Question

A point moves along a circle with a velocity $v=kt$, where $k=0.5\text{m}/{\text{s}}^2$. Find the acceleration of the point at the moment when it has covered the $n^{th}$ fraction of the circle after the beginning of motion, where $n=\frac{1}{10}$.

• A. $0.5\text{m}/{\text{s}}^2$
• B. $0.3\text{m}/{\text{s}}^2$
• C. $0.8\text{m}/{\text{s}}^2$
• D. $1.1\text{m}/{\text{s}}^2$

Answer 1

The correct option is C $0.8\text{m}/{\text{s}}^2$

Given, velocity, $v=kt$

$v=\frac{t}{2}(\because k=0.5)$

Tangential acceleration,

$a_t=\frac{dv}{dt}=\frac{1}{2}=0.5\text{m}/{\text{s}}^2$

Angular displacement in $n^{th}$ fraction is,

$\theta =2\pi n=\frac{2\pi }{10}=\frac{\pi }{5}[\because n=\frac{1}{10}]$

$\because \theta =\frac{1}{2}\alpha t^2$

$\Rightarrow \frac{\pi }{5}=\frac{1}{2}\left(\frac{a_t}{R}\right)\times t^2$

$\Rightarrow \frac{\pi }{5}=\frac{1}{2}\times 0.5\times \frac{t^2}{R}$

$\therefore \frac{t^2}{R}=\frac{4\pi }{5}......\left(1\right)$

Centripetal acceleration,

$a_c=\frac{v^2}{R}=\frac{(0.5t{)}^2}{R}$

Using eqn $\left(1\right)$ we get,

$a_c=\frac{4\pi (0.5{)}^2}{5}=\frac{\pi }{5}\text{m}/{\text{s}}^2$

$\therefore$Total acceleration,

$a=\sqrt{a_t^2+a_c^2}=\sqrt{(0.5{)}^2+(0.6{)}^2}\approx 0.8\text{m}/{\text{s}}^2$

Hence, option $\left(c\right)$ is the correct answer.