Question

A particle moves in a circle of radius $4\text{cm}$ at a speed given by $v=t^2$, where $v$ is in $\text{cm/s}$ and $t$ is in $\text{seconds}$. Total acceleration of the particle at $t=2\text{second}$ is $2\sqrt{n}{\text{cm/s}}^2$, then value of $n$ is

Answer 1

Given: $v=t^2;R=4\text{cm};t=2\text{s}$

Tangential acceleration will be

$a_t=\frac{dv}{dt}=2t=4{\text{cm/s}}^2$ (at $t=2\text{s})$

and centripetal acceleration will be

$a_c=\frac{v^2}{R}=\frac{t^4}{R}=\frac{2^4}{4}=4{\text{cm/s}}^2$

So, net acceleration of the particle is given by

$a_{net}=\sqrt{a_t^2+a_n^2}=\sqrt{4^2+4^2}$

$a_{net}=4\sqrt{2}=2\sqrt{8}{\text{cm/s}}^2$

$\therefore n=8$