Question

A point moves along a circle with a velocity $v=at,wherea=0.50m/s^2$. 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 $m/s^2$

$\begin{array}{c}Thevelocityoftheparticlev=at\\ So,\frac{dv}{dt}=w_t=a...................\left(1\right)\\ And{w}_n=\frac{v^2}{R}=\frac{a^2{t}^2}{R}\left(u\sin gv=at\right)............\left(2\right)\\ Froms=\int vdt\\ 0.2\pi R\eta =\frac{t^2}{R}..............\left(3\right)\\ Fromeq\left(2\right)and\left(3\right)w_n=4\pi a\eta \\ Hencew=\sqrt{w_t^2+w_n^2}\\ =\sqrt{a^2+{\left(4\pi a\eta \right)}^2}\\ =a\sqrt{1+16{\pi }^2{\eta }^2}\\ =0.8m/s^2\end{array}$