Question

A point P is moving on a circular path in anti clock wise direction as shown in adjoining diagram. The motion of P is such that it covers length s = t³ + 5, where s is in meters and t in seconds. Radius of the path is 20 m. When t = 2 s, then acceleration of P is approximately:

(a) 1.3 ms^-2

(b) 1.2 ms^-2

(c) 7.2 ms^-2

(d) 4 ms^-2

Answer 1

Correct option is : (d) 4 ms^-2

Given: S = t³ + 5; r = 20 m

Linear velocity of the particle,

\(v = \frac{ds}{dt} = \frac{d}{dt}[t^3 + 5] = 3t^2\)

Velocity at t = 2s, v = 3(t)² = 3 x (2)²

= 3 x 4 = 12 ms^-1

Linear acceleration a =  \(\frac{dv}{dt} = \frac{d}{dt}(3t^2) = 6t\)

At t = 2s, a_T = 6 x 2 = 12 ms^-2

Radial acceleration

\(a_c = \frac{v^2}{r} = \frac{12\times12}{20} = 7.4\ ms^{-1}\)

∴ Resultant acceleration

\(a = \sqrt{a_c^{2} + a_T^{2}} = \sqrt{(7.4)^2 + (12)^2}\)

or a = 14 ms^-2

Hence, option (d) is correct.