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

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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

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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