Question

A car accelerates from rest at a constant rate of 3m/s square for some time then it retards at constant rate of 6m/s square and comes to rest . If the total time for which it remains in motion is 3 second so calculate the how many distance will cover by the car

Answer 1

Explanation:

C

4m/s

Let v,t

1

,t

2

be the maximum velocity attained, time for which car accelerated and the time for which car retarded respectively

⇒t

1

=(v−u)/a=(v−0)/2=v/2

⇒t

2

=(v−u)/a=(0−v)/(−4)=v/4

Now it is given that,

⇒t

1

+t

2

=3

⇒v/2+v/4=3

⇒v=4m/s

Hence correct answer is option C

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