Question

A point moves along a circle with a velocity v = at, where a = 0.50 $m/s^2$. Find the total acceleration of the point at the moment when it has covered the $n^{th}$ (n = 0.10) fraction of the circle after beginning of the motion.

• A. 0.8 $m/s^2$
• B. 0.3 $m/s^2$
• C. 0.2 $m/s^2$
• D. 0.50 $m/s^2$

Answer 1

The correct option is A 0.8 $m/s^2$

$\begin{array}{c}Thevelocityoftheparticlev=at\\ So,\frac{dv}{dt}=w_t=a-------\left(1\right)\\ and{w}_R=\frac{v^2}{R}=\frac{a^2{t}^2}{R}\left(u\sin gv=at\right)-----\left(2\right)\\ froms=\int vdt\\ 2\pi R\eta =a{\int }_0^{t}vdt=\frac{1}{2}a{t}^2\\ So,\frac{4\pi \eta }{a}=\frac{t^2}{R}-------\left(3\right)\\ fromequation\left(2\right)and\left(3\right)\\ w_R=4\pi a\eta \\ then,w=\sqrt{w_t^2+w_{\pi }^2}\\ =\sqrt{a^2+{\left(4\pi a\eta \right)}^2}=a\sqrt{1+16{\pi }^2{\eta }^2}=0.8m/\end{array}$

Hence,

option $\left(A\right)$ is correct answer.