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

  • #1
Heisenberg7
101
18
Homework Statement
A runner accelerates from rest through a measured track distance in time with acceleration (constant). If runner was to increase his acceleration by differential amount , what is the change in the time required for the run?
Relevant Equations
According to the problem statement: Now, the distance covered is given by, Differentiating, 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 ), 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 ), right?
 
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  • #2
Heisenberg7 said:
Differentiating,
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.
 
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  • #3
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 instead of . 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?
 
  • #4
Heisenberg7 said:
Oops, my bad. It should indeed be instead of .
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.
 
  • #5
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
 
  • #6
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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  • #7



Is it OK?
 
  • #8
anuttarasammyak said:



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