Question

A car accelerates from rest at a constant rate of 6 m/st for sometime after which it
decelerates at a constant rate 9 m/s to come to rest. If the total time of motion is
30 sec, then the maximum velocity acquired by the car is given by
(A) 54 m/s (B) 108 m/s (C) 27 m/s (D) 36 m/s (E) 45 m/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

Hence correct answer is option C

Explanation: