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NB: This is about velocity against distance not against time and $a$=acceleration

I know for a fact that for $a<0$, the gradient is negative and keeps decreasing, $a=0$ gradient 0. What happens when $a>0$ and can you explain this in terms of equations or any method for when $a<0$ and $a>0$ why it produce this result. I tried using $v^2 = u^2 + 2as$ but it gets confusing for me.

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closed as off-topic by ZeroTheHero, John Rennie, Jon Custer, Yashas, sammy gerbil Apr 14 '17 at 12:32

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If you dealing with constant acceleration then $v^2=u^2+2as$ can be written as $v^2=2a\cdot s + u^2$ which is of the form of the general equation of a straight line $y=mx+c$ where if $v^2$ is plotted against $s$ the gradient is $2a$ and the intercept on the $v^2$ axis is $u^2$.

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  • $\begingroup$ Oh Ic thats quite helpful. I was having trouble doing it without demos thats why its hard for me to visualize. Now that I know that the answer is so simple I want to punch myself. Thanks $\endgroup$ – Mew Leenutaphong Apr 13 '17 at 7:35
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The relationship between a, v (velocity), and d (distance travelled) is in time; you travel a distance d = d(t), where t is the amount of time it takes you to travel that distance.

In other words, how much the velocity varies with distance is a function of how it would vary over that time.

The equation $v^2 = u^2 + 2as$ takes advantage of the relationship between distance and time to relate $a$, $v$, and $d$ (in this equation $s$) directly. It says that if you start at velocity = $u$ and move a distance of $s$ with an acceleration of $a$ you will end up at a velocity $v$. Set $u$ equal to zero (we can do this if we are assuming there are no velocity dependent forces acting on the object). The equation is now $v^2 = 2as$, which is equivalent to $$v = \sqrt{2as}$$

That is a direct relationship expressing how velocity varies over distance given an a. When a > 0, velocity will increase.

When a = 0, velocity will not change. When a < 0, velocity will increase in the opposite direction - i.e., decrease in the "positive" direction. (Note, since we can not take the sqrt of a negative number, we can simply switch which way we consider the positive direction, get the resultant velocity, and 'map it back' to the original positive direction by slapping a negative on it.)

This is just one way to think of it. I am sure there are more mathematically satisfying answers than changing coordinate systems to deal with negative acceleration, but the result will be the same.

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I'm not sure exactly what you're asking for, but here is one way to look at this. The relationship is definitely not obvious and it seems like I can't avoid talking about time at some point.

Let's flip the question (assuming $u=0$ and $a>0$): when the object has gotten to velocity $v$, how far has it gone? Well the time it took is $t=v/a$, and the average velocity from start to finish is $v/2$ because acceleration was constant.

Therefore, the object has traveled a distance of $s = (v/2)t = (v/2)(v/a) = v^{2}/2a$ and so $v = \sqrt{2as}$.

Similarly, you can think about the general problem to get the general formula. Let's say you have constant acceleration, initial velocity $u$, and final velocity $v$. The average speed is $(v+u)/2$, and the total time to get from velocity $u$ to velocity $v$ is $t=(v-u)/a$, assuming constant acceleration. Hence, the distance traveled is $s = \left(\frac{v+u}{2}\right)t = \left(\frac{v+u}{2}\right)\left(\frac{v-u}{a}\right)$. Finally, this gives $2as = v^{2}-u^{2}$.

All of this reasoning actually works for $a<0$ just as well. The only other case here is $a=0$, which should be simple.

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