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
$a_T=a_N\Rightarrow \frac{dv}{dt}=\frac{v^2}{r}\Rightarrow \frac{d}{dt}\left(3c{t}^2\right)=\frac{4^2}{4}\Rightarrow 6ct=4\Rightarrow 6c\times \frac{2}{\sqrt{3c}}=4\Rightarrow \sqrt{c}=\frac{4\sqrt{3}}{12}\Rightarrow c=\frac{48}{144}=0.3333m/s^3$