  Question

# Instantaneous velocity of an object (starting from origin) varies with time as v=α−βt2(where α and β are positive constants). If x and a are the position and acceleration of the object as a function of time. Then,x=αt−13βt3x=αt−12βt2α=−βtMaximum positive displacement from origin =2α3√αβ

Solution

## The correct options are A x=αt−13βt3 D Maximum positive displacement from origin =2α3√αβAcceleration of the object is given as a=dvdt=ddt(α−βt2)=−2βt For displacement we use v=dxdt=α−βt2  ......(i) Integrating with proper limits, we get x∫0dx=t∫0(α−βt2) Or x=αt−13βt3  ......(ii) At t=0, v=α Velocity is positive in the beginning and acceleration is negative, so the velocity will keep on decreasing until it becomes 0 and then it reverses its direction. Displacement in the positive direction will be maximum when v becomes zero. 0=α−βt2 ⇒t=√αβ  ......(iii) Putting this in equ. (ii), we get  xmax=α√αβ−13β(αβ)3/2 ⇒Maximum positive displacement from the origin ⇒xmax=23α√αβ  Suggest corrections   