SP015 Tutorial Notes Ch 02

CHAPTER 2:
Kinematics of linear motion

(5 houCrHs)APTER 2:

Kinematics of linear motion
(5 hours)

1

2.1 Linear Motion (C3 & C4)

c) Determine the distance travelled, displacement,
velocity and acceleration from appropriate graphs.

2

Graphs for linear motion ( s – t )

Gradient of s-t graph = v  s2 s1  s
t2  t1 t

No Type of Graph Description
Motion

s(m)

1 Uniform Velocity is constant.
Motion
t (s)

0

s (m) The gradient increases, velocity

Non

2 Uniform increases  Object is under

Motion 0 acceleration.

t(s)

s(m) B C -From point A to C the gradient

Non decreases, velocity decreases 

3 Uniform Object is under deceleration.

Motion A - Gradient at B to C is zero,

0 t(s) velocity = 0

3

Graphs for linear motion ( v – t )

Gradient of v-t graph = a  v2v1  v Area under graph=Displacement
t2  t1 t

No Type of Graph Description
Motion

v (ms-1)

1 Uniform Gradient = Acceleration=0
Motion
0 t (s)

-1
v (ms )
Non
Velocity increases uniformly 
2 Uniform Constant acceleration

Motion 0 t (s)

-1
v (ms )
Non
3 Uniform Velocity decreases uniformly 

Motion 0 Constant deceleration

t (s) 4

Graphs for linear motion ( a – t )

Area under graph = velocity

No Type of Graph Description
Motion
a (ms-2)

1 Uniform Acceleration is 0
Motion
0 t (s)
a (ms-2)
Non
2 Uniform Object is under acceleration (positive
value).
Motion 0
t (s)

-2
a (ms )

Non Object is under deceleration
3 Uniform 0 t (s) (negative value).

Motion

5

Type of Displacement Velocity acceleration
Motion
v a
Uniform x
Motion 0 0t 0t
v a
t
t t
Non x 0 0
Uniform 0 v
Motion 0t
0t
Non x t 6
Uniform 0 t
Motion

Example 1

A toy train moves slowly along a straight track according to the

displacement, s against time, t graph in figure below.

s (cm)

10
8
6
4
2

0 2 4 6 8 10 12 14 t (s)

a. Explain qualitatively the motion of the toy train.
b. Sketch a velocity (cm s-1) against time (s) graph.
c. Determine the average velocity for the whole journey.
d. Calculate the instantaneous velocity at t = 12 s.

Solution

a. 0 to 6 s : The train moves at a constant velocity of 0.67 cm s-1 .
6 to 10 s : The train stops.

10 to 14 s : The train moves in the same direction at a constant
velocity of 1.50 cm s-1 .

b.

v (cm s1)

1.50

0.68

0 2 4 6 8 10 12 14 t (s)

c. vav  s2  s1
t2  t1

vav  10  0
14  0

vav  0.714 cm s1

d. v  instantane ous velocity at 12 s
v  s12
t12
v 8
12

v  0.67 cm s1 9

Example 2

A velocity-time (v-t) graph in figure below shows the
motion of a lift.

v (m s 1)

4
2

0 5 10 15 20 25 30 35 40 45 50 t (s)

-2
-4

a. Describe qualitatively the motion of the lift.

b. Sketch a graph of acceleration (m s-1) against time (s).
c. Determine the total distance travelled by the lift and its

displacement.
d. Calculate the average acceleration between 20 s to 40 s.

Solution

a. 0 to 5 s : Lift moves upward from rest with a constant
acceleration of 0.4 m s2.

5 to 15 s : The velocity of the lift increases from 2 m s-1
to 4 m s1 but the acceleration decreasing to
0.2 m s-2 .

15 to 20 s : Lift moving with constant velocity of 4 m s-1 .
20 to 25 s : Lift decelerates at a constant rate of 0.8 m s-2 .
25 to 30 s : Lift at rest or stationary.
30 to 35 s : Lift moves downward with a constant

acceleration of 0.8 m s-2.
35 to 40 s : Lift moving downward with constant velocity

of 4 m s-1 .
40 to 50 s : Lift decelerates at a constant rate of 0.4 m s-2

and comes to rest.

Solution

b. a (m s2)

0.8
0.6
0.4
0.2

0 5 10 15 20 25 30 35 40 45 50 t (s)

-0.2
-0.4
-0.6
-0.8

Solution

c. i. v (m s 1)

4

2 A1 A2 A3
5
0 10 15 20 25 30 A435 40 45 50 t (s)
-2 A5

-4

Total distance  area under the graph of v-t

 A1  A2  A3  A4  A5

Total distance  1 25 1 2  410 1 5 104 1 54 1 15  54

22 2 22

Total distance 115 m

Solution

c. ii. Displaceme nt  area under the graph of v-t

 A1  A2  A3  A4  A5

Displacement  1 25 1 2  410 1 5 104
22 2

1 5 4 1 15  5 4

22

d. Displaceme nt  15 m

aav  v2  v1
aav  t2 4t14

40  20

aav  0.4 m s2

2.2 Uniformly Accelerated Motion

At the end of this chapter, students should be able
to:

 Apply equations of motion with uniform
acceleration:

v  u  at

s  ut  1 at2
2

v2  u2  2as

15

Example 3

A plane on a runway takes 16.2 s over a distance of 1200 m to take off from

rest. Assuming constant acceleration during take off, calculate :

a. the speed on leaving the ground,

b. the acceleration during take off.

Solution   ? v?
a

u0

a. Use s  1200 m
t  16.2 s
v  148 m s1
s  1 u  vt

2

1200  1 0  v16.2

2

16

Solution

b. By using the equation of linear motion,

v2  u2  2as

1482  0  2a1200

a  9.13 m s2

OR

s  ut  1 at2
2

1200  0  1 a16.22

2
a  9.14 m s2

Example 4

The driver of a pickup truck going 100 km h–1 applies the

brakes, giving the truck a uniform deceleration of 6.50 m s–2

while it travels 20.0m.

(a) What is the speed of the truck in kilometers per hour at

the end of this distance ?

(b) How much time has elapsed ?

Solution
Given :
a = – 6.5 m s–2 ( deceleration )

s = 20 m

u = 100 km hour –1  1010k0mm  
 100 km 1h
Since, u( in m s-1)  1h 3600 s

 27.78 m s-1

Solution

(a) Final velocity , v = ?

v2  u 2  2as

v2  (27.78)2  2(6.5)(20)
v2  511.7284
v  22.62 ms 1

Convert to km h–1

 (km h-1)   22.62 m  1km  3600 s 
Since, v in 1s 1000 m 1h

 81.43 km h-1

19

Solution

(b) Assume : time elapsed, t

v  u  at

22.62  27.78  (6.5)t

 6.5t  5.16

t  5.16
6.5

 0.794 s

20

2.3 Projectile Motion

a) Solve problems on projectile motion launched at an
angle θ= 90o (free fall)

v  u  gt

s  ut  1 gt 2
2

v2  u2  2gs

21

Example 1

A student drops a ball from the top of a tall building, it takes
2.8 s for the ball to reach the ground.
(a) What was the ball’s speed just before hitting the

ground ?
(b) What is the height of the building ?

Solution

Given :
u = 0 m s–1 (dropped) ;
t = 2.8 s ;
free fall motion, g = 9.81 m s–2

22

Solution (a) v  u  gt -ve : the direction of
v  0  (9.81)(2.8) velocity is downward

v   27.47 m s-1

(b) s  ut  1 gt2
2
s  0  1 (9.81)(2.8)2
2

s   38.46 m - ve : the direction of the
displacement is downword

Height of the building, h = 38.46 m

23

Example 2

A stone is thrown vertically downward at an initial speed
of 14 m s–1 from a height of 65 m above the ground.
(a) How far does the stone travel in 2 s ?
(b) What is its velocity just before it hits the ground ?

Solution
Given : u = –14 m s–1 ; g = 9.81 m s–2 ; t = 2 s

(a) Using free fall equation :

s  ut  1 gt2
2
s  (14)(2)  1 (9.81)(2) 2

2

s  28 19.62

s   47.62 m 24

(b) Assume the velocity just before hitting the ground = v

v2  u2  2gs

v2  (14)2  2(9.81)(65)

where s   47.62 m

v2  1471.3

   38.36 m s-1
v

25

Example 3

A ball is thrown from the top of a building is given an
initial velocity of 10.0 m s1 straight upward. The building
is 30.0 m high and the ball just misses the edge of the roof
on its way down, as shown in figure below. Calculate B

a. the maximum height of the stone from point A.

b. the time taken from point A to C. u =10.0 m s1 C
c. the time taken from point A to D.
A

d. the velocity of the stone when it reaches point D.
(Given g = 9.81 m s2)

30.0 m

D

26

Solution

a. At the maximum height, H, v= 0 and u = 10.0 m s1 thus

B v2  u2  2gs

u 0  10.02  29.81H

AC H  5.10 m

30.0 m b. From point A to C, the vertical displacement, s= 0 m thus
D
s  u t  1 gt 2
2

0  10.0t  1 9.81t 2

2

t  2.03 s

27

Solution

c. From point A to D, the vertical displacement, s = 30.0 m

B thus s  ut  1 gt 2

u  30.0  2  1 9.81t 2

10.0t 2

AC 4.91t 2 10.0t  30.0  0

a bc

30.0 m By using t   b  b2  4ac Time
D 2a don’t
have
t  3.69 s OR 1.66 s negative
value.

28

Solution

B d. Time taken from A to D is t = 3.69 s thus

u v  u  gt

AC v  10.0 9.813.69

30.0 m v  26.2 m s1

D OR
From A to D, s = 30.0 m

v 2  u 2  2gs

v2  10.02  29.81 30.0

v  26.2 m s1 or v  26.2 m s1
Therefore the ball’s velocity at D is v  26.2 m s1

29

2.3 Projectile Motion

b) Solve problems on projectile motion launched at
an angle θ= 0o

30

Projectile motion refers to the motion of an object
projected into the air at an angle.

the path of motion : parabolic arc 31

A motion where object travels at uniform velocity in horizontal direction;
at the same time undergoing acceleration in downward direction under
the influence of gravity.
Horizontal & vertical motions are independent and discussed separately in calculation.

The initial velocity, u is resolved into horizontal and vertical components :

ux  u cos
uy  u sin 

32

Example 4 y

H
u

O  = 60.0 vP 1x x

R v v v1y

1
2x

Q

v2 y
v2

Figure below shows a ball thrown by athlete with an initial
speed , u = 200 m s – 1 and makes an angle, θ = 60.00 to

the horizontal. Determine :

33

a. the position of the ball, the magnitude and direction
of its velocity, when t = 2.0 s.
b. the time taken for the ball reaches the maximum
height, H and calculate the value of H.

c. the horizontal range, R.

d. the magnitude and direction of its velocity when the
ball reaches the ground (point P).

(Given g = 9.81 m s – 2 )

34

Solution

The component of initial velocity :

ux  200 cos60.0  100 m s1

uy  200 sin 60.0  173 m s1

a) Position of the wall when t =2 .0 s

Horizontal component :

sx  uxt  1002.00

sx  200 m from point O

Vertical component 1: gt2
2
sy  uyt 

sy  1732.00  1 9.812.002
2
sy  326 m abovethe ground

Therefore the position of the ball is (200 m , 326 m) 35

Solution

a) magnitude and direction of the ball’s velocity at t =2.0 s

Horizontal component : vx  ux  100 m s1

Vertical component : vy  uy  gt

vy  173 9.812.00

vy  153 m s1

Magnitude, v vx2  v 2  1002  1532  183 m s1
y

Direction, θ  tan1 vy   tan1 153 
vx  100 

θ  56.8 from positive x-axis 36

Solution

b. i. At maximum height, H :

vy  0

Thus the time taken to reach maximum height is given by

v y  u1y73gt 9.81t

0

t  17.6 s

ii. Apply

sy  u1y7t31127g.6t   1 9.8117.62
H 
H  1525 m 2

37

Solution 38

c. Flight time = 2x(the time taken to reach the maximum height) :

t  217.6  35.2 s

Hence the horizontal range, R is

sx  uxt  10035.2  3520 m

d. When the ball reaches point P thus sy  0

The velocity of the ball at point P,

Horizontal component : v1x  ux  100 m s1
Vertical component : v1y  u y  gt

v1y  173  9.8135.2

v1y  172 m s1

Magnitude,

   v1  v12x  v12y 
100 2  172 2

Direction, v1  200 m s1

θ  tan 1 v1y   tan 1 172 
v1x  100 
θ  60.0

39

Example 6

A transport plane travelling at a constant velocity of 50 m s– 2 at an
altitude of 300 m releases a parcel when directly above a point X on
level ground. Calculate :
a. the flight time of the parcel.
b. the velocity of impact of the parcel.
c. the distance from X to the point of impact.

(Given g = 9.8 m s – 2 )

u  50 m s1

300 m d 40
X

Solution

The parcel’s velocity = plane’s velocity

u  50 m s1
Thus ux  u  50 m s1 and u y  0 m s1

a. The vertical displacement is given by

sy  300 m

Thus the flight time of the parcel’s is

 sy  uyt  1 gt2 2
300  0  1 2
2
9.81t

t  7.82 s

41

Solution

b. The component of velocity of impact of parcel :

Horizontal component : vx  ux  50 m s1

Vertical component : vy  uy  gt

vy  0  9.817.82

vy  76.7 m s1

Magnitude, v v 2  v 2  502   76.72
Direction, x y

v  91.6 m s1

θ  tan561.9vv xy   tan 1  76.7 
 50 
θ 

42

c) the distance from X to the point of impact.

sx  uxt  507.82  391 m

43