Q:The motion of a particle along a straight line is described by equation

x = 8 + 12t – t^{3 } . where, x is in metre and t in sec. The retardation of the particle when its velocity becomes zero is

(a) 24 ms^{–2}

(b) zero

(c) 6 ms^{–2}

(d) 12 ms^{–2}

**Ans: (d)**

Solution: x = 8 + 12t – t^{3 }

Differentiating w.r.t. time

$ \displaystyle \frac{dx}{dt} = 0 + 12-3t^2 $

$ \displaystyle v = 12-3t^2 $

When v = 0 ; t = 2 sec

Again differentiating with time

$ \displaystyle \frac{dv}{dt} = 0-6t $

$ \displaystyle a = -6t $

a = -12 m/s^{2}