Page 50 - text book form physics kssm 2020
P. 50
Solution:
Velocity, v / m s –1 Table 2.5
20 Time Displacement Acceleration
18
16 0 s – 10 s Displacement Acceleration
= area A = gradient of
14
12 = 10 × 10 the graph
–2
= 100 m = 0 m s
10
8 10 s – 30 s Displacement Acceleration
6 A B C = area B = 16 – 10
1
4 = (10 + 16)(20) 30 – 10 –2
2
2 = 260 m = 0.3 m s
0 Time,
5 10152025 3035 40 45 50 t / s 30 s – 50 s Displacement Acceleration
= area C 0 – 16
Figure 2.32 =
1 50 – 30
= × 20 × 16 –2
Displacement after 30 s = 100 + 260 2 = –0.8 m s
= 360 m = 160 m
Displacement after 50 s = 100 + 260 + 160
= 520 m
(a) Displacement-time graph (b) Acceleration-time graph
Acceleration, a / m s –2
Displacement, s / m
0.3
0.2
520
500 0.1
0 Time, t / s
400 –0.1 10 20 30 40 50
360 –0.2
300 –0.3
–0.4
200 –0.5
–0.6
100 –0.7
–0.8
0 Time, t / s –0.9
10 20 30 40 50
Figure 2.34
Figure 2.33
44
44 2.2.5
2.2.5
