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This question is similar, but I am a bit out of practice on mathematics and struggle to apply the steps to my problem.
How can the $\triangle$$t$ be calculated for a given $\triangle$$v$ when acceleration changes as a function of $v$?

Given:

$a(v) =- \frac{8.69+0.0252v^{2}}{200}$

My current goal is to find the time required to change from $v=70$ to $v=60$. (velocity: $\frac{km}{hr}$, acceleration: $\frac{km}{hr/s}$)

My approach:

I have tried to use the linked question to rewrite $a(v)$ as $v(t)$. Then I would set the equation equal to 70 and then 60 respectively and solve for $\triangle$$t$.

I do not understand where $U$ comes from in the linked problem.

Semi-related question, are the steps above necessary for expressing $a(v)$ in terms $a(t)$?

Cardinal
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    If the acceleration is always positive, why would the velocity ever decrease? – Vishu May 18 '20 at 19:38
  • $U$ “comes from” the equation that starts with “$U(q)=$“. That’s the definition of $U$. But note that the linked question is asking about distance, not time. Time is easier than distance and there is not so much need to define a function such as $U$. – David K May 18 '20 at 20:13

1 Answers1

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$$ \frac{dv}{dt}=a+b v^2\implies \frac{dv}{a+bv^2}=dt\implies \Delta t=\frac1{\sqrt{ab}}\left[\arctan\sqrt{\frac ab}v\right]_{v_1}^{v_2}. $$

user
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  • I have integrated as you have listed and all of those values are off by a factor of 3.6, which is the conversion factor between km/hr and m/s. I assume the math is right because of the predictability of the results but I do not know where the 3.6 conversion factor came from or went. I will convert a(v) to be a function of m/s instead of km/hr to see if your equation aligns. – Cardinal May 18 '20 at 20:14
  • When the conversion factor of 3.6 is applied the equation works perfectly. Now I need to figure out how to use this information for my larger problem. Thank you. – Cardinal May 18 '20 at 20:21