'SUVAT' Equations Proofs

There are 5 'SUVAT' equations: 


\[②:s=\frac{1}{2} (u+v)t\]


\[④:s=ut+\frac{1}{2} at^2\]

\[⑤:s=vt- \frac{1}{2} at^2\]


Proof using a velocity-time graph

Consider the following general velocity-time graph for constant acceleration:


  1. The gradient of the line is given by `\frac{v-u}{t}`, and by definition, the gradient is the acceleration (since both are the change in velocity over time). Therefore: \[a=\frac{v-u}{t}\] \[at=v-u\] \[\therefore v=u+at\] 
  2. By definition, the area under the graph is displacement (since displacement = velocity \(\times\) time). Therefore (using the area of a trapezium): \[s=\left(\frac{u+v}{2}\right) t\] \[\therefore s=\frac{1}{2} (u+v)t\] 
  3. Rearranging \(①\) gives: \[t=\frac{v-u}{a}\] Substituting this into \(②\): \[s=\left(\frac{u+v}{2}\right)\left(\frac{v-u}{a}\right)\] \[2as=(u+v)(v-u)\] \[2as=v^2-u^2\] \[\therefore v^2=u^2+2as\] 
  4. Substituting \(①\) into \(②\) gives: \[s=\left(\frac{u+u+at}{2}\right) t\] \[s=\left( \frac{2u}{2}+\frac{at}{2} \right) t\] \[\therefore s=ut+\frac{1}{2} at^2\] 
  5. Substituting \(①\) into \(④\) gives: \[s=(v-at)t+\frac{1}{2}at^2\] \[s=vt-at^2+\frac{1}{2} at^2\] \[\therefore s=vt-\frac{1}{2} at^2\]  


Proof using integration

We can derive equations \(①\) and \(④\) using integration.


  1. Acceleration is the rate of change of velocity with respect to time, so: \[v=\int a \hspace{1 mm} dt\] \[=at+c\] when \(t=0\) ,  \(v=u\) so substituting this in gives: \[u=0+c\] \[\therefore c=u\] \[\therefore v=u+at\]
  2. Velocitiy is the rate of change of displacement with respect to time, so: \[s=\int v \hspace{1mm} dt\] \[=\int (u+at)\hspace{1mm}dt\] \[=ut+\frac{1}{2} at^2+c\] \(s=0\) when \(t=0\) meaning that \(c=0\) \[\therefore s=ut+\frac{1}{2} at^2\]
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