Question

A particle is moving on a circular path of radius $3cm$ with a time varying speed of $v=3tcm/s$, where $t$ is in seconds. The magnitude of tangential acceleration and the total acceleration of the particle at $t=1s$ is

• A. $3cm/s^2,3cm/s^2$
• B. $3\sqrt{2}cm/s^2,3\sqrt{2}cm/s^2$
• C. $\frac{3}{2}cm/s^2,3cm/s^2$
• D. $3cm/s^2,3\sqrt{2}cm/s^2$

Answer 1

The correct option is D $3cm/s^2,3\sqrt{2}cm/s^2$

Given,
Radius of the circular path, $r=3cm$
Speed, $v=3tcm/s$
Total acceleration $\overrightarrow{a}=\overrightarrow{a_c}+\overrightarrow{a_t}$
$|\overrightarrow{a}|=\sqrt{a_c^2+a_t^2}$
Centripetal acceleration, $\left(a_c\right)=\frac{v^2}{r}=\frac{9t^2}{3}=\frac{9t^2}{3}=3t^2=3cm/s^2$
Tangential acceleration, $a_t=\frac{dv}{dt}=\frac{d}{dt}\left(3t\right)=3cm/s^2$
Then, $|a|=\sqrt{(3{)}^2+(3{)}^2}=3\sqrt{2}cm/s^2$