Q: A point P moves in a counter-clockwise direction on a circular path as shown in the figure. The moment of P is such that it sweeps out a length s = t³ + 5, where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when, t = 2 s is nearly, is

(a) 13 ms^{-2}

(b) 12 ms^{-2}

(c) 7.2 ms^{-2}

(d) 14 ms^{-2}

Ans: (d)

$ \displaystyle v = \frac{ds}{dt} = 3t^2 = 3\times 2^2 = 12m/s $

$ \displaystyle a_t = \frac{dv}{dt} = 6t = 6\times 2 = 12m/s^2 $

$ \displaystyle a_c = \frac{v^2}{R} = \frac{144}{20} = 7.2 m/s^2 $

$\displaystyle a = \sqrt{(12)^2 + (7.2)^2} \approx 14 m/s^2$