Question

A particle moves in straight line such that its displacement $x$ at time $t$ is $x^2=t^2+1.$ Acceleration of particle at $t=\frac{3}{4}sec$ is.

• A. $\frac{125}{64}$ $m/s^2$
• B. $\frac{64}{125}$ $m/s^2$
• C. $\frac{25}{16}$ $m/s^2$
• D. $\frac{16}{25}$ $m/s^2$

Answer 1

The correct option is B $\frac{64}{125}$ $m/s^2$

$x=\sqrt{t^2+1}\Rightarrow v=\frac{dx}{dt}=\frac{t}{\sqrt{t^2+1}}\Rightarrow a=\frac{dv}{dt}=\frac{\sqrt{t^2+1}-\frac{t^2}{\sqrt{t^2+1}}}{t^2+1}$

$a=\frac{1}{(t^2+1{)}^{\frac{3}{2}}}$
$\Rightarrow att=\frac{3}{4}a=\frac{1}{(\frac{9}{16}+1{)}^{\frac{3}{2}}}=(\frac{4}{5}{)}^3=\frac{64}{125}$