Find the change in the time required, if acceleration increases by the differential amount 'da'

[Heisenberg7]

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?

 

[haruspex]

Differentiating, $$0 = \frac{da}{dt}t^2 + 2at \implies t^2 da+ 2 a t dt = 0 \implies -t da = 2 a dt \implies dt = - \frac{t}{2a} da$$

Why are you differentiating wrt t? The question is how a change in a affects T (not t). So you need to find T as a function of a.

 

[Heisenberg7]

Why are you differentiating wrt t? The question is how a change in a affects T (not t). So you need to find T as a function of a.

Oops, my bad. It should indeed be $T$ instead of $t$. Anyway, could you answer my 2 questions:

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?

 

[haruspex]

Oops, my bad. It should indeed be $T$ instead of $t$.

Ok, but differentiating wrt it is wrong either way. What should you have differentiated and wrt what?

the object would end up with a higher velocity at the end (after the increase of 𝑑𝑎), right?

Yes.

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?

Yes.

 

[Heisenberg7]

Ok, but differentiating wrt it is wrong either way. What should you have differentiated and wrt what?

Hmm, I'm not sure really. Could you elaborate? I simply followed the blueprint of this solution:

 

[haruspex]

Hmm, I'm not sure really. Could you elaborate? I simply followed the blueprint of this solution:
View attachment 348676

Hmm.. yes, that is valid. Sorry for the noise.
I think I was thrown by the t/T confusion.

 

[anuttarasammyak]

$$L=\frac{a}{2}T^2$$
$$T=\sqrt{\frac{2L}{a}}$$
$$dT=-\sqrt{\frac{L}{2}}a^{-\frac{3}{2}}da$$
Is it OK?

 

[Heisenberg7]

$$L=\frac{a}{2}T^2$$
$$T=\sqrt{\frac{2L}{a}}$$
$$dT=-\sqrt{\frac{L}{2}}a^{-\frac{3}{2}}da$$
Is it OK?

I think so. But I only need it with respect to $T$, $a$ and $da$. $L$ is a surplus (we can get rid of this by just plugging in for $L$).