Question

A car accelerates from rest at a constant rate of 2 ms 2 for some time. Then it retards at a constant
rate of 4 ms and comes to rest. It remains in motion for 6 s.​

Answer 1

Answer:

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