Question

A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t₁ = 5.00 s, it is at point (5.00 m, 6.00 m) with velocity (3π m/s)$\hat{j}$ and acceleration in the positive x direction. At time t₂ = 10.0 s, it has velocity (-3π m/s)$\hat{i}$​ and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path if t₂ - t₁ is less than one period?

• A. (15 m, 6 m)
• B. (-5 m, 6)
• C. (15 m, 0)
• D. (-5 m, 0)

Answer 1

The correct option is A

(15 m, 6 m)

We know the body is undergoing uniform circular motion,

$\therefore$ only possible path is

As acceleration should always be directed toward the centre

To calculate the position of the centre, we need the radius, to calculate radius,$v=\omega r$ we can use the relation,as we can calculate $\omega$
$\Delta \theta =\frac{3\pi }{2},\Delta t=5s$
=$\frac{3\pi }{2\times 5}=\frac{3\pi }{10}{S}^{-1}$
$usingv=\omega r$
$r=\frac{v}{\omega }$
$|\overrightarrow{v}|=3\pi m/sgiven$
$\therefore r=\frac{3\pi }{\frac{3\pi }{10}}=\frac{30\pi }{3\pi }=10m$
$\therefore ifr=10m$, coordinates of the centre (15 m,6m)