Displacement (x) of a particle is related to time (t) as x = at + bt2– ct3where a, b and are constants of motion. The velocity of the particle when its acceleration is zero is given by(A)a + b2/c(B)a + b2/2c(C)a + b2/3c(D) a + b2/4cPlease help as I am facing problem with this question.......

velocity=dx/dt=a+2bt-3ct²

acceleration =dv/dt=2b-6ct,

when a=0,t=b/3c

velocity=a+2bt-3ct²,where t=b/3c

Hi Akshay,

The velocity is rate of change of displacement.

And Acceleration is rate of change of velocity.

So v = dx/dt = a+2bt-3ct².

And acc = dv/dt = 2b-6ct.

So when acc = 0, we have 2b=6ct or t=b/3c.

At this time, the velocity v = a+2b(b/3c)-3c(b/3c)² = a + b²/3c.

And hence option (C).

All the best.

Regards,

Ashwin (IIT Madras).

x=at+bt²-ct³,

then, v=a+2bt-3ct²(at given situation)

        a=2b-6ct=0 

      2b=6ct .......(1)

      v= a+(6ct)t - 3 ct²

       =   a+6ct²-3ct²

       =    a+3ct²  =a+ 3*b/3t*t² 

       =    a+ 3ct * t

       = a+ bt

       = a+ 2b²/6c

       =  a+ b²/3c

     HENCE OPTION (C) IS THE CORRECT ANSWER!!!  IF U LIKE IT PLZ APPROVE IT!!!