COMEDK2021PhysicsMotion in One Dimension
The displacement x of a particle varies with time t, x=a e^ -p t +b e^ q t , where a, b, p and q are positive constant. The velocity of the particle will
Options
- Ago on increasing forever
- Bbe independent of p and q
- Cdrop to zero when p=q
- Dgo on decreasing with time
Correct answer
A. go on increasing forever
Step-by-step solution
The displacement of the particle is given by x = a e^ -pt + b e^ qt , where a, b, p, q > 0 . The velocity v is the time derivative of displacement x : v = dx dt = d dt (a e^ -pt + b e^ qt ) v = -ap e^ -pt + bq e^ qt To determine the behavior of velocity with time, we find the acceleration a_ acc by differentiating v with respect to t : a_ acc = dv dt = d dt (-ap e^ -pt + bq e^ qt ) a_ acc = ap^2 e^ -pt + bq^2 e^ qt Since a, b, p, q > 0 , the terms ap^2 e^ -pt and bq^2 e^ qt are always positive for all t 0 . Therefo