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Question: Derive the first equation of motion by graphical method....

Derive the first equation of motion by graphical method.

Explanation

Solution

Hint So to derive the first equation of motion we can derive it by using the graph made for the velocity-time relation and it will be for the moving object which having uuas the initial velocity, vv as a final velocity, and aaas an acceleration of the body.

Complete Step By Step Solution

So from the above graph, we see that,
From AAto BBin the time duration tt at a uniform rate, there is a change in the velocity of the body. Here, at the time tt the total time will be OCOC and the final velocity will be BCBC.
From BB to OCOC a perpendicular will be drawn and similarly from AAto DD parallel line will be drawn and one another there is the line which is perpendicular drawn from BB to OEOE and this line is represented in the form of dotted lines.
So, from the graph we get,
Initial velocity, u=OAu = OA
Final velocity, v=BCv = BC
And as we know that, from the graph
BC=BD+DCBC = BD + DC
Therefore, we can say that
BC=BD+DCBC = BD + DC
And also since, DC=OADC = OA
Therefore,
BC=BD+OABC = BD + OA
So from the above, we can write it as
V=BD+u\Rightarrow V = BD + u , sinceOA=uOA = u.
So let the above equation be 1st{1^{st}}the equation.
Now we see that the acceleration aawill be equal to the slope of the velocity-time graph.
Therefore,
a=a = Slope of line ABAB
And also
a=BDAD\Rightarrow a = \dfrac{{BD}}{{AD}}
Since we see from the graph that theAD=AC=tAD = AC = t, therefore the above equation will look like
BD=atBD = at, let make this equation the second equation.
So by combining the two-equation we have got, we will get
v=u+atv = u + at
The above is the first equation of motion.

Note So we see the derivation of the first equation of motion which is derived by the use of the graphical method. There are two other methods with which it can be derived, the first one is by using the algebraic method and the other one is by using the calculus method.