Question

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derive second law of motion S=ut+1/2at^.... please ❤​

Answer 1

Answer:

Derive second law of motion:

Explanation:

So far we have the equations: v = u + at and s = ½ (u+v) t

The second can be useful for finding displacement. However, if we don't know the final velocity, it's useless.

We could first work out v from "v = u + at". But let's be radical and substitute "v = u + at" into the second equation!

s = ½ (u + v) × t

s = ½ (u + u + at) × t

s = ut + ½ at²

A third equation, useful for finding displacement when you know the initial velocity, time and acceleration.

#BAL

Answer 2

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$\huge\mathcal\pink{Añswer:-}$

☘ Correct Question:-

⛄ Derive the second equation of motion.

☘ Derivation of the Second equation of Motion --Position-time relation [ Graphical method] :-

⛄ s = ut + 1/2 at²

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⛄ Here,

OA = u = initial velocity.

OE = v = final velocity.

OC = t = time taken.

Now,

S = Area under graph AB.

=> s = Area under trapezium OABC.

or

s= Area of ∆ABD + area of the rectangle OADC.

=> s = 1/2 × b × h + l × b.

Now,

On substituting the values from the graph, we get,

s = 1/2 × AD × BD + OC × OA

=> s= 1/2 × AD × (BC - DC) + OC × OA

=> s = 1/2 × t × (v - u) + t × u

=> s = 1/2 × (v- u) t + ut .......→ (i)

From 1st equation of the motion, we know that,

v = u + at

=> at = v - u .....→ (ii)

Now,

On substituting the equation (ii) in equation (i), we get,

s = 1/2 × (v-u)t + ut

=> s = 1/2 × (at)t + ut

=> s = 1/2 × at² + ut

Now, on reordering the terms, we get,

☞ s = ut + 1/2 at²

[2nd Equation of motion]

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