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## Homework Statement

A particle is moving along a straight line such that its acceleration is defined as a=(-2v)m/s^2.

If v=20 m/s when s=0 and t=0, determine the particle's velocity as a function of position and the distance the particle moves before it stops.

The answer is v=(20-2s)m/s ; s=10m

## Homework Equations

a=[tex]\frac{dv}{dt}[/tex]

a=[tex]\frac{dv}{ds}[/tex]*[tex]\frac{ds}{dt}[/tex] which becomes...

ads=vdv

## The Attempt at a Solution

given: a=(-2v) m/s

v=20 m/s

t=0

s=0

solution attempt:

a=[tex]\frac{dv}{dt}[/tex]

-2v*dt=dv

dt=-[tex]\frac{dv}{-2v}[/tex]

[tex]\int dt[/tex] = [tex]\int\frac{dv}{-2v}[/tex]

That's as far as I can get. If I evaluate the left hand side from 0 to t, I get t.

If I evaluate the left, I get messed up. If I pull out the -[tex]\frac{1}{2}[/tex],

I am left with [tex]\frac{1}{v}[/tex]. I think the integral of that is ln v.

I need some direction on this one please.