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I was trying to solve a simple problem where it was given that the velocity of a box on a smooth surface is:

$$ v = 0.05s^{3/2} $$

Where s is the number of meters that the box travels from rest. The question was: what is the acceleration of the box when the distance traveled is 10m. Now there are many ways to solve this problem, but I took the following approach:

$$ ds/dt = 0.05s^{3/2} $$

$$ \frac{ds}{0.05s^{3/2}} = dt $$

Integrate from 0 to s and 0 to t for the left and right hand sides, respectively

$$ -40s^{-1/2} = t $$

$$ s^{1/2} = -40/t $$

$$ s(t) = \frac{1600}{t^2} $$

$$ v(t) = \frac{-3200}{t^3} $$

$$ a(t) = \frac{9600}{t^4} $$

Let $T$ be the time to reach $10$ m

$$ 10 = \frac{1600}{T^2} $$

$$ T = \sqrt{160} $$

$$ a(s = 10m) = \frac{9600}{T^4} $$

However, I stopped here because according to s(t), I saw that as time increases, the distance traveled from the origin decreases, which is inconsistent with the intuition that the given velocity equation gives me, which is that distance traveled should increase over time.

Where did my thinking go wrong?

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    $\begingroup$ Have you tried a different approach? The question does not mention time, so there is no need to calculate how acceleration varies with time. $\endgroup$ – sammy gerbil May 3 '16 at 19:55
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Differentiate $v$ with respect to time.
On the left hand side you will get the acceleration $a$.
On the right hand side you will get an expression which includes $s$ and $\frac {ds}{dt}$ which is $v$ which is ?

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The approach really fails, because the proposed motion is kind of broken.

Including the integration constant, the distance traveled as a function of time becomes:

$s(t) = \frac{1600}{(t+C)^2}$

Since we want $s=0$ when $t=0$, the constant C becomes infinity. This means that, the motion can not really start from rest -- it takes infinite time to do so. This is evident from the problem -- at $s=0$, the speed is zero. Not only is the speed zero, the acceleration is also (you will find if you solve the problem) is zero. All higher derivatives are also zero! The damn thing will never move.

But...

The problem can still be solved. Even if it takes infinite time (ugh) we can find what the acceleration is when the distance traveled is a given $s$.

Now, here is the trick you need:

You know $v(s)$. You want $a(s)$.

$a(s) = \frac{dv(s)}{dt}$

$ a(s) = \frac{dv(s)}{ds} \cdot \frac{ds}{dt}$

Proceeding from here leads to the answer, which is not difficult to obtain. That is, you will be able to calculate $a(s=10)$. What you still will not be able to find out is when that happens... Because it never happens.

I would classify this as a non-kosher problem, if this is intended as a physics problem. The motion never starts as described. As such, the "when" of having traveled $10m$ is non-existent. You could not measure this even in principle as an experiment, which takes this into the realm of philosophy rather than physics.

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