# How does velocity vary with DISTANCE travelled (in a straight line motion) if $a<0$, $a=0$ and $a>0$? [closed]

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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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 $u^2$.

• 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 – Mew Leenutaphong Apr 13 '17 at 7:35

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.

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.