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For a particle moving along x - axis, the acceleration a of the particle in terms of its x - coordinate 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 maximum distance of the particle along the positive x-direction from origin will be

A
1 m
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B
2 m
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C
3 m
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D
4 m
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Solution

The correct option is B 2 m
The velocity initially is in positive direction and acceleration is in negative direction, so the velocity will keep on decreasing till the velocity becomes 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 .....(i)
In order to find the maximum distance travelled by the particle along positive x-direction, we need to put v=0 in eqn. (i)
0=9x22+18
On solving, we get
x=2 m

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