# Kinematics problem

1. Aug 30, 2008

### naggy

A particle moves along a straight line with accel. a = 2$$\sqrt{v}$$ where v = $$\frac{dx}{dt}$$ is the velocity. At t = 2s, the particle is located at x= (64/3)m with velocity v = 16m/s. Find the location and acceleration at time t = 3s (hint: integrate)

Now, if the differential equation is dv/dt = 2$$\sqrt{v}$$. I don´t know how to solve such an equatoin (because of the square root). But then I thought v must be constant for the integration to work, but that would mean that the acceleration would be zero and that makes no sense. I need help.

2. Aug 30, 2008

### naggy

never mind, I figured it out as soon as I clicked submit post.

dv over the sqrt(v) = 2dt

No w8, this makes no sense, if you integrate you get 2sqrt(v) = 2t + c

Last edited: Aug 30, 2008
3. Aug 31, 2008

### Nolgosphere

Square both sides of that expression and integrate again. You have enough data to find the constants.

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