Question

derive equation of motion graphically for a particle having uniform acceleration moving along straight line

answer plzz​

Answer 1

Answer:

Given:

Particle is moving along a straight line with uniform Acceleration .

To find :

Equations of motion using graph.

Calculation:

First let's draw the Acceleration vs Time graph. Since acceleration os constant, it will be straight line parallel to x -axis.

Now, change in velocity is the area under acc-time graph :

$v - u \: = at$

$= > v = u + at$

So 1st equation of motion :

$\boxed{ \red{ \sf{ \huge{ v = u + at}}}}$

Now velocity time graph will be linear :

Displacement will be equal to area under velocity-time graph :

$s = area \: of \: trapezium$

$= > s = \frac{1}{2} (u + v)t$

$= > s = \frac{1}{2} (u + u + at)t$

$= > s = ut + \frac{1}{2} a {t}^{2}$

2nd equation is:

$\boxed{ \huge{ \red{ \sf{s = ut + \frac{1}{2} a {t}^{2} }}}}$

Again ,

$s = \frac{1}{2} (u + v)t$

$= > s = \frac{1}{2} (u + v)( \dfrac{v - u}{a} )$

$= > s = \dfrac{ {v}^{2} - {u}^{2} }{2a}$

$= > {v}^{2} = {u}^{2} + 2as$

So 3rd equation is:

$\boxed{ \red{ \huge{ \sf{{v}^{2} = {u}^{2} + 2as}}}}$

Answer 2

QUESTION :-

derive equation of motion graphically for a particle having uniform acceleration moving along straight line.

SOLUTION :-

=> [Note:- Refer to the attachment ]

=>A particle start to move from A with velocity u and a constant acceleration acting on particle is a after time t, particle reach the point B where velocity of particle is v.

=> According , to the definition of acceleration,

=> a = (v-u) /t

=> .°. at = (v-u)

=> .°. v = u + at --------(1)

=> Now, we know that, displacement is equal to area of under the velocity time graph.

=> .°. So, Area of AOEB = displacement

=> .°. s = ½ ×[AO + BE]× OE

=> s = ½ × ( u + v) × t

=> s = ½ × (u+u+at)× t --------( °.° v = u + at)

=> s = ½ × (2u + at) × t

=> .°. s = ut+½×at²

=> [ Note:- Squaring equ. (1)]

=> then, we get, v² = (u+ at)²

=>v²= u²+ a²t²+2uat

=>v²= u²+2a(½at²+ut)

=>v²= u² + 2a(ut+½at²)

=>v²= u² + 2as ----------------[ °.° s = ut+½at²]

=> .°. v² = u²+2as

ANSWER :-

(1) v = u + at

(2)s = ut+½×at²

(3) v² = u²+2as