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I realize this might be an absurd questions to ask directly here, but I cannot seem to wrap my head around it. There is no information about the time, and I have not studied any equations when the acceleration is varying linearly. This questions was given by our professor without any explanations.

I want to know if it is possible to solve this. If so how? It seems to be a simple problem and yet I cant see a simple solution. I have tried drawing Acceleration time, and A-V graphs but dont seem to have any results. I also tried integrating but couldn't find a solution. Any method to proceed or a solution would be very helpful. Thanks!

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  • $\begingroup$ is the question more suited to math stack exchange? $\endgroup$
    – jim
    Commented Oct 30, 2021 at 8:28
  • $\begingroup$ @jim I think the OPs main issue is that he doesn't know, we can write $\frac{dv}{dt}=v\frac{dv}{dx}$ $\endgroup$
    – Nakshatra Gangopadhay
    Commented Oct 30, 2021 at 8:50

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You just need to find a suitable function that relates position and acceleration.

Then using Newton's second law, we know :

$$\frac{dp}{dt}=m\frac{dv}{dt}=m\frac{dv}{dx}\frac{dx}{dt}=mv\frac{dv}{dx}=ma$$

Thus, $$vdv=adx=a(x)dx$$

If you know $a$ as a function of $x$, which you should be able to derive, then you can integrate from $v_i$ to $v_f$ on LHS, and from initial position to final position on RHS.

As you can see, we don't need an idea of time anymore. Just need to find acceleration as a function of position that satisfies that at $40mm$ the acceleration is $2$ and at $120mm$ it is $4$

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  • $\begingroup$ How do I find a function that relates Acceleration and position that fits the criteria? the best i can do is $a = 50 x$ but its still not right. $\endgroup$
    – Krishnaraj PT
    Commented Nov 4, 2021 at 10:24
  • $\begingroup$ @KrishnarajPT use the $x$ coordinate to denote position and $y$ to denote acceleration. At $x=40$, $a=y=2$. You can write this as $(40,2)$ on a graph. Similarly the other point is $(120,4)$. The function that you are looking for is the straight line connecting these two points. You know how to find the equation of a straight line between two points. $$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$$ $\endgroup$
    – Nakshatra Gangopadhay
    Commented Nov 4, 2021 at 14:27
  • $\begingroup$ Yess, I got it. The answer also makes sense. I am not good in Calculus, and so it took a while, but it makes sense now. Thanks for the time and effort $\endgroup$
    – Krishnaraj PT
    Commented Nov 5, 2021 at 12:25

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