Question

A particle moves in a straight line and its position x at time t is given by $x^2=2+t$. Its acceleration is given by.

• A. $\frac{-2}{x^3}$
• B. $-\frac{1}{4x^3}$
• C. $-\frac{1}{4x^2}$
• D. $\frac{1}{x^2}$

Answer 1

The correct option is B $-\frac{1}{4x^3}$

Given $x^2=2+t$ Now differentiating this equation with respect to time we get ,

$2x\frac{dx}{dt}=1v=\frac{dx}{dt}=\frac{1}{2x}$

Now again differentiating equation A with respect to dx we get $v\frac{dv}{dx}.\frac{dx}{dt}=\frac{1}{2x}dx$

$v\frac{dv}{dt}=-\frac{1}{2x^2}$

$\frac{dv}{dt}=-\frac{1}{2x^2}\left(\frac{1}{2x}\right)=-\frac{1}{4x^3}$