\begin{align} v&=u+at\\ s&=ut+\frac{1}{2}at^2\\ v^2&=u^2+2as\\ s&=\frac{(v+u)t}{2} \end{align} I have just started with learning acceleration in school and I don't really understand why these equations are the way they are. I know how to apply them and use them but not why they exist.

  • 4
    $\begingroup$ Welcome to stackexchange! Here it is preferred to type your equations in 'mathjax' as opposed to posting pictures. Take a look at the edit I just made. $\endgroup$ Nov 1 '20 at 9:52
  • $\begingroup$ Not really related to the question, but I think that it is very important that you want to understand why these equations work instead of blindly applying them. $\endgroup$
    – Jonas
    Nov 1 '20 at 11:00
  • $\begingroup$ @randomlad do you want proof for those equations ? $\endgroup$
    – Ankit
    Nov 1 '20 at 11:53

For constant acceleration, $a$ is defined as the change in velocity per unit time, $$a = \frac{v-u}{t}.$$ Rearrange to get your first equation.

Distance travelled is average velocity times time. $$ s = \frac{v + u}2 t.$$ That is your last equation. Multiply these two equations together,

$$a s = \frac{v-u}{t} \frac{v + u}2 t = \frac{v^2-u^2}2,$$ and rearrange to get your third equation.

Substitute your first equation into your last equation, $$ s= \frac{ ((u+at) + u)t}2 = \frac{ 2ut+at^2} 2 ,$$ and simplify to get your second equation.

(Note: There are really only two independent equations here. The definition of average acceleration, and the definition average velocity, applied to the case of constant acceleration. The rest is just algebraic manipulation to produce four (or sometimes five) equations in a form useful to apply to problems).

  • $\begingroup$ how did you derive or assumed that the average velocity is $\frac{v+u}{2}$ ? $\endgroup$
    – Ankit
    Nov 1 '20 at 11:51
  • $\begingroup$ @Ankit, we have uniform acceleration, so the average velocity is half way between initial and final velocity. One way to see this is to plot a velocity time graph. The graph is a straight line. $\endgroup$ Nov 1 '20 at 12:22

You know how velocity is defined as the derivative of position? $$v(t)=\frac{ds}{dt}\tag{1}$$ Acceleration is defined the same way as the derivative of velocity $$a(t)=\frac{dv}{dt}\tag{2}$$ In physics you often know the acceleration and you want to know the position or velocity as a function of time. Why? Because acceleration is related to forces by $F=ma$. If you've not heard of this equation you will probably encounter it soon. For example gravity on earth is described by $F=mg$ with $g$ a constant. Using this you can solve for acceleration to get $$ma=mg\\ a=g\tag{3}$$ In other words the acceleration is just a constant. Mathematically you can say this as $a(t)=a$. Usually it is pretty hard to find the position/velocity when you know the acceleration but when the acceleration is constant it is doable. This is one of the main reasons why constant acceleration is so important. It is easy. If you use $a(t)=a$ in equation (2) you get \begin{align} \frac{dv}{dt}&=a\\ \int\frac{dv}{dt}dt&=\int a\,dt\\ v(t)&=a\cdot t+u\tag{4} \end{align} We integrated both sides over $t$. The integral of the derivative of a function gives you the function back. Integrating gives you an integration constant on one side. I called this $u$. For $t=0$ you get $v(t)=u$ so $u$ is just the velocity at time zero.

We can do this again to solve for position \begin{align} \frac{ds}{dt}&=v(t)\\ \int\frac{ds}{dt}dt&=\int (a\cdot t+u)\,dt\\ s(t)&=\frac{1}{2}a\cdot t^2+u\cdot t+s_0\tag{5} \end{align} Here $s_0$ is another integration constant that represents the position at time zero. We could keep it but it's not really interesting since the starting position is arbitrary. So we can set $s_0=0$. This gives us two of your equations $$v=u+at\\ s=ut+\frac 1 2 at^2$$ The other two equations are just clever ways of combining these two.

If you have any more questions let me know.


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