleration is zero
Practice Questions
(a) I only
(b) I and II only
(c) I and III only
(d) III only
(e) II and III only
Understanding
Understanding
Which of the following is/are true?
If an object’s acceleration is constant, then it must
move in a straight line.
II. If an object’s acceleration is zero, then its speed must
remain constant.
III. If an object’s speed remains constant, then its
acceleration must be zero.
I.
A baseball is thrown straight upward. What is the ball’s
acceleration at its highest point?
Gravity is always turned on.
(a) I and II only
(b) I and III only
(c) II only
(d) III only
(e) II and III only
(a) 0
(b) ½ g [down]
(c) g [down]
(d) ½ g [up]
(e) g [up]
1
9/3/2009
Understanding
Understanding
A rock is dropped off a cliff and strikes the ground with an
impact of 30 m/s. How high was the cliff?
How long would it take a car, starting from rest and
accelerating uniformly in a straight line at 5 m/s 2, to cover
a distance of 200m?
Hint:
Hint:
1
s  v0t  at 2
2
s 
(a) 15.0 m
(b) 20 m
(c) 30 m
(d) 46 m
(e) 60 m
Understanding
A stone is thrown horizontally with an initial speed of 10 m/s from a
bridge. If air resistance could be ignored, how long would it take the
stone to strike the water 80 m below the bridge?
Hint:
We need only think
about the vertical
components.
(a) 1 s
(b) 2 s
(c) 4 s
(d) 6 s
(e) 8 s
v 2f  vi2  2a  s 
v 2f  vi2  2a  s 
 m 1 m 
200m   0  t   5 2  t 2
 s  2 s 
400m
t2 
m
5 2
s
t  8.9s
(a) 9.0 s
(b) 10.5 s
(c) 12.0 s
(d) 15.5 s
(e) 20.0 s
We need both the
initial and final velocity
in the equation
1
s  v0t  at 2
2
m
 m 1
80m   0  t   9.8 2  t 2
s 
 s  2
160m
t2 
m
9.8 2
s
t  4.0s
v 2f  vi2
2a
2
 m  m
 30    0 
s  s

m

2  9.8 2 
s 

 45.9m
2
Understanding
A soccer ball, at rest on the ground, is kicked with an initial velocity of 10
m/s at a launch angle of 30 0. Calculate its total flight time, assuming air
resistance is negligible?
Hint:
We need only think
about the vertical time
to travel up to its max
height, then double it.
v f  vi  at
t
(a) 0.5 s
(b) 1 s
(c) 1.7 s
(d) 2 s
(e) 4 s
v f  vi
a
 m  m

0

  10 sin  30  
s  s


m

 9.8 2 
s 

 0.51s
2
9/3/2009
Understanding
Understanding
A stone is thrown horizontally with an initial speed of 30 m/s from a
bridge. Find the stone’s total speed when it enters the water 4 seconds
later?
Hint:
We need only think
about the vertical time
to travel to find vertical
speed, then combine
with horizontal speed
v f  vi  at
m
m
v yf  0  9.8 2  4 s 
s
s
m
2
2
 39.2
v f   vx    v y 
s
m
2
2
vxf  30
m  m

s
  39.2 2    30 2 
s   s 

m
 49.4 2
s
(a) 30 m/ s
(b) 40 m/s
(c) 50 m/s
(d) 60 m/s
(e) 70 m/s
Which of the following statements is true concerning the
motion of an ideal projectile launched at an angle of 45 0
to the horizontal?
(a) The acceleration vector points opposite to the
velocity vector on the way up and in the same
direction as the velocity vector on the way down.
(b) The speed at the top of the trajectory is zero.
(c) The object’s total speed remains constant during the
entire flight.
(d) The horizontal speed decreases on the way down.
(e) The vertical speed decreases on the way up and
increases on the way down.
Understanding
Understanding
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
20 velocity (m/s)
20 velocity (m/s)
10
10
time (s)
1
2
3
4
5
6
time (s)
7
−10
−20
(a) Describe what happened to the car at time t=1 s
(b) How does the car’s average velocity between time t=0 sand t=1 s compare to its
average velocity between times t=1 s and t=5 s?
(c) What is the displacement of the car from time t=0 to time t=7s?
(d) Plot the car’s acceleration during this interval as a function of time.
(e) Plot the object’s position during this interval as a function of time. Assuming that
the car begins at s=0.
1
2
3
4
5
6
7
−10
−20
(a) Describe what happened to the car at time t=1 s
The car’s velocity remains at 20 m/s, but the acceleration changes from
positive to negative at this time (the foot leaves the accelerator)
3
9/3/2009
Understanding
Understanding
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
20 velocity (m/s)
20 velocity (m/s)
10
10
time (s)
1
2
3
4
5
6
time (s)
7
1
−10
2
3
4
5
6
7
−10
−20
−20
(b) How does the car’s average velocity between time t=0 sand t=1 s compare to its
average velocity between times t=1 s and t=5 s?
(c) What is the displacement of the car from time t=0 to time t=7s?
Displacement is the net area between the graph as the time axis
Between t=0 and t=1: the average velocity is ½(0 m/s + 20 m/s)=10 m/s
Between t=1 and t=5: the average velocity is ½(20m/s + 0m/s)=10 m/s
Area=1/2(5*20) – ½(2*10) =40m
Understanding
Understanding
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
50 velocity (m/s)
20 velocity (m/s)
20 velocity
(m/s) (m/s)
20 velocity
10
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
40
10
time (s)
1
1
2
2
3
3
4
5
4
6
5
7 time (s)
6
−10
10
time (s)
30
1
2
3
4
5
6
7
7
20
−10
−10
−20
10
−20
−20
time (s)
1
(d) Plot the car’s acceleration during this interval as a function of time.
2
3
4
5
6
7
(e) Plot the object’s position during this interval as a function of time. Assuming that
the car begins at s=0.
4
9/3/2009
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0
toward a castle wall 220m away. The height of the wall is 30 m. Assume that
effects due to the air are negligible.
(a) Show that the cannonball will strike the wall.
(b) How long will it take for the cannonball to strike the wall?
(c) At what height above the base of the wall will the cannonball strike?
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0
toward a castle wall 220m away. The height of the wall is 30 m. Assume that
effects due to the air are negligible.
(a) Show that the cannonball will strike the wall.
R
x  v0 xt
v02 sin  2 
g
2
 m
 50  sin  2  40  
s
R
m

 9.8 2 
s 

 251m
x
t
v0 x

x
v0 cos  
We need now only show that the
height of the ball is below 30 m
when the horizontal displacement
is 220m
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0
toward a castle wall 220m away. The height of the wall is 30 m. Assume that
effects due to the air are negligible.
b) How long will it take for the cannonball to strike the wall?
1
y  v0 yt  gt 2
2

 1 

x
x
  v0 sin    
  g 

 v0 cos    2  v0 cos   
gx 2
 x tan   
2v0 2 cos 2  
m
2

 9.8 2   220m 
s 
  220m  tan  40   
2
 m
2  50  cos 2  40 
 s
 23m
2
This is less than 30 m, so contact
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40 0
toward a castle wall 220m away. The height of the wall is 30 m. Assume that
effects due to the air are negligible.
c) At what height above the base of the wall will the cannonball strike?
x
t
v0 x
From (a) we have 23 m
x

v0 cos  
220m
 m
50

 cos  40 
 s
 5.7 s

5
9/3/2009
Understanding
Understanding
A physics student is driving home after class. The car is travelling at 14.7 m/s
when it approaches an intersection. The student estimates that he is 20.0 m
from the entrance to the intersection (10.0 m wide) when the traffic light changes
from green to yellow. The light will change from yellow to red in 3.00 seconds.
The maximum safe deceleration of the car is 4.00 m/s2 while the maximum
acceleration of the car is 2.00 m/s2. Should the physics student a) decelerate
and stop or b) accelerate and travel through the intersection? Note: there is a
police car directly behind the student.
A physics student is driving home after class. The car is travelling at 14.7 m/s when it
approaches an intersection. The student estimates that he is 20.0 m from the entrance to
the intersection (10.0 m wide) when the traffic light changes from green to yellow. The light
will change from yellow to red in 3.00 seconds. The maximum safe deceleration of the car
is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics
student a) decelerate and stop or b) accelerate and travel through the intersection? Note:
there is a police car directly behind the student.
Let’s try braking first and see how far the car travels before stopping.
Data Inventory
m
a  4.0 2
s
xi  0 m
vi  14.7
m
s
t  3.0 s
vf  0
xf  ?
Understanding
A physics student is driving home after class. The car is travelling at 14.7 m/s when it
approaches an intersection. The student estimates that he is 20.0 m from the entrance to
the intersection (10.0 m wide) when the traffic light changes from green to yellow. The light
will change from yellow to red in 3.00 seconds. The maximum safe deceleration of the car
is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics
student a) decelerate and stop or b) accelerate and travel through the intersection? Note:
there is a police car directly behind the student.
Now let’s try accelerating and see how far the car travels before the light turns red.
Data Inventory
a  2.0
m
s2
xi  0 m
vi  14.7
vf  ?
t  3.0 s
xf  ?
m
s
m
s
1
x f  xi  vit  at 2
2
m
1 m 
2

x f  0m  14.7   3.0s    4 2   3.0s 
s
2 s 

 0m  44.1m 18.0m
We shall use:
 26.1m
Therefore the student will travel 6.1 m into the intersection
(not good with police car behind him)
Understanding
A car traveling at a constant speed of 30 m/s passes a highway patrol police car
which is at rest. The police officer accelerates at a constant rate of 3.0 m/s 2 and
maintains this rate of acceleration until he pulls next to the speeding car.
Assume that the police car starts to move at the moment the speeder passes
the car. Determine:
a) The time required for the police officer to catch the speeder?
b) The distance travelled during the chase?
1 2
We shall use: x f  xi  vit  at
2
m
1
m
2

x f  0m  14.7   3.0s    2.0 2   3.0s 
s
2
s 

 0m  44.1m  9.0m
 53.1m
Therefore the student will travel about 23 m past the
intersection before the light turns red.
6
9/3/2009
Understanding
A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is
at rest. The police officer accelerates at a constant rate of 3.0 m/s 2 and maintains this rate
of acceleration until he pulls next to the speeding car. Assume that the police car starts to
move at the moment the speeder passes the car. Determine:
a) The time required for the police officer to catch the speeder?
Plan of attack: If we can find a position function for both the motorist and the
police car in terms of time, then we can set both functions equal to each other
(same position) and solve for time
Data Inventory motorist
m
s2
xi  0 m
a0
m
s
m
v f  30
s
t ?
vi  30
Understanding
Let’s set then equal and solve
m 2
 m 
 30  t   1.5 2  t
 s  s 
m
 m 
0   30  t  1.5 2  t 2
 s  s 
 m 
m 
0  t  30  1.5 2  t 
 s  s  
Therefore
t 0
or
m 
m
30  1.5 2  t  0
s  s 
Data Inventory Police
x f  xi  vt
m
s2
xi  0 m
a  3.0
 m
x f  0m   30  t
 s
 m
x f   30  t
 s
vi  0
m
s
vf  ?
t ?
xf  ?
xf  ?
1
x f  xi  vit  at 2
2
m
 m 1
x f  0m   0  t   3.0 2  t 2
s 
 s  2
m

x f  1.5 2  t 2
 s 
Understanding
Therefore the police car catches up with
the speeder at 20s (the 0s is when the
car initially passes the police car)
m
m

1.5 2  t  30
s
 s 
m
30
s
t
m
1.5 2
s
 20 s
Understanding
A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is
at rest. The police officer accelerates at a constant rate of 3.0 m/s 2 and maintains this rate
of acceleration until he pulls next to the speeding car. Assume that the police car starts to
move at the moment the speeder passes the car. Determine:
A stone is thrown vertically upward from the edge of a building 19.6 m high with
an initial velocity of 14.7 m/s. The stone just misses the building on the way
down and strikes the street below. Determine:
b) The distance travelled during the chase?
a) The time of flight?
b) The velocity of the stone just before it strikes the ground ?
Since we know the time (20s) from part a), we need only plug it into the
distance formula from either car. [let’s use the speeder ]
x f  vt
 m
  30   20s 
 s
 600m
7
9/3/2009
Understanding
Understanding
A stone is thrown vertically upward from the edge of a building 19.6 m high with
an initial velocity of 14.7 m/s. The stone just misses the building on the way
down and strikes the street below. Determine:
A stone is thrown vertically upward from the edge of a building 19.6 m high with
an initial velocity of 14.7 m/s. The stone just misses the building on the way
down and strikes the street below. Determine:
a) The time of flight?
b) The velocity of the stone just before it strikes the ground ?
m
Given: vi  14.7
s
m
a  9.80 2
s
xi  19.6m
1
x  xi  vit  at 2
Work: f
2
1 2
0  at  vit  xi  x f
2
t
x f  0m
Required:
t
Relationship: x f  xi  vit 

1 2
at
2
Given:
Therefore
1 
vi  vi2  4  a   xi  x f 
2 
1 
2 a 
2 
v  vi2  2a  xi  x f 
t  1s
t  4s
x f  0m
t  4s
Flight time is 4 s
a
2
m
s
m
a  9.80 2
s
xi  19.6m
vi  14.7
m
m
m



 14.7   14.7   2  9.8 2  19.6m  0m 
s
s
s 




m
9.8 2
s
Required:
vf
Relationship:
v f  vi  at
Work:
v f  vi  at
m 
m

 14.7    9.80 2   4.0s 
s 
s 

m
 24.5
s
The negative value in the velocity indicates that
the stone is travelling downward.
`
8