- #1

Heisenberg7

- 99

- 18

- Homework Statement
- A runner accelerates from rest through a measured track distance in time ##T## with acceleration ##a## (constant). If runner was to increase his acceleration by differential amount ##da##, what is the change in the time required for the run?

- Relevant Equations
- ##a = \frac{dv}{dt}##

According to the problem statement: $$a = \frac{dv}{dt} = const \implies dt = \frac{dv}{a} \implies \int_{0}^{T} \,dt = \frac{1}{a} \int_{0}^{v_f} \,dv \implies T = \frac{v_f}{a}$$ Now, the distance covered is given by, $$L = \int_{0}^{T} v \,dt \implies L = \frac{1}{a} \int_{0}^{v_f} v \,dv \implies L = \frac{v_f^2}{2a} \implies L = \frac{a}{2} \frac{v_f^2}{a^2} \implies L = \frac{aT^2}{2} \implies 2L = aT^2 = const$$ Differentiating, $$0 = \frac{da}{dT}T^2 + 2aT \implies T^2 da+ 2 a T dT = 0 \implies -T da = 2 a dT \implies dT = - \frac{T}{2a} da$$ Now, I would like to know if this is correct and obviously, the object would end up with a higher velocity at the end (after the increase of ##da##), right? The actual problem was about power, but I reformulated it. Now, for the sake of it, let's say that it was power instead of acceleration. It would also mean a higher velocity at the end (after the increase of ##dP##), right?

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