Page 182 - Physics

Physics for Secondary Schools

Example 8.6 1
= 48s 20m/s ´ ´
1. A car starts from rest and accelerates 2
uniformly at a rate of 4 m/s for = 480 m
2
5 seconds. It maintains a constant Therefore, the total distance
velocity for 20 seconds. The covered is 480 m.
brakes are then applied, and the car
decelerates for 3 s. Find: If the acceleration of an object is constant,
(a) The maximum velocity attained. the velocity will change uniformly, that
(b) The total distance covered. is, by the same amount each second. In

Solution that case, the average velocity is equal to
(a) From the equation of acceleration, the sum of the initial and fi nal velocities
D v divided by two.
a =
t Therefore, for a uniformly accelerated
But
a = 4 m/s 2 body.
t = 5 s Average velocity = u + v
Δv = at 2
= 4 m/s ×5 s Where, u is the initial velocity and v is
2
= 20 m/s the fi nal velocity.

Therefore, the maximum velocity Task 8.3
attained is 20 m/s.
(b) By graphical method. Using Figure 8.17, describe how the
Figure 8.16 shows the graph of object was moving. Include where the
velocity (m/s) against time (s). original position of the object, whether
it was moving or not, the direction it
was moving, the sign (direction) of
30 the velocity and whether the velocity
Velocity (m/s) 20 P Q was constant, increasing or decreasing.
corresponding
Then
sketch
the
10
velocity-time graph in Figure 8.18.
S
O 5 T 10 15 20 25 R 30
Time (s)
Figure 8.16 2
The area under the graph represents the Displacement (m) 1 3
distance covered by the car. Therefore,
1 Time (s)
Area = (PQ OR+ ) PS´ 4
2

1
= (20s 28s+ ) 20 m/s´
2 Figure 8.17

176
Student’s Book Form One

Physics Form 1 Final.indd 176 16/10/2024 20:58