Question

Proove the third equation of motion by mathematical formula

Answer 1

Consider a body of mass “m” having initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. This is the first equation of motion. ... This is the third equation of motion.

Answer 2

TO PROVE :-

• Third equation of motion by mathematical formula.

PROOF :-

From the velocity - time graph shown in the attachment, Distance (s) covered by the object in time (t), moving under uniform acceleration (a) is given by the area enclosed with in the trapezium OABC. From the graph that is,

$\\ \\ : \implies \displaystyle \sf \: Distance = Area \: of \: trapezium \: OABC.$

$: \implies \displaystyle \sf \:s = \dfrac{1}{2} \: \bigg(OA + BC \bigg) \times OC$

$: \implies \displaystyle \sf \:s =\dfrac{\bigg(OA + BC \bigg) \times OC }{2}$

Now , substitute OA as (u) , BC as (v) and OC = t.

$\\ \\ : \implies \displaystyle \sf \:s = \dfrac{ \bigg(u + v \bigg) \times t}{2} \: \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \: Equation \ 1 \bigg \rgroup$

As we know that,

$\\ \\ \dashrightarrow \displaystyle \sf a \: = \dfrac{\bigg(v - u \bigg)}{t}$

$\dashrightarrow \displaystyle \sf \: t = \dfrac{\bigg(v - u \bigg)}{a} \: \: \: \: \: \: \: \: \: \: \bigg \lgroup Equation \ 2\bigg \rgroup$

using Equation 1 and Equation 2,

$\\ \\ \dashrightarrow \displaystyle \sf \: s = \dfrac { \bigg(v + u \bigg) \bigg(v - u \bigg)}{2a}$

$\dashrightarrow \displaystyle \sf \: s = \frac{v ^{2} - u ^{2} }{2a}$

$\dashrightarrow\underline{ \boxed{ \displaystyle \sf \: v ^{2} - u ^{2} = 2as}} \: \: \: \: \: \: \: \bigg \lgroup \displaystyle \sf 3 \: Equation\bigg \rgroup$