Question

The distance of a particle moving on a circle of radius 12m measured from a fixed point on the circle and measured along the circle is given by s=2t m. The ratio of its tangential to centripetal acceleration at=2s is

Answer 1

distance covered by the particle in circle is given by equation

$s = 2t$

now we can find the tangential speed using this

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

now we will substitute the value of "s" in it

$v_t = \frac{d}{dt}(2s)$

$v_t = 2 m/s$

now from above we can say that tangential acceleration must be ZERO as the tangential speed is constant here and not changing with time

so tangential acceleration is zero

now to find the centripetal acceleration we can use

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

here we will plug in value of tangential speed and radius R

$a_c = \frac{2^2}{12}$

$a_c = \frac{1}{3}m/s^2$

now in order to find the ratio of tangential acceleration and centripetal acceleration we can say it is ZERO as tangential acceleration is zero here.

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