Question

The displacement of a particle moving along a straight line varies with time as t = x^2 + x . If v is velocity , then acceleration will be
(a) -4v^3
(b) -2v^3
(c) 4v^3
(d) 2v^2

Answer 1

1=2xv+v
differentiatig w.r.t. t and dx/dt=v
1=v(1+2x)
v=1/(1+2x)
a =dv/dt=-1/(1+2x)^2 *2 =-2v^2

Answer 2

Given:

                        $t = x^2 + x$

In order to find the acceleration, we have to differentiate the above equation with respect to time,

                        $1 = 2x \times \frac { dx }{ dt }  + \frac { dx }{ dt }$

We know that, differentiation of displacement with respect to time will give velocity,           dx/dt = v

                        1 = 2xv + v

                        1 = v (2x+1)

                        v = 1/ (2x+1)

To find acceleration which is the rate of change of velocity, we have to differentiate the above equation again. That is differentiation of velocity with respect to time,

                        $\frac { dv }{ dt } =2 \times[\frac { 1 }{ 2x\quad +\quad 1 } ]^2$

which is = $2 \times v^2$

Hence option d is the answer.