Question

Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line​

Answer 1

Explanation:

so ,we know that a=v-u/t

so,

v-u=at or

v=u+at

Answer 2

Answer:

        See outline, beginning of particle's movement from A with speed 'u' and 'a' accelerating steadily following up on particle is a. after time t, particle arrive at the point B where speed of particle is v.  

From meaning of acceleration, speed, time graphs slope gives acceleration,  

So acceleration, $a = \frac{(v - u)}{t}$

Or then again, at = v - u ⇒[ v = u + at ].....(1)  

Once more, we know, distance is equivalent to slope under the speed time graph.  

Slope of AOEB = distance, s $= \frac{1}{2} [AO + BE ] \times OE$

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

Then again, s $= \frac{1}{2} (u + u + at) \times t$ [from eq. (1) ]  

or then again, $s = \frac {1}{2} (2u + at) \times t$

$[ s = ut + \frac{1}{2} at^2 ]$......(2)  

Figuring out condition (1),  

$v^2 = (u + at)^2 = u^2 + a^2t^2 + 2uat \\= u^2 + 2a(\frac{1}{2} at^2 + ut)$

$= u^2 + 2a(ut + \frac{1}{2} at^2) \\= u^2 + 2as$ [ from eq . (2) ]

⇒$[v^2 = u^2 + 2as ]$.......(3)  

The equations of motions are derived.