Question

A point P revolves on a circular path of radius 20 m figure. The motion of P covers the distance S = t³ + 5, where S is measured in meters and t in seconds. When t = 2 s, then acceleration of particle P is nearly: 

(a) 14 ms^-2

(b) 13 ms^-2

(c) 12 ms^-2

(d) 7.2 ms^-2

Answer 1

Correct option is : (a) 14 ms^-2

Distance covered by point P

S = t³ + 5

\(\therefore\ speed\ v = \frac{dS}{dt} = \frac{d}{dt}[t^3 + 5]\)

or v = 3t²

Speed of point P at t = 2s,

v = 3 x (2)² = 12 ms^-1

∴ Tangential acceleration.

∴ Option (a) is correct.