Deriving Equations of Motion – Types and Importance Points

Deriving Equations of Motion

Introduction

The equation of motion is mostly derived from the definition of the quantity, which describes the motion and the time related to motion.

The equations to be derived are:

v = u + at

s = ut + 1/2at²

v²-u² = 2as

Deriving the Equation in Two Types

1. Algebraic method
2. Graphical method

Deriving the equation v = u + at using the algebraic method:

We know that acceleration is the ratio of change of velocity with time; hence, we have

Deriving the equation v=u+at using the graphical method:

 Important Points

1. In the above graph, the y-axis represents velocity, and the x-axis represents time.
2. We can see initial velocity u on the y-axis and some points on v in the graph.
3. Therefore, the graph represents a v-t graph, and the slope of the chart is acceleration.
4. The slope and the equation are derived as follows:

Deriving the equation s = ut + 1/2 at² using the algebraic method:

we know that displacement = velocity x time

From the first derivation, we know v = u + at

Deriving the equation s = ut + 1/2 at² using graphical method

Important Motion

In the above graph, the total distance traveled in the graph is given by the area of RPQS

The area under RPQS is further divided into the area of triangle RQS and the area of the rectangle RTOS.

Hence, the distance traveled is,

S = area of RQS + area RTOS

S = ½ (RS x SQ) + RT x TO

S=1/2 (t x at) + u x t

S = ut +1/2 at²

Deriving the equation v² – u²  = 2as using the algebraic method:

Deriving the equation v² – u² = 2as using the graphical method:

Important Points

S =1/2(sum of opposite sides) x height

S = (RS + PQ) x TO

The above equation becomes

S = 1/2(u + v) x (v – u)/a

Rearranging the equation, we get

S = 1/2(v + u) x (v – u) x a

S = (v² – u²)/2a

Therefore,

v² – u² = 2as