Question

I WANT DERIVATION OF S =UT+1/2 AT^2​

Answer 1

refer to the image attached for your answer

Answer 2

Answer:-

From graph:-

BC = v -u.

OD = AC = t.

OA = u.

# refer the attachment for diagram

As you know area under velocity time graph is equal to displacement.

Now,

$s = area \: of \: triangle + area \: of \: rectangle$

$s = \frac{1}{2} \times ac \times bc + od \times dc$

Substituting the values.

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

as

$(v - u) = at$

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

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

rearranging

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

Hence derived the second equation of motion.