Question

a particle accelerates from the rest with an acceleration of 3m/s2 and then immediately retards at rate of 2m/s2 and finally comes to rest after traveling a distance of 60m along straight line path. the maximum speed of the particle during it's motion can be​

Answer 1

Explanation:

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