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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 ?

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  1. When there is no wind, the speed of the runner relatively to air is the same as the speed of the runner relatively to the ground.

  2. When there is some wind, the speed of the air relatively to the ground and the speed of the runner relatively to the ground add up, if he runs against the wind.

It is easy to demonstrate ad absurdum: imagine that, in another experiment, the runner runs in the opposite direction, and moreover that he runs at the speed of the wind, then the speed of the runner relatively to the air/wind is naught, and therefore he should feel no friction due to the air.

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