Question : Let a=2x; initially, the particle is at x=2m and is moving with 3 m/s. Then find out v at x=5m.
Doubt by Vanshika
Solution :
a=2x (Given)
Initially when t=0 sec
then
x = 2 m
v = 3 m/s
We know,
a = dv/dt
a = (dv/dt)×(dx/dx)
a = (dv/dx)×(dx/dt)
a = (dv/dx)×(v)
adx = vdv
vdv = adx
Integrating both sides
a = (dv/dt)×(dx/dx)
a = (dv/dx)×(dx/dt)
a = (dv/dx)×(v)
adx = vdv
vdv = adx
Integrating both sides
3∫vvdv = 2∫52xdx
[v²/2]3v = 2[x²/2]25 = [x²]25
[v²/2-(3)²/2]=[5²-2²]
v²/2-9/2=25-4
[v²/2-(3)²/2]=[5²-2²]
v²/2-9/2=25-4
v²/2=21+4.5
v²/2=25.5
v²=25.5×2
v²=51
v=√51 = 7.14 m/s
v²/2=25.5
v²=25.5×2
v²=51
v=√51 = 7.14 m/s