a point 'p' moves in countrclock wise direction on a circular path .the movement of 'p' such that it sweeps a length of ''s=t(cube)+5 where 's'in mtrs and 't' in secnds .the radius of path is 20 m . the acceleration of 'p' when t=2 sec is nearly

a)14 b)13 c)12 d)7.2 ans:

 s = t³ + 5

  velocity =dr/dt   &  tengential accletarion (a_T) = dv/dt

so

     v = 3t²

     a_T= 6t

when a particle is in circular motion then , centripital accleration(a_c) acts along center & tangential accleration acts along tangent perpendicular to a_c...

 net accleration = (aT² ₊ a_c²)^1/2  ...........1

a_c = v²/r    

at t=2 sec  ,    v=12m/s

so a_c = (12)²/r =144/20 = 7.2m/s

now at t=2sec

        a_T = 12m/s²

putting a_{T &}  a_c in eq 1

 we get total acc = (12² + 7.2²)^1/2 = 13.99 =14m/s²

14.differentiate s to get v.v square/r is radial acceleration and dv/dt is tangential one.root over the square of both the accelerations give you the answer