Derive the following equation of motion by the graphical method : v2 = u2 + 2as, where the symbols have their usual meanings.
In the given figure, the distance travelled (s) by a body in time (t) is given by the area of the figure OABC which is a trapezium.
Distance travelld = Area of the trapezium OABC
So, Area of trapezium OABC,
= `"(Sum of parallel sides)(Height)"/2`
Now, (OA + CB) = u + v and (OC) = t.
Putting these values in the above relation, we get:
`s = ((u+v)/2)t` ....(1)
Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
v = u + at
`t = "v-u"/a`
Now, put this value of t in equation (1), we get:
`s = (((u+v)(v-u))/(2a))`
On further simplification,
2as = v2 – u2
Finally the third equation of motion.
`v^2 = u^2 + 2as`
(s) - Displacement
(u) - Initial velocity
(a) - Acceleration
(v) - Final velocity
(t) - Time taken
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