Question

# For a particle moving along x - axis, the acceleration a of the particle in terms of its x - coordinate is given by a=âˆ’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 mThe 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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