Question

A particle is constrained to move in a circle of radius $300\text{cm}$. Its linear speed increases with time as $v=2t$, where $t$ is in second and $v$ is in $\text{m/s}$. The acceleration of the particle at $t=3\text{s}$ is

• A. $12{\text{m/s}}^2$
• B. $2{\text{m/s}}^2$
• C. $14{\text{m/s}}^2$
• D. $12.16{\text{m/s}}^2$

Answer 1

The correct option is D $12.16{\text{m/s}}^2$

Note that the particle is not undergoing uniform circular motion because its speed varies with time. The speed of the particle at $t=3\text{s}$ is $v=2t=2\left(3\right)=6\text{m/s}$. The centripetal acceleration of the particle at this instant is
$a_c=\frac{v^2}{r}=\frac{(6{)}^2}{3}=12{\text{m/s}}^2$.
The tangential acceleration of the particle at $t=3\text{s}$ is
$a_t=\frac{dv}{dt}=2{\text{m/s}}^2$.

The centripetal acceleration of the particle is radially inward (in - $\hat{r}$ direction) and tangential acceleration is along the tangent (in $\hat{\theta }$ direction). Thus, the acceleration of the particle at t = 3 s is
$\overrightarrow{a}=a_c(-\hat{r})+a_t\hat{\theta }=-12\hat{r}+2\hat{\theta }.$
Note that $\hat{r}$ and $\hat{\theta }$ are perpendicular to each other.
Thus, the magnitude of acceleration is
$|\overrightarrow{a}|=\sqrt{a_c^2+a_t^2}=\sqrt{(12{)}^2+(2{)}^2}=12.16{\text{m/s}}^2$.