Extraneous Negative and positive velocities

Question

Consider the following question:

A rock falls from a 500m high cliff and is accelerated downwards at $10~ \text{ms}^{-2}$. What is the velocity at which it hits the ground if we ignore air resistance friction etc.?

Using $v_f^2 - v_i^2 = 2ad$, we have $v_f^2 - 0^2 = 2(-10)(-500)$. Then, we have $\pm \sqrt{10000} = \pm 100 \text{ms}^{-1}$. Intuitively one would think that only the negative answer is a real answer and the positive answer is extraneous. Perhaps it is the case. Can someone please explain why $+100~\text{ms}^{-1}$ is extraneous (if it is) because from a mathematical standpoint, it does not seem to be the case.

Answer 1

This has to do with what happens if you run time in reverse. That is we transform $t\rightarrow -t$. Let's see how that changes quantities. $v$ goes as $x/t$, so inserting $-t$, we see $v$ must become negative, so $v\rightarrow -v$. Acceleration, $a$ goes as $x/t^2$, so putting in $-t$, we see acceleration is unchanged, $a\rightarrow a$.

Now if we insert these into your equation, $v_f^2 - v_i^2 = 2ad$, we get $$(-v_f)^2 - (-v_i)^2 = 2ad,$$ but if we square the bracketed terms, we get back $$v_f^2 - v_i^2 = 2ad.$$

That means that equation describes the same situation with time going forward, and backwards. It's what we would call a symmetry of the problem. Like if you rotate a circle, the circle is unchanged, if we change $t\rightarrow -t$, nothing is changed in the equation, this means the equation must be able to give you the answer for forward time, and for reverse time. If you look at the solutions, you see we have $v_f$, and $-v_f$. These correspond to time running forwards, and time running backwards, just as we worked out earlier, $v\rightarrow -v$ under this transformation.

In Newtonian mechanics, the direction of time is arbitrary, so you need to then choose which answer corresponds to your direction of time

Answer 2

Suppose we have two cars. Car A starts at rest and accelerates at 10 ms⁻², and has a displacement of +500m. Car B also starts at rest but accelerates at -10 ms⁻², and has a displacement of -500m. When you plug in the numbers for Car A, you get 2ad = 10000, and when you plug in the numbers for Car B, you also get 10000. The formula has no way of distinguishing those two circumstances, so it gives the answer to both.

If you look at the derivation of the formula, it works by taking the displacement, which is vt, and multiplying it by acceleration, which is v/t. This cancels out t, giving you a formula that you can use even if you don't know t. But it also means that a negative velocity will give both a negative displacement and a negative acceleration. The two negatives cancel out, making negative velocities indistinguishable from positive.

Note that this formula is related to the formula for kinetic energy; multiply both sides by m/2, and you get mv²/2 = mad. Since F= ma, and Fd = work, we have that mv²/2 = work. But work is a scalar, not a vector, so it can only tell you what the magnitude of the velocity is. It can't tell you what direction something is going; the work done to change an object's velocity from v to 0 is the same as the work to change it from -v to 0.

So, no, the positive solution isn't mathematically extraneous; the formula doesn't distinguish between the two solutions. v= +100 and v = -100 are both consistent with ad = 10000. It's only by looking at other facts that you can determine the sign of the velocity.