Question

A point moves in a straight line so that its displacement x m at time t sec is given by $x^2=1+t^2$ Its acceleration in m/${\sec }^2$ at a time t sec is

• A. $1/x^3$
• B. $-t/x^3$
• C. $\frac{1}{x}-\frac{t^2}{x^3}$
• D. $\frac{1}{x}-\frac{1}{x^2}$

Answer 1

The correct option is A $1/x^3$

$x^2=t^2+1$

taking it differentiation-

$\begin{array}{c}2x\times \left(\frac{dx}{dt}\right)=2t\\ x\left(\frac{dx}{dt}\right)=t\end{array}$

again taking differentiation for acceleration

$\begin{array}{c}x\left(\frac{d^{2}x}{d{t}^2}\right)+{\left(\frac{dx}{dt}\right)}^2=1\\ x\left(\frac{d^{2}x}{d{t}^2}\right)=1-{\left(\frac{dx}{dt}\right)}^2\\ x\left(\frac{d^{2}x}{d{t}^2}\right)=\frac{1-{\left(\frac{t}{x}\right)}^2}{x}\\ \left(\frac{d^{2}x}{d{t}^2}\right)=\frac{1}{x}-\frac{t^2}{x^3}\\ put{t}^2=x^2-1\\ \left(\frac{d^{2}x}{d{t}^2}\right)=\frac{1}{x^3}\end{array}$

Hence,Option $A$ is correct.