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shreyas Grade: 12
A particle moves along the X-axis according to the law x=at2-bt3,where a and b are positive constants.At moment t=0,force acting on the particle is equal to F0.Find the value of force Fx at points of turn and at moment when particle again turn up at point x=0.
one month ago

Answers : (1)

1078 Points
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
Fx when particle turn = m*(ax at t = 2a/3b) = m*(2a – 4a) = –2ma \hat{i}
Fx when particle again turn up at initial point = m*(ax at t = 2a/3b) = m*(2a – 6a) = -4ma\hat{i}
Hope it clears.
one month ago
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