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Question: A particle moves rectilinearly. Its displacement x at time t is given by \(x^{2} = at^{2} + b\)where...

A particle moves rectilinearly. Its displacement x at time t is given by x2=at2+bx^{2} = at^{2} + bwhere a and b are constants. Its acceleration at time t is proportional to

A

1x3\frac{1}{x^{3}}

B

1x−1x2\frac{1}{x} - \frac{1}{x^{2}}

C

−tx2- \frac{t}{x^{2}}

D

1x−t2x3\frac{1}{x} - \frac{t^{2}}{x^{3}}

Answer

1x3\frac{1}{x^{3}}

Explanation

Solution

Given : x2=at2+bx^{2} = at^{2} + b …… (i)

Differentiating w.r.t. t on both sides, we get2xdxdt=2at2x\frac{dx}{dt} = 2at

xv=atxv = at (∵v=dxdt)\left( \because v = \frac{dx}{dt} \right)

Again differentiating w.r.t. t on both sides, we get

xdvdt+vdxdt=ax \frac { d v } { d t } + v \frac { d x } { d t } = a \quad or xdvdt=a−v2\quad x \frac { d v } { d t } = a - v ^ { 2 } dvdt=a−v2x=a−(atx)2x=a−a2t2x2x=a(x2−at2)x3\frac{dv}{dt} = \frac{a - v^{2}}{x} = \frac{a - \left( \frac{at}{x} \right)^{2}}{x} = \frac{a - \frac{a^{2}t^{2}}{x^{2}}}{x} = \frac{a(x^{2} - at^{2})}{x^{3}}

dvdt=abx3\frac{dv}{dt} = \frac{ab}{x^{3}} (Using (i))

Or Accelerations ∝1x3\propto \frac{1}{x^{3}}