A sledge is pushed in a straight line. Assume surface is smooth. When the sledge is x distance away from the start the magnitude of its acceleration is given by $0.08e^{-4x}$ and is going in the same direction of the sledge's motion.

I need to show that $v = 0.2 \sqrt{1-e^{-4x}}$

Now I 'know' that to get $v$ from a you need to integrate a. So I did that and I got: $v=-0.02e^{-4x}$.

Are these two things one and the same? If not, can someone explain why?

  • $\begingroup$ possible duplicate of How to get distance when acceleration is not constant? $\endgroup$ – ja72 Aug 11 '15 at 17:50
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    $\begingroup$ I disagree with the "duplicate" marking. The problem here is that acceleration is given as a function of position, not time. The duped question and answer are only tangentially relevant. $\endgroup$ – Floris Aug 11 '15 at 19:03

To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information.

Since they give you the solution and you just have to prove it is correct, you should start from the velocity information and see if you can get from there to acceleration - noting that

$$a=\frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt}=\frac{dv}{dx}\cdot v$$

Now we know $v(x)$ so we can differentiate that with respect to $x$; then we multiply by $v$ and should get the expression for $a$ that was given.

$$\begin{align}\\ v &= 0.2\sqrt{1-e^{-4x}}\\ a &= 0.2\cdot \frac12 \cdot\left(1-e^{-4x}\right)^{-\frac12}\cdot 4e^{-4x}\cdot 0.2\sqrt{1-e^{-4x}}\\ &= 0.08 e^{-4x}\end{align}$$

Q.E.D. - no integration required.

  • $\begingroup$ Totally a basic math problem. However, you deserve one up-vote :) $\endgroup$ – user36790 Aug 11 '15 at 16:07
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    $\begingroup$ @user36790 - thanks. Sometimes, physics problems reduce to "basic maths". Note the handle of the OP: "MathsNoob2015". So... $\endgroup$ – Floris Aug 11 '15 at 16:08
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    $\begingroup$ Literally, at this time we require some humour! lol:) Actually, the upvote isn't only for answering but also for noticing & caring for a novice question, which you always dilligently do & that I like the most. Keep it up with your charitable work & one vote of mine was, is & always will be reserved for you:) $\endgroup$ – user36790 Aug 11 '15 at 16:13
  • $\begingroup$ Hi Floris. Thanks for your answer. You've gone from the thing I am supposed to prove back to the thing they've given. I'm supposed to start from a and work my way to v. I tried to do that and my answer is v=sqrt(0.04-0.04e^-4x), though I don't think that is the same as what they want. $\endgroup$ – MathsNoob2015 Aug 12 '15 at 16:03
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    $\begingroup$ I think my approach is valid. "Given A, prove that the velocity is X" - if the derivative of X is A, that is a valid proof. You don't have to integrate. Now if they said "find the velocity" you would have no choice but to integrate - but that's not what is asked. By the way, $\sqrt{0.04 - 0.04 e^{-4x}} = \sqrt{0.04}\sqrt{1-e^{-4x}}=0.2\sqrt{1-e^{-4x}}=v$ - so maybe you actually have the solution... $\endgroup$ – Floris Aug 12 '15 at 18:02

How did you integrate acceleration to get velocity? Note that

$\Delta v = \int_{t_i}^{t_f} a(t) dt$

But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as

$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$

You can use this to set up your integral so that you are able to integrate a(x) w.r.t. x. But notice the "v" hanging around. You won't just get v on the other side of the equation.

Next, be careful of initial conditions. You haven't told us what the initial conditions are. But I infer that the sledge must start from rest because it starts at $x=0$ and if you plug $x=0$ into the expression you are supposed to prove you get $v(0) = 0$. So you need to pay attention to the fact that you are doing a definite integral, not simply an antiderivative.

  • $\begingroup$ We both spotted the same issue, at almost exactly the same time. But I think that differentiating the velocity expression is easier than integrating the acceleration (as you say, you have to worry about constants of integration then). $\endgroup$ – Floris Aug 11 '15 at 16:02
  • $\begingroup$ Thanks for the tips. I approached it by doing a=vdv/dx. Eventually I got v=sqrt(0.04-0.04e^-4x). If I plug x=0 into that I get 0, as expected (you were right about your inference). Is that equivalent to what I am supposed to show? The sqrt of 0.04 is 0.2, so I think I follow that... $\endgroup$ – MathsNoob2015 Aug 11 '15 at 18:53
  • $\begingroup$ @gleedadswell, suppose particle starts from x=a and v(a) is not equal to zero, then how to do it. $\endgroup$ – 2.0 Sep 20 '19 at 10:06
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    $\begingroup$ @math Note that when you write a = v (dv/dx) the dv/dx implies that once you carry out the integral (and one additional step to fully isolate v) you will have v as a function of x. But as usual with an integration, you can get a family of functions. So if you know a, and v(a), you can put those in as the lower limits of your integrals and you will pick out the member of the family which satisfies that initial condition. The answer that the OP was trying to prove comes from assuming that the initial x and initial v (the lower limits on the integrals on each side of the eqn) are both zero. $\endgroup$ – gleedadswell Sep 22 '19 at 23:05

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