Question

A particle moves in a circle of radius $3.5\text{cm}$ at a speed given by $v=8t$, where $v$ is in $\text{cm/s}$ and $t$ in seconds. Find the tangential acceleration at $t=2\text{s}$. (in ${\text{cm/s}}^2)$

• A. $32{\text{cm/s}}^2$
• B. $24{\text{cm/s}}^2$
• C. $16{\text{cm/s}}^2$
• D. $8{\text{cm/s}}^2$

Answer 1

The correct option is D $8{\text{cm/s}}^2$

The tangential acceleration $\left(a_t\right)$ is the rate of change of speed of particle moving on a circular path.
$\because$ speed of particle is given by $v=8t\text{cm}$
$\therefore a_t=\frac{dv}{dt}$
$\Rightarrow a_t=\frac{d}{dt}\left(8t\right)=8{\text{cm/s}}^2$
Here, tangential acceleration $\left(a_t\right)$ is constant, it will not change with time.
Hence, tangential acceleration at $t=2\text{s}$ is $8{\text{cm/s}}^2$.