When a body starts from rest and starts accelerating in positive direction and moves certain positive distance, its final velocity is given by : $$v^2 = u^2 + 2as$$ Solving it gives two values of $v$, one positive and one negative. But how can a body accelerating in positive direction and starting from rest have a negative direction of its velocity? Am I missing something?
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3$\begingroup$ Its not unusual when you solve a quadratic equation, you get two values. Both of them are there because they satisfy the equation but you have to take care of the problem being solved. Not both of the values will always be useful for solving the question. $\endgroup$– user8718165Commented May 8, 2019 at 14:58
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$\begingroup$ The equations of motion for a body with constant acceleration assume that it always has been accelerating with acceleration $a$. For a given displacement $s$, one solution has $t\ge 0$ and the other has $t \le 0$ (as long as $u^2+2as \ge 0$). Often we are only interested in the motion for $t \ge 0$ so we discard the "mirror image" solution, $\endgroup$– gandalf61Commented May 8, 2019 at 16:42
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3 Answers
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There are two solutions for
$$v^2 = v_i^2 + 2a\Delta s,$$
one positive and one negative. You can give a meaningful interpretation to both solutions. For example, for $v_i = 0 \textrm{ m/s}$, $a = 0.5 \textrm{ m/s}^2$, and $\Delta s = 2.0 \textrm{ m}$, we get
$$v = \pm \sqrt{2} \textrm{ m/s}.$$
The positive solution is easy to interpret. It is the speed of the particle at some time after the starting point. The negative solution can be interpreted in a symmetric way: it is the speed of the particle some time before the starting point.
So just imagine a particle moving in the negative direction while slowing down, coming to rest, and then moving in the positive direction while speeding up, with constant acceleration the entire time.
(Don't get hung up on 'starting from rest'. 'Starting from' just means that your analysis of the particle starts with initial values of position, velocity, etc. It does not need to imply that the particle had those same values for all times prior to your analysis of it.)
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$\begingroup$ I felt myself getting convinced by your reasoning, but this statement confused me: "So just imagine a particle moving in the negative direction while slowing down, coming to rest, and then moving in the positive direction while speeding up, with constant acceleration the entire time." Where in this is the final velocity negative? $\endgroup$ Commented May 8, 2019 at 15:22
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$\begingroup$ The positive solution is the 'final' velocity. But not final right? We're not assuming the particle instantly stops at that point and remains at rest for all eternity. So if we assume it continues to move at constant $a$, we can calculate its 'post-final' velocity at some later point without worrying about how we've contradicted the word 'final'. In the same way, the negative solution is the 'pre-initial' velocity. My goal was just to show that the negative solution is not entirely meaningless. $\endgroup$ Commented May 8, 2019 at 15:32
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$\begingroup$ Fair enough, you're basically saying if you played the problem in reverse, but didn't stop when $v=0$, then $v$ will be negative and if you let it run into the negative zone for $2m$, then we get a speed of $-\sqrt{2}m/s$. Or you could also see it as from $-\sqrt{2}$ to $0$ is the same as that from $0$ to $\sqrt{2}$ since the gradient is constant and goes through the origin - so long as we take absolute areas. $\endgroup$ Commented May 8, 2019 at 16:28
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$\begingroup$ To clarify further, this is happening because in the equations above you don't have time anywhere so there is no way to distinguish past from future. The way the equation $v^2=u^2+2a\Delta s$ is written, you are simply asking what velocities can be associated when a uniformly accelerating particle is at a certain distance $\Delta s$ from the origin. The answer is indeed both positive and negative velocity depending on whether you evolve into future or into past. $\endgroup$ Commented May 8, 2019 at 17:07
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$\begingroup$ So it's like, it's accelerating in negative direction negatively(retarding), so the acceleration is positive, distance is positive and since moving in negative direction before my analysis, all conditions are met. Am I right? $\endgroup$ Commented May 9, 2019 at 4:18
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Your reasoning is not logically valid.
Your claim is the folowing:
"The answer to the problem is a solution to this equation." $$\implies$$ "Every solution to this equation is an answer to the problem."
I see this as analogous to the following.
Given $x = 5$, we could reason that,
$$x = 5\implies x^2 = 25 \implies x = \pm\sqrt{25}= \pm5$$
But clearly $x \ne-5$, so even though $x = -5$ is a solution the equation $x^2 = 25$, we are wrong to conclude that it is true, given all our information (namely that $x = 5$ !).
My point is that it is often the case that you can reason for the equations to imply that there is additional solutions but, as is the case in maths as well as physics, we often have to return to the original conditions and question whether a solution is valid (e.g. "$f(x)$ is defined for all $x\in \mathbb{R}$" but then we find an solution to the equation $f(x) = 0$ but $x$ is complex - so we discard it).
In your case, a negative velocity would not cause the body to move a positive distance, so it is to be discarded.
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Maths is the language of physics not the other way around!
After this being said, you are right that velocity should have negative value even though a is positive, but ask yourselves does this solution makes practical sense?
Obviously no, so we discard the negative value.
There will be many scenarios where there would solutions that do not make any physical sense, just like 'hjjhk' is a word made of alphabets but does not have any meaning in English.