Question

deduce the following equation of motion ( a) s= ut+1/2 at² (b) v² =u² + 2 as

Answer 1

$s = avg.v \times t \\ s = \frac{v + u}{2} \times t \\ s = \frac{u + at + u}{2} \times t \\ s = \frac{2u + at}{2} \times t \\ s = \frac{2ut + a {t}^{2} }{2} \\ s = ut + \frac{1}{2}a {t}^{2} ...... \:\\ \\ s = avg.v \times t \\ s = \frac{v + u}{2} \times \frac{v - u}{a} \\ s = \frac{(v + u)(v - u)}{2a} \\ 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as...... \: \: \:$

Answer 2

Answer:

I) Consider a body which starts with initial velocity u,and due to uniform acceleration a, it's final velocity becomes v after time't.There , it's average velocity is given by

$average \: velocity \ = \frac{initial \: velocity \: + \: final \: velocity}{2}$

$= \frac{u + v}{2}$

Therefore, distance covered by the body in time t is given by distance s= Average velocity× time

or

$s = \frac{u + v}{2 \: } \: \times 2 \\ or \: s = \frac{u + (u + at)}{2} \times t \\ s = \frac{2ut +at^{2} }{2 } \\ or \: s \: = ut + \frac{1}{2} {at}^{2}$