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Question

Instantaneous velocity of an object (starting from origin) varies with time as v=αβt2(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,
  1. x=αt13βt3
  2. x=αt12βt2
  3. α=βt
  4. Maximum positive displacement from origin =2α3αβ


Solution

The correct options are
A x=αt13βt3
D Maximum positive displacement from origin =2α3αβ
Acceleration of the object is given as
a=dvdt=ddt(αβt2)=2βt
For displacement we use
v=dxdt=αβt2  ......(i)
Integrating with proper limits, we get
x0dx=t0(αβt2)
Or x=αt13βt3  ......(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=αβt2
t=αβ  ......(iii)
Putting this in equ. (ii), we get 
xmax=ααβ13β(αβ)3/2
Maximum positive displacement from the origin
xmax=23ααβ

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