- Introduction
- |
- Speed and Velocity
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- acceleration
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- Equations of uniformly accelerated motion
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- Graphical representation of motion
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- Equations of motion by graphical method
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- Uniform circular motion

- We already know about equations of motion when an object moves along straight line with uniform acceleration. We already know how to derive them but these equations can also be derived by graphical method.

- Consider the velocity-time graph of an object that moves under uniform acceleration as shown below in the figure 7.
- From this graph, you can see that initial velocity of the object is \(u\) (at point A) and then it increases to \(v\) (at point B) in time \(t\). The velocity changes at a uniform rate \(a\).
- Again from figure it is clear that time \(t\) is represented by OC , initial velocity \(u\) by OA and final velocity of object after time \(t\) by BC.

- From graph as given in figure 7 it is clear that \(BC = BD + DC = BD + OA\).

So we have

v=BD+u (1) - We should now find out the value of BD. From the velocity-time graph (Fig. 7), the acceleration of the object is given by

which gives, \(BD = at\)

putting this value of BD in equation 1 we get

\(v = u + at\)

which is the equation for velocity time relation.

- Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 7, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
- Thus, the distance s travelled by the object is given by

*s*= area OABC (which is a trapezium)

*s*= area of the rectangle OADC + area of the triangle ABD

So,

\(s = OA \times OC + \frac{1}{2}\left) {AD \times BD} \right)\)

Substituting \(OA = u\), \(OC = AD = t\) and \(BD = at\), we get

\(s = \left( {u \times t} \right) + \frac{1}{2} \times \left( {t \times at} \right)\)

or,

\(s = ut + \frac{1}{2}a{t^2}\)

which is the equation of position time relation

- Again consider graph in figure 7. We know that distance travelled s by a body in time t is given by the area under line AB which is area of trapezium OABC. So we have

Since \(OA+CB=u+v\) and \(OC=t\), we thus have

\(s = \frac{{\left( {u + v} \right)t}}{2}\)

From velocity time relation

\(t = \frac{{v - u}}{a}\)

putting this t in equation for s we get

\(s = \frac{{\left( {u + v} \right)}}{2}\left( {\frac{{v - u}}{a}} \right)\)

or we have

\({v^2} = {u^2} + 2as\)

which is equation for position velocity relation.

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