Khimraj
Last Activity: 7 Years ago
Let mass of particle is m.
x = at2 – bt3
particle will be at x= 0 at time
0 =t2(a-bt)
t = 0, a/b sec
velocity of paerticle
vx= dv/dx = 2at -3bt2
particle will turn when velocity of particle will be zero at a point another then starting point
0 = t(2a-3bt)
t = 2a/3b
accelaration of particle
ax= dvx/dt = 2a – 6bt
F
x when particle turn = m*(a
x at t = 2a/3b) = m*(2a – 4a) = –2ma

F
x when particle again turn up at initial point = m*(a
x at t = 2a/3b) = m*(2a – 6a) = -4ma

Hope it clears.