Question : The displacement of a particle is zero at t=0 and it is x at t=t. It starts moving in the positive x direction with a velocity which varies as v=k√x, where k is a constant. Show that the velocity is directly proportional to time.
Doubt by Kartik and Krish
Solution :
At t=0, x=0
At t=t, x=x
At t=0, x=0
At t=t, x=x
Here limit are giving for time and displacement only, this is why we will reduce the given relation (v=k√x) in terms of time and displacement before integrating.
v=k√x
dx/dt = k√x
dx/√x=kdt
dx/(x)1/2 = kdt
x-1/2 dx=kdt
Integrating both sides within proper limit.
dx/dt = k√x
dx/√x=kdt
dx/(x)1/2 = kdt
x-1/2 dx=kdt
Integrating both sides within proper limit.
Squaring both sides
Now differentiating both sides w.r.t time.
Clearly v∝t
Hence, the velocity is directly proportional to time.