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Physics Form 4 Chapter 2 Force and Motion I

Acceleration and Deceleration Solution
25 30 35 25 30 35 25 30 35 25 30 35 –1
20 40 20 40 20 40 20 40 Initial velocity, u = 10 m s
15 45 15 45 15 45 15 45 –1
10 50 10 50 10 50 10 50 Final velocity, v = 20 m s
5 –1 55 5 –1 55 5 –1 55 5 –1 55
m s m s m s m s
0 60 0 60 0 60 0 60 Time taken, t = 2.5 s
20 – 10
Acceleration, a =
Figure 2.7 2.5
= 4.0 m s –2
1. Figure 2.7 shows a car moving along a straight
line. The speedometer of the car shows that it
is moving with increasing velocity. The car is EXAMPLE 2.4
accelerating. A car travelling at
24 m s slowed down
–1
2. (a) Acceleration is defined as the rate of when the traffic light
change of velocity. turned red. After
Chapter
2 Acceleration, a = Change of velocity undergoing uniform Figure 2.10
deceleration for 4 s, it
Time taken
stopped in front of the traffic light. Calculate the
(b) The formula for an object travelling with magnitude of the deceleration.
a uniform acceleration, a is given by: Solution

Initial velocity, u = 24 m s –1
Final velocity, v = 0 m s –1
Time taken = 4 s
where u is the initial velocity of the object, Using the formula a = v – u = 0 – 24 = –6.0 m s –2
v is its final velocity and t is the time taken. t 4
–2
(c) The SI unit for acceleration is m s . The negative value shows that the car is decelerating.
–2
(d) Acceleration is a vector quantity. Hence, the magnitude of deceleration is 6.0 m s .
25 30 35 25 30 35 25 30 35 25 30 35
20 40 20 40 20 40 20 40 EXAMPLE 2.5
15 45 15 45 15 45 15 45
10 50 10 50 10 50 10 50
5 m s –1 55 5 m s –1 55 5 m s –1 55 5 m s –1 55
0 60 0 60 0 60 0 60 (a) A car moves from
BERHENTI
rest and accelerates
uniformly to a
Figure 2.8 velocity of 20 m s Figure 2.11
–1
3. (a) An object is said to be undergoing a in 5 seconds. What
deceleration or retardation when it is is the acceleration of the car?
slowing down. The rate of change of (b) After that, the car slows down at a uniform
velocity of the object then has a negative rate until it stops in 5 seconds. What is the
value. acceleration of the car?
(b) Figure 2.8 shows a car decelerating. Solution
The speedometer of the car shows that (a) Initial velocity, u = 0 m s –1
it is moving with decreasing velocity. Final velocity, v = 20 m s –1
Time taken, t = 5 s
v – u 20 – 0
EXAMPLE 2.3 Therefore a = t = 5 = 4 m s –2

A van accelerates uniformly (b) Initial velocity, u = 20 m s –1
from a velocity of 10 m s Final velocity, v = 0 m s –1
–1
to 20 m s in 2.5 s. What is Time taken, t = 5 s 0 – 20
–1
v – u
the acceleration of the van? Therefore, a = = = –4 m s –2
Figure 2.9 t 5
–2
The car decelerates with a magnitude of 4 m s .
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