# Maximum velocity

Marioqwe

## Homework Statement

A particle starts from rest at the origin and is given an acceleration a=k/(x+4)^2, where a and x are expressed in m/s2 and m, respectively, and k is a constant. Knowing that the velocity of the particle is 4 m/s when x=8m, determine (a) the value of k, (b) the position of the particle when v = 4.5 m/s, (c) the maximum velocity of the particle.

## The Attempt at a Solution

I calculated part (a) and (b) easily. However, I am having some troubles with part (c).
The maximum velocity occurs when the acceleration is equal to zero because there is either a minimum or maximum right? So I set

0 = k/(x+4)^2

which cannot be.
What am I doing wrong?

Homework Helper
write:

$$v\frac{dv}{dx}=\frac{k}{(x+4)^{2}}$$

Integrate to get v as a function of x and then write v=dx/dt so find x as a function of t then differentiate again and examine the limit as t tends to infinity.

Marioqwe
I am getting a very ugly integral. But you said to get v as function of x right?
Then

v = dx/dt ---> dt = (1/v)dx

and integrate the above to get the position x as a function of time? And then take the limit of that function as t goes to infinity? Why would that give me the maximum velocity? Is it because the graph of x(t) has a vertical asymptote?

Homework Helper
What do you get for the integral?