Page 14 - Focus TG4 KSSM (Physics) Terbitan Penerbitan Pelangi Sdn Bhd

Physics Form 4 Chapter 2 Force and Motion I

Acceleration-Time Graphs
EXAMPLE 2.11
Table 2.3
Graph of a against t Explanation Figure 2.31 shows the velocity-time graph of a
motorcycle along a straight road.
(a)
a / m s –2 v / m s –1
Object moves with positive 25.0 A B
a uniform acceleration.
20.0
0 t /s 15.0
(b) 10.0
a / m s –2 5.0
Object moves with negative C t /s
uniform acceleration or 0 5.0 10.0 15.0 20.0 25.0
0 t /s deceleration. –5.0
–10.0 D
Chapter
2 –a Figure 2.31

EXAMPLE 2.10 (a) Explain the state of motion of the motorcycle
represented by the portion of the graph
Figure 2.30 shows a displacement-time graph of a (i) OA (ii) AB
runner during a warming up session. (iii) BC (iv) CD
(b) Determine the displacement of the motorcycle
s /m
C (i) after 20.0 s.
s 1
(ii) after 25.0 s.
s A B
2 (c) Sketch a acceleration-time graph for the
motorcycle during the first 25.0 seconds.
Solution
D
0 t /s
t t t t
1 2 3 4 (a) (i) OA
Figure 2.30 Gradient of graph = 25.0 – 0
10.0 – 0
Explain the state of motion of the runner at the = 2.5 m s –2
portion of the graph represented by (a) OA (b) AB The motorcycle moves with a constant
(c) BC (d) CD.
acceleration of 2.5 m s .
–2
Solution (ii) AB
(a) OA The motorcycle moves with a constant
–1
The gradient of the graph is positive and velocity of 25.0 m s .
constant. The runner runs with a uniform velocity.
(b) AB (iii) BC: (0 – 25.0)
The displacement does not change. The runner Gradient =
is not moving. (20.0 – 15.0)
(c) BC = –5.0 m s –2
The gradient of the graph is increasing. The The motorcycle moves with a deceleration
–2
runner runs with increasing velocity, or the with a magnitude of 5.0 m s until it stops
runner accelerates. at the time t = 20.0 s.
(d) CD (iv) CD
The gradient of the graph is negative and The motorcycle moves with a negative
constant. The runner is running with a uniform uniform velocity. This means the
velocity but in the opposite direction. The motorcycle is now moving in the opposite
runner goes back to his original position. direction, on its way back.

30

02 FOC PHYSICS F4 3P.indd 30 29/01/2020 1:39 PM