A point `p` moves in counter - clockwise direction on a circular path as shown in the figure . The movement of 'p' is such that it sweeps out in the figure . The movement of 'p' is such that it sweeps out a length `s = t^(3) + 5 ` , where `s` is in metres and ` t` is in seconds . The radius of the path is `20 m` . The acceleration of 'P' when ` t = 2 s` is nearly .

Options

A. `12 m//s`
B. ` 7 .2 m//s^2`
C. ` 7. 2 m//s^2
D. ` 13 m//s^2`

Answer

C

Explanation

` s=t^2 , r= 20 20 m`
velocity, ` v=(ds)/(dt) = 3 t^2`
When, ` t= 2 s, v = 3 xx 2^2 =12 ms^(-1)`
.
Tangential acceleration, ` a_T = (dv)/(dt) = 6 t`
when, ` t=2 s, a_T = 6 xx 2 = 12 m//s^2`
Centripetal acceleration,
` a_c = v^2/r = (12^2)/(20) = 7.2 m//s^2`
Efdfective acceleration,
` a= sqrt( a_T^2 + a_c^2) = sqrt (12^2 + (7.2)^@) = 14 m//s^@`.