The three equations of motion v = u + at; s = ut + (1/2) at2 and v2 = u2 + 2as can be derived with the help of graphs as described below.
Derive v = u + at by Graphical Method
The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE
Now, Initial velocity of the body, u | = | OA ------- (1) |
And, Final velocity of the body, v | = | BC -------- (2) |
But from the graph BC | = | BD + DC |
Therefore, v | = | BD + DC -------- (3) |
Again DC | = | OA |
So, v | = | BD + OA |
Now, From equation (1), OA | = | u |
So, v | = | BD + u --------- (4) |
Derive s = ut + (1/2) at2 by Graphical Method
Distance travelled | = | Area of figure OABC |
= | Area of rectangle OADC + Area of triangle ABD | |
We will now find out the area of the rectangle OADC and the area of the triangle ABD. | ||
(i) Area of rectangle OADC | = | OA × OC |
= | u × t | |
= | ut ...... (5) | |
(ii) Area of triangle ABD | = | (1/2) × Area of rectangle AEBD |
= | (1/2) × AD × BD | |
= | (1/2) × t × at (because AD = t and BD = at) | |
= | (1/2) at2 ------ (6) | |
So, Distance travelled, s | = | Area of rectangle OADC + Area of triangle ABD |
Derive v2 = u2 + 2asby Graphical Method