Solve Maximum Velocity: Determine a, x, k, v for Particle

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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.

Homework Equations





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?
 
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write:

<br /> v\frac{dv}{dx}=\frac{k}{(x+4)^{2}}<br />

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.
 
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?
 
What do you get for the integral?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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