Question

The distance of a particle moving on a circle of radius 12 m measured from a point on the circle and measured along the circle is given by s=2t^3 ( in meters) the ratio of its tangential to centripetal acceleration at t =2s

Answer 1

given that distance covered by the particle is

$s = 2t^3$

now the tangential speed of the particle will be given by

$\frac{ds}{dt} = v_t$

$v_t = 6t^2$

now to find the tangential acceleration we can say

$a_t = \frac{dv_t}{dt}$

$a_t = 12t$

centripetal acceleration is given by

$a_c = \frac{v_t^2}{R}$

$a_c = \frac{(6t^2)^2}{12}$

$a_c = 3t^4$

now in order to find the ratio of tangential and centripetal acceleration

$\frac{a_t}{a_c} = \frac{12t}{3t^4}$

$\frac{a_t}{a_c} = \frac{4}{t^3}$

put t = 2 s

$\frac{a_t}{a_c} = \frac{4}{2^3} = \frac{1}{2}$

so the ratio of two acceleration at given instant of time will be 1:2

Answer 2

Centripetal acceleration of the particle is given by,

ac = v^2/r = v^2/12 -- (1)

and distance measured s = a t^3

ds/dt = v = 3 a t^2 and d^2s/dt^2  = at = 6at

The ratio of ac to at is given by,

ac/at = v^2/12/6at = 3at^2/72at = t/24

Therefore ratio of these two acceleration at t = 2s is given by,

ac/at = 2/24 = 1/12 or 1:12