Question

Derive the third equation of the motion-

Answer 1

Hint: The third equation of motion is the relationship between final velocity ‘v’, initial velocity ‘u’ , acceleration ‘and distance travelled ‘s’. We will start by considering that a body is moving with uniform acceleration. We will plot the v-t (velocity-time) graph for the body and using the graph we will reach the equation.

Complete step-by-step solution:
Let us first know a few terms:
1) initial velocity is the velocity with which the body first starts moving.
2) final velocity- velocity the body attains after completing its motion.
Let us start by considering a body, moving on a plane with uniform acceleration ’a’. let ‘u’ be its initial velocity ’v’ be its final velocity, ‘t’ be time travelled and ‘s’ be distance travelled. And ‘a’ is the acceleration .
We will plot a( v-t) graph to derive this equation:
Look at the following graph,

Here: OA=CD=u
BC=v
AND, OC=AD=t
As total height is v, then the height BD=BC-BD.
That is BD=v-u
We know that the slope of the v-t graph gives acceleration.
Hence, $$a={\displaystyle \frac{BD}{AD}}$$
$$a={\displaystyle \frac{BC-CD}{OC}}$$
$$a={\displaystyle \frac{v-u}{t}}$$
$$t={\displaystyle \frac{v-u}{a}}-----------\left(1\right)$$
Distance travelled by the body in time t= area enclosed by v-t graph.
We see that there is a trapezium formed with height ‘t’ and sum of parallel sides as ‘u’ and ‘v’ .
Hence the distance travelled by the body =area of the trapezium in the graph.
X=area of trapezium
$$x={\displaystyle \frac{1}{2}}\times OC\times \left(OA+BC\right)$$
$$x={\displaystyle \frac{1}{2}}\times t\times \left(u+v\right)$$
Now substituting value of t from equation 1, we get
$$x={\displaystyle \frac{1}{2}}\times \left({\displaystyle \frac{v-u}{a}}\right)\times \left(u+v\right)$$
$$x={\displaystyle \frac{1}{2a}}\times \left(v-u\right)\times \left(u+v\right)$$
$$x={\displaystyle \frac{1}{2a}}\times \left(v^2-u^2\right)$$
Or, $$x2a=\left(v^2-u^2\right)$$
Substituting x=s we get :
$$2as=\left(v^2-u^2\right)$$
$$\therefore v^2=u^2+2as$$
Hence derived.

Note: The slope under the (v-t) graph gives acceleration.
The area under the curve of a (v-t) graph gives the displacement.
Remember to write BD=v-u , otherwise the whole derivation could go wrong.
Also all the equations of motion can be applied only when the acceleration is constant.