# What are the 3 equations of motion?

Solution,

1) First equation of Motion:

V = u + at
Derivation:
Consider a body of mass “m” having initial velocity “u”.Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”.

Now we know that:
Acceleration = change in velocity/Time taken

=> Acceleration = Final velocity-Initial velocity / time taken

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

or v = u + at
This is the first equation of motion.
—————————————-
(2) Second equation of motion:
s = ut + $\frac{1}{2}a{t}^{2}$

Derivation:
Let the distance travelled by the body be “s”.
We know that
Distance = Average velocity X Time

Also, Average velocity =$\frac{v+u}{2}$

.: s(t) =$\left(\frac{u+v}{2}\right)t$ …….eq.(1)

Again we know that:
v = u + at
substituting this value of “v” in eq.(2), we get

$s=\left(\frac{u+u+at}{2}\right)t=\left(\frac{2u+at}{2}\right)t=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}$

This is the 2nd equation of motion.
……………………………………………………………
(3) Third equation of Motion
${v}^{2}={u}^{2}+2as$

Derivation
We know that
V = u + at
=> v-u = at
or t = $\frac{v-u}{a}$ ………..eq.(3)

Also we know that
Distance = average velocity X Time

$s=\left(\frac{v-u}{2}\right)×\left(\frac{v+u}{a}\right)=\frac{{v}^{2}-{u}^{2}}{2a}\phantom{\rule{0ex}{0ex}}2as={v}^{2}-{u}^{2}$

$or\phantom{\rule{0ex}{0ex}}{v}^{2}={u}^{2}+2as$

This is the third equation of motion.

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