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For a particle moving along x - axis, the acceleration a of the particle in terms of its x- coordinate x is given by a=9x, where x is in meters and a is in m/s2. Take acceleration, velocity and displacement in positive x - direction as positive. The initial velocity of particle at x=0 is u=+6 m/s. The velocity of particle at x=2 m will be
  1. +62 m/s
  2. 62 m/s
  3. 72 m/s
  4. 0


Solution

The correct option is D 0
Given a=9x
We know that, a=dvdt=vdvdx
vdvdx=9x
v dv=9x dx
On integrating both sides, we get
v22=9x22+C
At x=0,v=u=+6 m/s
C=18
v22=9x22+18
At x=2 m,
v22=9(2)22+18
v=0 at x=2 m
The answer is (d)

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