Derivations for equations of motion

• Equation For Velocity-Time Relation:

Consider the velocity-time graph of an object that moves under uniform acceleration. Initial velocity of the object is u and then it increases to v in time t. The velocity changes at a uniform rate a.

The perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t.

Let us draw AD parallel to OC. From the graph, we observe that BC = BD + DC = BD + OA

Substituting    BC = v and OA = u,

we get v = BD + u                                                                                                      i.

or BD  = v – u                                                                                                                            ii.

From the velocity-time graph,

The acceleration of the object is given by a = Change in velocity/time taken

= BD/AD = BD/OC

Substituting OC = t, we get

a = BD/t

BD = at                                                                                                                          iii.

Using the equations ii. & iii. We get,

v = u + at

• Equation For Position - Time Relation:

Let us consider that the object has travelled a distance s in time t under uniform acceleration a. 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)

= area of the rectangle OADC + area of the triangle ABDM.

= OA × OC + 1/2 (AD × BD)

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

s = u × t + 1/2 (t×at)

or

s = u t + 1/2 a t²

• Equation For Position - Velocity Relation:

From the velocity-time graph, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is,

s = area of the trapezium OABC = (OA + BC) X OC/2

Substituting OA = u, BC = v and OC = t,

We get

s= (u+v) t/2

From the velocity-time relation

We get

t= v - u/a

We have

s = (v + u) x (v – u)/2a

or 2as = v² – u²