Question

Proove that 2nd equation of motion

Answer 1

Answer:-

Let the distance travelled by the object be “s”.

We know that,

→Distance = Average velocity × time

→Average velocity = (u+v)/2

Then s=(u+v)/2×t

According to first equation if motion,

v = u + at

substituting the value of v in above equation

s = (u+u+at)/2 x t

→s = (2u+at)/2 × t

→s = (2ut+at²)/2

→s = 2ut/2 + at²/2

→s=ut+1/2at²

Hence derived.

Answer 2

Explanation:

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,

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

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

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