Question

A point moves along $x$ axis. Its position at time $t$ is given by $x^2$=$t^2$+$1$. Its acceleration at time $t$ is:

• A. $1/x^3$
• B. $1/x^4$
• C. $-1/x^3$
• D. $1/x$

Answer 1

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

Given, $x^2=t^2+1$

or $x=(t^2+1{)}^{1/2}$

So, $\frac{dx}{dt}=\left(1/2\right)\frac{2t}{(t^2+1{)}^{1/2}}=\frac{t}{\sqrt{t^2+1}}$

or acceleration ,$a=\frac{d^{2}x}{d{t}^2}=t(-1/2)\frac{2t}{(t^2+1{)}^{3/2}}+\frac{1}{\sqrt{t^2+1}}=\frac{-t^2}{x^3}+\frac{1}{x}=\frac{1-x^2}{x^3}+\frac{1}{x}=1/x^3$