Question

Derive the following

1. v = u + at
2. s = ut + 1/2 at²
3. v² = u² + 2as

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Answer 1

Explanation:

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Answer 2

To do :-

Derivation of the following

1. v = u + at

2. s = ut + 1/2 at²

3. v² = u² + 2as

Let's start

Let a body

moves with an initial velocity ‘u’ final velocity ‘v’ and

acceleration ‘a’ during time ‘t’. Suppose its motion is represented

by velocity time graph. It is clear from the graph that the body has

an initial velocity ‘u’ at A and changes at a uniform rate from A to

A to B in time t seconds. The final velocity of the body becomes v and

v and it is equal to BC in the graph.

Now Initial velocity (u) = OA = CD

Final velocity (v) = BC

Time (t) =AD = OC

Change in velocity = v – u

= BC–CD = BD

Since Acceleration

$a = \frac{v - u}{t}$

$=>at = v - u$

$=>v = u + at --- \sf \: (First \: Equation \: of \: motion)$

When the velocity of the body remains constant say ‘u’ throughout the motion.

Then distance covered = Area of rectangle OADC

=> S = OA x AD [ Length × Breadth ]

=> S = u x t

But if velocity goes on increasing constantly with time, the distance covered = area of trapezium OABD

$S = ( OA × AD ) + \frac{1}{2} \: AD × BD$

$\sf \: S = u×t + \frac{1}{2} \: t \times (v - u)$

$S=ut + \frac{1}{2} t \times at$

$\sf \: S=ut + \frac{1}{2} a {t}^{2} - - - (Second \: Equation \: of \: motion)$

Now, distance travelled by body = area of trapezium OABD

$S= \frac{1}{2} (OA + BC ) × OC$

$S= \frac{1}{2} (u+ v ) × t$

As we know ,

$v - u = at$

$\frac{v - u}{a} = t$

So we can write

$S= \frac{1}{2} (OA + BC ) ×( \frac{v - u}{a} )$

$S= \frac{(v + u)(v - u)}{2a}$. As OA = u and BC = v

$2as = {v}^{2} - {u}^{2}$

${v}^{2} - {u}^{2} = 2as- - - (Third \: Equation \: of \: motion)$