Question

A particle is moving in a circle of radius $R$ = $1m$ with constant speed $v=4\text{m/s}$. The ratio of displacement to acceleration of the foot of the perpendicular drawn from the particle on the diameter of the circle is:

• A. $\frac{1}{16}{\text{s}}^2$
• B. $\frac{1}{2}{\text{s}}^2$
• C. $2{\text{s}}^2$
• D. $16{\text{s}}^2$

Answer 1

The correct option is A $\frac{1}{16}{\text{s}}^2$

$\text{Displacement}=x=OQ=R\cos \theta$

$\text{Acceleration}=\frac{d^{2}x}{d{t}^2}=\frac{d^2\left(R\cos \theta \right)}{d{t}^2}=-w^{2}Rcos\theta$

$\frac{\text{Displacement}}{\text{Acceleration}}=\frac{R\cos \theta }{{\omega }^{2}R\cos \theta }=\frac{1}{w^2}$

$v=\omega R\Rightarrow \omega =\frac{v}{R}=\frac{4\text{m/s}}{1\text{m}}=4\text{rad/sec}$

$\frac{\text{Displacement}}{\text{Acceleration}}=\frac{1}{16}{\text{s}}^2$