Question

A point moves along a circle of radius $4\text{m}$. The distance $x$ is related to time by $x=c{t}^3$. What should be the value of $c$, so that the tangential acceleration is equal to the normal acceleration when its linear velocity is $4\text{m/s}$?

• A. $0.288{\text{m/s}}^2$
• B. $0.288{\text{m/s}}^3$
• C. $0.333{\text{m/s}}^2$
• D. $0.333{\text{m/s}}^3$

Answer 1

The correct option is D $0.333{\text{m/s}}^3$

Given, $x=c{t}^3\Rightarrow \frac{dx}{dt}=3c{t}^2$
When the linear velocity $v=4\text{m/s}$, then
$3c{t}^2=4\Rightarrow t=\frac{2}{\sqrt{3c}}$
When $v=4\text{m/s}$, given that
$a_T=a_N\Rightarrow \frac{dv}{dt}=\frac{v^2}{r}$
$\Rightarrow \frac{d\left(3c{t}^2\right)}{dt}=\frac{4^2}{4}=4$
$\Rightarrow 6ct=4\Rightarrow 6c\times \frac{2}{\sqrt{3c}}=4$
$\Rightarrow \sqrt{c}=\frac{4\sqrt{3}}{12}$
$\Rightarrow c=\frac{1}{3}=0.333{\text{m/s}}^3$