Question

Prove : s=ut +1/2 at square , where s is the distance

Answer 1

S = ut 1 / 2 at²

Here ,

»S = distance

»u = initial velocity

»v = final velocity

»a = acceleration

»t = seconds

As per in the attachment your equation has been proved , there are some special cases also ,

SPECIAL CASES

(i) If initially the body is at rest then , u= 0

→ S= 1 / 2 at²

(ii) a = -g , when body rises up vertically , then

→S = ut- 1 / 2 gt²

(iii) acceleration when body is freely falling ,

→S = ut + 1 / 2 gt²

Answer 2

$\bold{\underline{Answer\: \colon}}$

$\bold{ To \: Prove \: \colon \: s \: = \: ut + \frac{1}{2} at^{2} }$

$\bold{\large{Solution\: \colon }}$

1) v = ( u + at )

Let

s = distance (m)

u = initial time (m\sec)

t = time taken (sec)

a = acceleration ( m\sec²)

Using graphical method we finding the derivation of second formula of motion

Distance travelled = Area of trapezium ABCE


$\bold{\large{S \: = \: \frac{1}{2} (AB+CE)\times AE}}$


$\bold{\large{\implies \frac{1}{2}[u + u + at]\times t }}$

......{from equation 1}

$\bold{\large{\implies [\frac{1}{2}(2u + at )] \times t }}$


$\bold{\large{\implies \frac{2ut}{2} + \frac{1}{2} at^{2}}}$


$\bold{\large{\implies ut \: + \: \frac{1}{2} at^{2}}}$


$\bold{\large{Thanks}}$