Question

The position x of a particle varies with time t as x=at^2 - bt^3 , where a and b are constant. What will be the velocity of ground with respect to particle at t=ls?​

Answer 1

Answer:

a/3b

Explanation:

x = at2 − bt3  

Velocity = (dx/dt) = 2at − 6bt

and acceleration = d2x / dt2 = 2a − 6dt

Asseleatin will be zero if  2a − 6bt = 0 ⇒ t = 2a / 6b = a / 3b.