Question

Derive:
S= ut + at^2

Answer 1

We need to do derivation of 2nd Equation Of Kinematics : $\large\sf{s = ut + \dfrac{1}{2}a {t}^{2} }$

Now, Solution :-

We know that,

Average Velocity = $\large\sf{ \dfrac{u + v}{2} }$

And,

Distance = Average Velocity × Time.

Distance = $\large\sf{ \dfrac{u + v}{2} \times t }$

And, From the First Equation,

Final Velocity = u + at.

Now, Substitute the values.

$\dashrightarrow\sf {s = \dfrac{u + u + at}{2} \times t}$

$\dashrightarrow\sf{s = \dfrac{2u + at}{2} \times t}$

$\dashrightarrow\sf{s = \dfrac{\cancel{2}ut}{\cancel{2}} } + \dfrac{ {at}^{2} }{2}$

$\boxed{\dashrightarrow\bf\red{s = ut + \dfrac{ {at}^{2} }{2} }}$

Hence, Derived.

Answer 2

Derivation of Second Equation of Motion by Algebraic Method

Velocity is defined as the rate of change of displacement. This is mathematically represented as:

$\boxed{ \large{velocity = \: \frac{displacement}{time}}}$

Rearranging, we get

$\boxed{ \large{displacement = velocity \: \times time}}$

If the velocity is not constant then in the above equation we can use average velocity in the place of velocity and rewrite the equation as follows:

$\boxed{ \large{displacement \: = \frac{initial \: velocity + final \: velocity}{2} \times time}}$

Substituting the above equations with the notations used in the derivation of the first equation of motion, we get

s=u+v2×t

From the first equation of motion, we know that v = u + at. Putting this value of v in the above equation, we get

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

$s = \frac{u(u + at))}{2} \times t \\ s = \frac{2u + at}{2} \times t$

On further simplification, the equation becomes:

$\boxed{ \huge{ \orange{⛄ : s = ut + \frac{1}{2} {at}^{2} }}}$