Question

can somebody explain me this equation
v^2-u^2=2as

Answer 1

So, see the graph and then understand .The motion third equation

To Derive

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

Distance travelled, s = Area of trapezium OABC

s = (sum of parallel sides) x Height /2

s = (OA+CB) x OC / 2

Now OA + CB = u + v and OC = t

s = (u + v) x t / 2

Now eliminate t

we know that a = v - u / t

at = v - u

t = v-u /a

Now put the value of t

s = (u + v) (v - u) / 2a

Now rearrange,

s = (v + u) (v-u) /2a

we know that

$(a + b)(a - b) = a ^{2} - b ^{2}$

So,

s = v^2 - u ^2 /2a

2as = v ^2 - u^2

So we derive the third equation of motion

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

Hope it's help you

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