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

AI Thread Summary
The discussion focuses on determining the change in time required when acceleration increases by a differential amount. The initial calculations show that time T is related to acceleration a and final velocity v_f, with the relationship T = v_f/a. The participants clarify that the differentiation should be with respect to T rather than t, confirming that an increase in acceleration (or power) results in a higher final velocity. The final consensus is that the differentiation should focus on T, a, and da, while L can be eliminated from the equations for simplicity. The conversation concludes with agreement on the correct approach to the problem.
Heisenberg7
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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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Heisenberg7 said:
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.
 
haruspex said:
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?
 
Heisenberg7 said:
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?
Heisenberg7 said:
the object would end up with a higher velocity at the end (after the increase of 𝑑𝑎), right?
Yes.
Heisenberg7 said:
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.
 
haruspex said:
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:
1721558363378.png
 
Heisenberg7 said:
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.
 
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L=\frac{a}{2}T^2
T=\sqrt{\frac{2L}{a}}
dT=-\sqrt{\frac{L}{2}}a^{-\frac{3}{2}}da
Is it OK?
 
anuttarasammyak said:
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##).
 
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