Equations of Motion by Graphical Method

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations of motion. There are three such equations. The equations were summarized by Issac Newton. These are –

Here,
u = velocity at start
v = final velocity
t = time period
a = acceleration
s = distance covered by the object

The above three equations can be derived from graphical method.

Uniform and Non-Uniform Motion

(A) Derivation of First Equation of Motion
Suppose an object has uniform acceleration ‘a’ and its starting velocity is ‘u’. When we plot velocity -time graph with time along x-axis, the velocity-time graph comes a straight line as the increase in velocity is uniform across uniform time intervals.  The graph is shown in fig. below. In the fig., the velocity increases by BC during the time interval of AB. The velocity at instant t = 0 is u. After interval t the velocity becomes v. By the definition of acceleration we know
that it is ratio of change in velocity to the time interval. Therefore, in the graph the change
in velocity is BC and the time interval is AB. The measurement of BC along velocity axis is v – u.
The measurement of AB along the time axis is t.

(B) Derivation of Second Equation of Motion

In fig.13, the distance travelled along the straight line in time interval t is the area under the graph
from 0 to t. The area is made up of two parts-
(i) rectangle 0tBA and (ii) triangle ABC.

Area of rectangle otBA = length along time axis x length along velocity axis
= t x u = ut

Area of triangle ABC = 1/2 x BC x AB

Here,
BC = v – u ;
v – u = at     ;
AB = t

therefore, area of triangle ABC = (1/2) at²

The distance traveled ‘s’ = area of rectangle otBA + area of triangle ABC

Uniform and Non-Uniform Motion

(C) Derivation of Third Equation of Motion

In fig.13, the distance traveled along the straight line in time interval t is the area under the graph from 0 to t. The area is of a trapezoid 0tCA.

=     (5+30) x 50 / 2

=      875 m