Page 183 - Physics

Linear motion

0 m/s + 30 m/s
v v = 2
a
Velocity (m/s) From, vv = s 15 m/s

a
v =
Time (s)
you get t
Figure 8.18
s = vt
Therefore, in OB, the object’s
Consider an object that accelerates from displacement is:
zero to 30 m/s in 8 s and then continues at s = 15 m/s × 8 s
a constant velocity of 30 m/s for another s = 120 m
10 s. A velocity–time graph of this motion Now consider triangle OAB in Figure
is shown in Figure 8.19. 8.19. The length of its base is 8 s,

and its height is 30 m/s. The area of a
Acceleration
35 triangle is calculated using:
1
B B 2 C C A =
30 2
1
Velocity (m/s) 20 1 = ´ 2 8 s 30 m/s´
25

= 120 m

15
In BC the average velocity is 30 m/s;
From,
10 s
v =
t
5
s = v t
a
O = 30 m/s × 10 s
0 5 A 10 15 D 20 25 = 300 m
Time (s)
You should verify that the area of the
Figure 8.19: A velocity–time graph
rectangle ABCD is equal to the object’s
In OB, the average velocity is obtained displacement in BC of the motion in
through: u + v Figure 8.19.
v a v = Area
2 of ABCD = AB BC´
´ 30 m/s ´10 s 30=
where, v a = is the average velocity 10 s 30 m/s =30 m/s = 10 s´ = 10 s ´ m/s
u = 0m/s , 300 m=
v = 30m/s Displacement = 300 m

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Physics Form 1 Final.indd 177 16/10/2024 20:58