Velocity (v) time (t) graph of a car starting from rest is shown in Fig. 2(b). 13. Draw acceleration (a) -time (t) graph for the motion of the car and find total distance travelled by car.
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Explanation

For the graph we bote that car starting from rest acclerates for time `0 to 5` s and attains a velcity `20 ms^(-1)` from `5 s to 10 s`. After that the motion of car is retarde for time `10 s to 14 s` and it comes to rest at time `t=14 s`.
For time `t0 to t=5a, the acceleration of car
`=slope of `v-t graph `(20-0)/5 =4 ms^(-20`.
For time `t=10 s =10s, velocity of car is constant, hence acceleration is zero.
For time `t=10 s to t=14 s`
Acceleration of car `=slpe of `v-t` graph or `a=(0-200/4 =- ms^(-2)`
Thus accleration (a) -time 9t) graph is of the type as shown in Fig. 2 (b) .14 .
Total distance travelled by car in `14 seconds `=Area OABC`
`=area of `Delta OAE +Area of squareabde`
`=(20 XX50/2 +(20XX5) + (20 XX4)/2`
`=50 +100 +40 =190 M`
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