Question

derive 2nd equation of motion​

Answer 1

DERIVATION OF SECOND EQUATION OF MOTION:

At $t = 0$, initial velocity $u = OA$

At $t = t$, final velocity $v = OC$

The distance $s$ traveled in time $t$ is equal to the area of the trapezium $OABD$.

$s = \frac{1}{2} \times (OA + DB) \times OD$

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

Since $v = u + at$

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

$s = ut + \frac{1}{2} at^{2}$  

Answer 2

An object is moving with uniform acceleration having initial velocity "u" and after time "t" second final velocity is "v". The distance travelled by an object is s in time "t" second.

DERIVATION OF SECOND EQUATION OF MOTION

Here,

Acceleration = Slope of the graph.

= (BD - CD)/AC

= BC /AC

where B is final velocity C is initial velocity and OD is time.

= (v - u)/t

Then acceleration is (v - u)/t i.e.

=> a = (v - u)/t

=> v - u = at

=> v = u + at ........... (i)

Therefore, we have derived first equation of motion.

Now ,

Displacement = Area under graph.

= 1/2 OD (OA + BD)

= 1/2 t ( u + v)

Put value of v by using equation (i)

= 1/2 t ( u + u + at)

= 1/2 t (2u + at)

= ( 1/2t × 2u) + ( 1/2t × at)

= ut + 1/2at²

So,

s = ut + 1/2at².

Therefore second equation of motion derived.