Derivation of Equations of Motion with the help of Graphs

Derivation of Equations of Motion with the help of Graphs

Derivation of Equations of motion with the help of graphs

(a) Graph in case of the equation, v = v0 + at

In this equation, there are two variables— one is t and the other one is v. A graph is drawn taking t on the X-axis and v on the Y-axis [Fig]. In the figure, normal PN is drawn from point P to Y-axis. Let consider that the final velocity at time t = v = OY. Now OY = OA + AY i.e., v = v0 + at.

Here, slope, a = PM/AM = AY/AM = AY/t

The equation v = v0 + at is obtained from the graph.

In case of motion of a body from rest, v0 = 0, a = constant.

So, v = 0 + constant x t

so, v ∞ t

That means, velocity is proportional to time.

(b) Graph for the equation, s = v0t + ½ at2

In figure, graph of ‘v’ versus ‘t’ is a straight line.

In the figure PN ┴ OX; AM ┴ PN.

Let initial velocity = v0, uniform acceleration = a, ON = t and distance travelled in time t = s.

Now, s = Area of OAPN

= area of the rectangle OAMN + area of the triangle AMP

= OA x ON + ½ AM PM

= v0t + ½ AM PM

Again, slope ‘a’ PM/AM = PM/t

so, PM = at

and, s = v0t + ½ t.at = v0t + ½ at2

So, the equation can be presented in graph.

In case of a body moving with uniform acceleration from rest, then,

s = 0 x t + ½ constant x t2

or, s = constant x t2

or, s ∞ t2

That means, displacement is proportional to time.

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