A particle moves for 8 seconds. It first accelerates from rest and then retards to rest. If the retardation be three times the acceleration, then time for which it accelerates will be
(A) 2s
(B) 3s
(C) 4s
(D) 6s

Hint:The total time is given for which the particle is in motion. At time t=0, the particle is at rest and then it accelerates and then de-accelerates and ultimately comes to at rest again. We can use equations of motion since no external force is acting on the system.

Complete step by step answer:
Let us assume that the particle accelerates for time=x seconds. Since, the total time for which it is in motion is 8 s, so the time for which it de-accelerates will be (8-x) s.
Given retardation is three times the acceleration.
For first part of the journey;
Initial velocity, u=0
Time taken, t= x
Acceleration= a
Using first equation of motion,
\[\begin{align}
& a=\dfrac{v-u}{t} \\
& {{v}_{1}}=ax \\
\end{align}\]-------(1)
Now for second part of the journey:
Initial velocity= \[{{v}_{1}}\]
Final velocity=0
Acceleration= 3a
Time= (8-x) s
Using first equation of motion,
\[\begin{align}
& a=\dfrac{v-u}{t} \\
& 3a=\dfrac{0-{{v}_{1}}}{8-x} \\
& 24a-3ax={{v}_{1}} \\
\end{align}\]
Substituting the value in equation (1) we get
\[\begin{align}
& 24a-3ax=ax \\
& 24a=4ax \\
& x=6 \\
\end{align}\]
So, the value of x comes out to be 6 s.
Hence, the correct option is (D).

Note:In this problem we have not taken the sign of negative in retardation because it is automatically adjusted in our equations. We have divided the time interval into two parts as per the demand of the question. These problems are easily solved using Newton’s equations of motion.