Question

Derive graphically the following equations of motion
1. v=u+at​

Answer 1

We are asked to derive v = u + at that is first equation of motion named velocity time relation.

Explanation:

Considering a open object that moves under uniform acceleration where u is not equal to zero from the graph we are able to see that the initial velocity of the object that is at point A and then it increased to final values that is at point B in the time, the velocity change at the uniform rate acceleration. In the graph the perpendicular line BC and BE are drawn from point B on the time and validity axis respectively.

According to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

~ Now let us derive velocity-time relation. Let's do it.

Firstly we can write BC as BD + DC. Now as BD have the velocity position and DC have the time position. Henceforth, we already know that

$\tt \Rightarrow Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{v-u}{t} \\ \\ \tt \Rightarrow a \: = \dfrac{BD}{t} \\ \\ \tt \Rightarrow at \: = BD$

As we write BD at the place of v-u henceforth,

$\tt \Rightarrow v - u \: = at \\ \\ \tt \Rightarrow v = \: u \: + at \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$