Question

speed of a particle moving on a circular path of radius 2m varying with time as v=2t^2 m/s net acceleration of the particle at t=2 s is​

Answer 1

Answer:

$8m/s^{2}$

Explanation:

The velocity is given by the expression $v(t) = 2t^2$

Acceleration is defined as the time derivative of velocity

a = dv/dt

here, v = 2t^2 m/s

therefore a = $d(2t^2)/dt$

= 2*2*t  (using the derivative rule of dx^n/dx = nx^(n-1)

= 4*t

given t = 2s,

acceleration after 2 seconds = 4*2 = 8m/s^2

Answer 2

Answer:

hope you UNDERSTOOD this answer