Question

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

- +6√2 m/s
- −6√2 m/s
- 72 m/s
- 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)

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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