# 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 nearlya)14         b)13       c)12    d)7.2             ans:

509 Points
12 years ago

s = t3 + 5

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

so

v = 3t2

aT= 6t

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

net accleration = (aT2 + ac2)1/2  ...........1

ac = v2/r

at t=2 sec  ,    v=12m/s

so ac = (12)2/r =144/20 = 7.2m/s

now at t=2sec

aT = 12m/s2

putting aT &  ac in eq 1

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

Shubahm Bhattacharya
16 Points
12 years ago

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