Question

No sooner car A starts from rest another car B moving
with constant speed crosses the car A. Their velocity
time graphs are shown in figure 2.23.
A
Velo (ms)
30
B
20
110
46
time (s
(1) What is the acceleration of the car A?
(ii) At t= 40 s, what is the distance between the cars?
(iv) When car A catches the car B?
[0.75 ms, 200 m, 60 s)
​

Answer 1

(i) Acceleration of car A can be obtained by taking the slope of velocity vs time graph of A.

⇒  $acceleration(a) = \frac{30-0}{40 - 0} =\frac{3}{4} = 0.75 \ m/s^2$

(ii) The distance travelled by the vehicles can be obtained by taking the area under the graph of velocity vs time.

so,

distance travelled by car A for 40 sec = 1/2 x 40 x 30 = 600 m

distance travelled by car B for 40 sec = 20 x 40 = 800 m

∴ the distance between cars A & B = 800 - 600 = 200 m

(iii) Let car A catches the car B in time t after 40 seconds

to catch the car B , A has to cover the distance between them and the total distance travelled must be equal to that of B after time 40 sec

    distance car A has to travel in time t = 200 + 30 x t

    distance car B travels after 40 sec = 20 x t

These two distances must be same

⇒ 20 x t = 200 + 30 x t

⇒ 10 x t = 200

⇒ t = 20 sec

so the car A catches car B after 20 from the first 40 sec

⇒ @ time t = 40 + 20 = 60 sec car A catches car B

HOPE THIS HELPS YOU !!