- #1

raphile

- 23

- 0

## Homework Statement

A particle moves in a straight line with displacement s from some fixed point on the line and velocity v.

If v is proportional to sqrt(s), find the behaviour of the acceleration with s; and if v=0 when t=0, find how v depends on time.

## The Attempt at a Solution

I put v = k*sqrt(s) and used a = v*dv/ds = k*sqrt(s)*(1/2)k*s^(-1/2) = (1/2)k^2. So it seems that the s disappears, and therefore the acceleration is constant. Is it correct to say this?

For the second part, I put dv/dt=(1/2)k^2 (the acceleration found above) and solved to give v(t) = (1/2)(k^2)t + c, and c=0 because v=0 when t=0. Hence my conclusion is that v varies with t, i.e. v and t are proportional. Is this right? Thanks.