# Third Equation of Motion

Let a body start with an initial velocity 'u' and attain a velocity 'v' in 't' seconds. Let the body move with uniform acceleration 'a' and let 's' be the distance travelled during the time of 't' seconds.

Average velocity =

Distance travelled = average velocity Ã— time

s = t

**\** v + u = â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

We know that v = u + at

v - u = at â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)

Multiplying equations (1) and (2) we get

(v + u) (v - u) = Ã— at

v^{2} - u^{2} = 2as â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

v^{2} = u^{2} + 2as

These equations completely describe the motion of a body under uniform acceleration. These equations are known as the equations of motion.

Equations of Motion by Graphical Method

The above three equations can be derived by graphical method.

Consider the velocity-time graph of an object moving under uniform acceleration.

To derive v = u + at (Velocity - Time relation)

In the above figure, â€˜uâ€™ is not equal to zero. The initial velocity â€˜uâ€™ of the object is at point A and increases to â€˜vâ€™, at point B, at a uniform rate in time â€˜tâ€™ sec. The acceleration â€˜aâ€™ is uniform. In the graph, OA represents the initial velocity u, BC represents the final velocity v, and OC represents time t. To complete the figure, draw perpendiculars CB to OC from C, BE from B to OE and draw AD parallel to OC. Now from the graph we see that,

BC = BD + DC = BD + OA.

But we know that BC = v and OA = u

Substituting, we get v = BD + u or BD = v â€“ uâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..

(1)The acceleration â€˜aâ€™ of the object is given by a BD / AD = BD / OC

But OC = t

Substituting, we get a = BD / t or

BD = atâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(2)

From equations 1 & 2, we get **v = u + at**

To derive s = ut + Â½ at^{2} (Position-Time relation)

Suppose the object travels a distance â€˜sâ€™ in time t sec. In the above figure. the distance travelled by the object is given by the area enclosed within OABC under the velocity-time graph AB.

Therefore distance travelled â€˜sâ€™ = area of OABC

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

= OA x OC + Â½ (AD x BD)

= u x t + Â½ (t x at)

Therefore, s = ut + Â½ at^{2}

To derive v^{2} â€“ u^{2} = 2as (Position - Velocity relation)

We have just seen that the distance travelled by a uniformly accelerated object, travelling in a straight line, is equal to or is given by the area of the space under the velocity-time graph (the area of the trapezium OABC).

That is, the distance travelled â€˜sâ€™ = area of the trapezium OABC

= (sum of parallel sides) x height / 2

= (OA + CB) x OC / 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. 1

But OA + CB = u + v and OC = t.

Substituting these values in equation. 1, we get

Distance travelled 's' = (u + v) x t / 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. 2

We know from the first equation of motion that t = v â€“ u / a. Substituting for t in equation 2, we get

s = (u + v) x (v â€“ u) / 2a

Therefore, 2as = v^{2} â€“ u^{2} or v^{2} = u^{2} + 2as