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

Answer 1

Answer: (B)

$x^{2}=t+2 \Rightarrow \frac{1}{x^{2}}=\frac{1}{t+2}$ ,..(i)
$\Rightarrow x=\sqrt{t+2}$
$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{\frac{1}{2}-1}$
$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{-\frac{1}{2}}$
$\Rightarrow \frac{d^{2} x}{d t^{2}}=\frac{1}{2}\left(-\frac{1}{2}\right)(t+2)^{\frac{1}{2}-1}$
$\Rightarrow a=-\frac{1}{4}(t+2)^{-\frac{3}{2}}$
$=-\frac{1}{4(t+2)} \times \frac{1}{(t+2)^{\frac{1}{2}}}$
$=-\frac{1}{4} \times \frac{1}{x^{2}} \times \frac{1}{x}$
$\Rightarrow a=-\frac{1}{4 x^{3}}$