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INNOVA JUNIOR COLLEGE
JC1 Physics (2009)
Tutorial 2: Kinematics Worksheet
Section 1 (Motion Graph)
The motion of an object along a straight line can be described by three types of graphs:
Graph
Slope
Instantaneous velocity
(1) Displacement – time
(s-t)
(2) Velocity – time
(v-t)
Area under the graph
ds
dt
v
Instantaneous acceleration
(3) Acceleration – time
(a-t)
dv
a
dt
–
Change in Displacement
v
ds
 s   v dt
dt
Change in velocity
–
a
dv
 v   a dt
dt
Of the three graphs, the velocity-time graph is the most versatile because each main feature
holds special significance:
 the area beneath the graph represents the change in displacement
 the tangent at any point on the graph gives the instantaneous acceleration.
1. A ball, thrown vertically upwards, rises to a height h and then falls to its starting point. Air
resistance may be taken as negligible. Which graph best shows the variation of kinetic
energy Ek of the ball with the distance s traveled?
Negligible air resistance:
 only force is weight (undergoing free fall) i.e. constant g at the surface of the Earth.
Thus for constant acceleration,
v = u + at
--- (1)
s = ut + ½ at2 --- (2)
v2 = u2 + 2as --- (3)
Eq (3) can be used to plot Ek vs s since Ek = ½ mv2
Multiplying ½ m to both side of Eq (3):
½ mv2 = ½ mu2 + m(-g)s
Ek = ks + c where k = - mg and c = ½ mu2
The graph is a downward sloping straight line with the y-intercept of ½ mu2.
However, Ek cannot be negative. Therefore, Ek = |ks + c|  Ans C.
1
Section 1 (Motion Graph)
2. vx is the velocity of a particle moving along the x-axis as shown in the graph below.
If the position of the particle along x-axis at t = 1.0 s is at x1 = 2.0 m, what is the position
of the particle along the x-axis at t = 6.0 s?
Given x1 = 2.0 m when t = 1.0 s, find x6 when t = 6.0 s.
The other information given is the velocity-time graph:
Thinking question: How is the velocity-time graph linked to position?
Answer:
The area under the velocity- time graph is the change in
displacement of the particle (not displacement)
Since x1 is given, the area under the velocity-time graph that is of concerned is
from t = 1.0 s to 6.0 s.
Change in displacement = Area under the velocity-time graph
Δx = (+ Area of ) + ( - Area of
)
Area of
is positive: the velocity is positive, thus leading to positive change in
displacement. Contrary, the area of
is negative: the velocity is negative, thus
leading to negative change in displacement.
Δx = (+ Area of ) + ( - Area of
 ½  2.0 1.0  -½  2.0  4.0 
= - 3.0 m
)
However the answer is looking for x6 and not Δx.
Therefore, Δx = x6 – x1
- 3.0 = x6 – (+ 2.0)
x6 = - 1.0 m
2
Section 1 (Motion Graph)
3. A rigid ball is released from rest at a distance above a hard surface. The velocity- time
graph of the ball over 3 bounces is as shown in the diagram below.
Which graph best represents the acceleration-time graph of the ball?
(Hint: what is the link between acceleration and velocity-time graph?)
4. A stone is thrown vertically upwards. A student plots the variation with time of its velocity.
What is the vertical displacement of the stone from its starting point after 5.0 s?
(Hint: what is the link between displacement and velocity-time graph?)
3
[+ 25 m]
Section 1 (Motion Graph)
5
The graph shows the displacement-time graph for 2 trains A and B running on parallel
tracks.
A
B
C
D
At time tB, both trains have the same velocity.
Both trains speed up all the time.
Both trains have the same velocity at some time before tB.
Somewhere on the graph, both trains have the same acceleration.
Which statement is correct?
For the incorrect statement, please amend it so that it will reflect the information given by
the graphs.
(Hint: how can displacement, velocity and acceleration be determined using a
displacement-time graph?)
6. A toy rocket is launched vertically from Earth with a constant acceleration. After some
time, the fuel is used up and the toy rocket falls freely back to Earth. Which one of the
velocity-time graphs best represents the journey? Neglect air resistance.
Explain the choice of the velocity-time graph, with reference to the rocket’s
velocity and acceleration before and after the fuel run out.
4
Section 2 (Linear Motion)
Section 2 (Linear Motion)
Consider an object undergoing a constant acceleration motion, its kinematics variable
can be obtained through:
v  u  at
--- (1)
--- (2)
s  ½ u  v  t
--- (3)
s  ut  ½ at 2
2
2
--- (4)
v = u + 2as
These derived equations of motion apply only to uniformly accelerated motion in a
straight line.
Very Important!!!
s, u, v and a are vector quantities! It is important to be consistent in the usage of the
signs (+/-). If the positive direction is taken to be that to the right, then
i. points on the left of the point of reference have negative values for s
ii. velocities directed to the left have negative values
iii. accelerations directed to the left have negative values
7
What are the necessary assumptions for the usage of the equations of motion?
I.
II.
III.
IV.
The particle must move in a straight line.
The particle must move with constant velocity.
The particle must move with increasing velocity.
The particle must move with constant acceleration.
8. Assuming a constant deceleration of 300 m s-2, how far will the front of a car collapse if
the car is travelling at a speed of 28 m s-2 just before impact and subsequently comes to
a complete stop?
Given: u = 28 m s-1
Deceleration = 300 m s-2 (slow down)  a = - 300 m s-2
Complete stop  v = 0 m s-1
s
Car

u
+ 28
v
0
a
- 300
t
?
Since it is a constant deceleration:
Using v2 = u2 + 2as
02 = 282 + 2(- 300)s
s = 1.31 m
5
Section 2 (Linear Motion)
9
A ball is thrown vertically upwards with a speed of 25 m s-1. How high can it rise?
[31.9 m]
The question is asking for HOW HIGH CAN it rises?
So the answer to be determined is s, v, a or t?
And what is so special with this point in the vertical motion?
What other kinematics information can be inferred in the question?
Ball
s
u
v
a
t
10 A helicopter is ascending vertically with a speed of 5.5 m s-1. At a height of 110 m above
the earth a package is dropped from a window. How much time does it take for the
package to reach the ground?
[5.33 s]
-1
Is the initial velocity of the package, u = 0 or 5.5 m s ?
Explanation:
……………………………………………………………………...………………
......…………………………………………………………………………………
Is the final velocity of the package just before it hit the ground, v = 0 m s-1? Or it has to
be determined?
Explanation:
……………………………………………………………………...………………
......…………………………………………………………………………………
Fill in the table below:
(Remember to put + / - to indicate the directions of these vectors.)
Package
s
u
v
a
t
6
Section 2 (Linear Motion)
11. A falling stone takes 0.30 s to travel past a window 2.2 m tall. From what height above
the top of the window did the stone fall from rest?
[1.75 m]
Hint: A sketch of the stone falling.
The time to travel past a window is from (2) to (3).
The question is asking about the displacement from (1) to (2).
(2)
0.30 s
(3)
Windows
(1)
Try to look at the motion from (1) to (2) to (3):
Thus fill in the blanks for point (2)  (3) and see how it help to solve for (1)  (2)
(Remember to put + / - to indicate the directions of these vectors.)
(2)  (3)
s
u
v
a
t
7
(1)  (2)
Section 3 (Projectile Motion)
Section 3 (Projectile Motion)
12 (a) Draw sample vectors to represent
the force(s) acting on the ball in the
positions shown below after it leaves
the throwers’ hand.
(b) Draw sample bold vectors to represent
the velocity of the ball in the positions
shown below after it leaves the throwers’
hand. With lighter vectors show the
horizontal and vertical components of
velocity for each position.
It is important to distinguish between force and velocity vectors. Force vectors combine with other force
vectors, and velocity vectors combine with other velocity vectors. But velocity vectors do not combine with
force vectors. So to avoid confusion, do not draw force and velocity vectors on the same diagram.
(c) Which velocity component in (b) remains constant? Why?
…………………………………………………………….……………………………………
……………………….…………………………………………………………………………
(d) Which velocity component changes along the path? Why?
…………………………………………………………….……………………………………
……………………….…………………………………………………………………………
(e) The ball thrown upwards has initial vertical and horizontal velocity of 30 m s -1 and
5 m s-1 respectively. The location of the ball is shown at 1.0 s intervals. Taking
g = 10 m s-2, fill in the values of the boxes for ascending velocity components and
resultant descending velocity.
30 m s-1
5 m s-1
For ease of calculation:
g = 10 m s-2
8
Section 3 (Projectile Motion)
For object undergoing free fall, ax = 0 and ay = - 9.81 m s-2
13 A cannon at the top of a 30 m high hill fires a shell at an angle of 30° upwards from the
horizontal with a speed of 50 m s-1. Taking air resistance to be negligible, what is the
angle to the horizontal at which the shell lands on the ground?
50 m s-1
(1)
30 °
30 m
(2)
Given: Negligible air resistance, (i.e. undergoing free fall) i.e. a = - 9.81 m s-2.
To find the angle of the landing, the direction of the velocity needs to be
determined. Thus the final horizontal and vertical velocities need to be found.
Based on the table,
vx = ux
(since ax = 0)
= 50 cos 30°
= 43.3 m s-1
12
s
x
?
y
- 30
u
50 cos 30°
50 sin 30°
vy2 = uy2 + 2aysy
vy2 = (50 sin 30°)2 + 2(- 9.81)(- 30)
vy = 34.8 m s-1
tan  
v
?
?
a
0
- 9.81
t
?
?
vy
vx
34.8
43.3
θ = 38.8 °
tan 
Food for thoughts:
- why the final velocity of the shell just before it hit the ground is not 0?
14. A projectile is fired from top of a cliff with a velocity of 150 m s -1 at 30° to the horizontal
as shown below. If the time of flight is 23 s, what is the height of the cliff?
[870 m]
(1)
(2)
sy = uyt + ½ ayt2
sy = (+ 150 sin 30°)(23) + ½ (- 9.81)(23)2
= 870 m
9
12
s
x
?
y

u
150 cos 30°
+ 150 sin 30°
v
?
?
a
0
- 9.81
t
23
23
Section 3 (Projectile Motion)
15. A plane is flying horizontally at a speed of 150 m s-1 over the sea. Suddenly a package is
released from the plane which is 200 m above the sea level. Neglecting air resistance
and assuming that the plane does not change its course, speed or altitude, what is the
horizontal displacement that the package travels?
[6120 m]
(1)
150 m s-1
200 m
(2)
The horizontal displacement cannot be directly obtained
from the x-component. There is a missing information t
that can also be obtained from the y-component.
12
s
x
y
u
v
a
t
16 A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the
ground, landing 10 m away as shown in the diagram.
(1)
(2)
What was the speed at take-off?
Given that sx = 10 m, sy = - 1.25 m (why negative)
Hint : Another info (the rider is moving horizontally)
[19.8 m s-1]
12
s
u
v
a
t
10
x
y