Speed skater slowing with drag force proportional to v²

[gaborfk]

Consider an object which the net force is a resistive force proportional to the square of its speed. For example: assume that the resistive force acting on a speed skater is F=-k*m*V^2, where k is a constant and m is the skater's mass. The skater crosses the finish line of a straight-line race with speed V(i) and the slows down by coasting on his skates. Show that his speed at time "t", any time after the finish line is equal to Vf=Vi/(1+Vi*k*t).

Any suggestions? Thank you in advance!

 

[arildno]

It's a separable differential equation.
Go ahead!

 

[gaborfk]

You mean mdv/dt= -mkV^2 cancel m, gives dv/dt= -kv^2, which in turn yields dv=-kv^2dt? But how does the Vi gets introduced?

Thank you

 

[arildno]

You get:
$$\frac{dv}{v^{2}}=-kdt$$
Right?
Vi enters as the initial condition in that v(0)=Vi

 

[gaborfk]

Thank you. I know I have to integrate both sides. Left side from Vi to V(t) and the right side from t to 0. I get -kt for the right side. But I am having trouble with the left.

 

[arildno]

Post what you've gotten so far! (The equation)

 

[gaborfk]

I got it. Thank you. 1/V(t)-1/Vi=-kt. Than simplify...

 

[arildno]

You have a sign flaw; you should simplify:
$$-\frac{1}{V(t)}+\frac{1}{V_{i}}=-kt$$