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 $-\frac{1}{4x^3}$

$x^2=2+t$

Differentiating with respect to t

$2xv=1$

Differentiating with respect to t

$2v^2+2xa=0$

$a=-\frac{v^2}{x}$

From first differentiation we have

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

Therefore

$a=-\frac{1}{4x^3}$