Question

12.
Prove
that V = u + at, s = ut + 1 at& V° = v*+ 2 as.​

Answer 1

Answer:

Ok wait Didi let me solve this

Answer 2

$\fcolorbox{red}{cyan}{Answer:-}$

Equation i)

$\bf\fbox\color{red}{v = u+at}$

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Proof :-

We know that :-

$a = \frac{v-u}{t} \\ => at = v-u \\ => \boxed{v = u+at}$

Hence Proved

Equation ii)

$\bf\color{blue}ut + \frac{1}{2}at²$

• u = Initial Velocity
• t = time taken
• a = acceleration
• s = Distance Travelled

Proof :-

We know that :-

$Average Velocity=\frac{Initial\:Velocity+Final\:Velocity}{2}$

Also :-

$Distance \: Travelled = Average Velocity \times t \\ => s = \frac{v-u}{2} \times t$

From First Equation :-

v = u + at

$s= \frac{u+u+at \times t}{2} \\ => s = \frac{2u+at \times t}{2} \\ => \frac{2ut+at²}{2} \\ =>\boxed{ s = ut + \frac{1}{2}at²}$

Hence Proved

Third Equation Of Motion

$\bf\fbox\color{orange}{v² = u² + 2as}$

• u = Initial Velocity
• v = Final Velocity
• a = Acceleration
• s = Distamce Travelled

Proof :-

From Second Equation :-

$s = ut + \frac{1}{2}at²$

And from First Equation :-

$v = u + at \\ => at = v - u \\ => t = \frac{ v - u}{a}$

Lets Put This Value of 't' in equation 2 :-

$s = \frac{u(v-u)}{a} + \frac{1}{2}a(\frac{v-u}{a})² </p><p> \\ => s = \frac{uv-u²}{a} + \frac{a(v²+u²-2uv}{2a²} \\ => s = \frac{uv-u²}{a} + \frac{v²+u²-2uv}{2a} \\ => s = \frac{2uv-2u²+v²+u² - 2uv}{2a} \\ => 2as = v² - u² \\ => \boxed{v² = u² + 2as}$

If it helped them follow