I have used dimensional analysis to show that Energy needed to overcome drag at constant speed is $E = C \times v^2 \times A \times p$ where C is a constant $v$ is the constant speed, $A$ is surface area and $p$ is air density. However I am struggling to show that doubling the energy expended in overcoming drag only improves your time by $30$%, which is what the question asked us to show.
I have tried:
Multiplying the E by 2 means that we can halve the drag force, then rewriting v as d/t...I get stuck here and do not know how to proceed. Maybe I'm missing something really simple :/
Thanks a lot :)
Doubling $E$ in the formula $E=Cv^2Ap$, where $C,A,p$ remain constant, amounts to multiplying speed $v$ by $\sqrt{2}$. This will result in the time-to-destination divided by $\sqrt{2}$. Since $1/\sqrt{2}\approx 0.7$, the new time to destination is about 70% of the old one; this is the gain of 30% to which the problem referred.