Question

A particle rotates in a circular path of radius $54m$ with varying speed $v=4t^2.$ Here $v$ is in m/s and $t$ in second. Find angle between velocity and acceleration at $t=3s$.

• A. ${45}^{\circ }$
• B. ${75}^{\circ }$
• C. ${90}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is A ${45}^{\circ }$

Given : $r=54$ m $t=3$ s

The velocity is given as $v=4t^2$ m/s

Radial acceleration $a_r=\frac{v^2}{r}=\frac{16t^4}{54}$

$\therefore$ $a_r{|}_{t=3}=\frac{16\times 3^4}{54}=24$ $m/s^2$

Tangential acceleration $a_t=\frac{dv}{dt}=8t$

$\therefore$ $a_t{|}_{t=3}=8\times 3=24$ $m/s^2$

Thus angle between $a$ and $v$, $tan\theta =\frac{a_r}{a_t}=\frac{24}{24}=1$

$⟹$ $\theta ={45}^o$