# A runner against the wind

We have a runner that is running against the wind.

The wind speed is $$10 m/s$$. The runner is using the same power when he is running against the wind and also when there is no wind, but we know, that when he is running against the wind, he needs 2 times more time than when there is no wind.

So the question is with what speed is he running?

For this problem we can assume that the runner is only affected by the wind resistance which is proportional to the square of the runner's speed. The wind resistance on the runner is thus: $$F_u = \frac{1}{2} Cu*\sigma*S*v^2 = K *v^2 \quad\text{"this should be known as quadratic law"}$$

So what we have done in lectures to solve this problem is to first see what are the forces if there is no wind: $$P = F_{wind}*v$$ and if there is wind : $$P' = F_{wind}'*v' = K ( v' + v_0)^2 * v$$ and $$P =Kv^3$$ And then we just say $$P = P'$$ and solve the equation for $$v$$. However my question is why did we pick $$(v' + v_0)^2$$ instead of just $$(v')^2$$. Can somebody please explain the process for a problem like this ?