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 of the particle measured while moving in circle is given by equation

$S = 2t$

now we can find the tangential speed of the particle by using the equation

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

now we can write

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

$v_t = 2m/s$

now again derivative of this tangential speed is known as tangential acceleration

As we can see above that tangential speed is constant here so tangential acceleration must be ZERO

now for centripetal acceleration we can use

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

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

$a_c = \frac{4t^2}{12} = \frac{t^2}{3}$

now we can say the ratio of two acceleration will be

$\frac{a_t}{a_c} = 0$

so the ratio will be ZERO as tangential acceleration is given ZERO

Answer 2

Here r is 12 m, and s is 2 t 3, and t is 2s.

The value of v is ds / dt which is 6 t 2.

Now the velocity after t is 2 s which is 24 m/s.

Now Ac will be v2 / r which comes down to 48.

And At is 12 t.

The final ratio comes to 24 / 48 which is 1 / 2