Question #743f5

Feb 5, 2018

Because they are derived,on the basis of the consideration that acceleration is constant.

Suppose, acceleration = $a$ =constant

So, $\frac{\mathrm{dv}}{\mathrm{dt}} = a$ (as,acceleration = rate of change in velocity)

Or, $\mathrm{dv} = a \mathrm{dt}$

Integrating from $v = u \to v$ and $t = 0 \to t$,
we get,

$v - u = a t$ or, $v = u + a t$

Again,acceleration can be written as, $v \frac{\mathrm{dv}}{\mathrm{dx}} = a$

or, $v \mathrm{dv} = a \mathrm{dx}$

Integrating from, $v = u \to v$ and $s = 0 \to s$ we get,

${v}^{2} - {u}^{2} = 2 a s$

Or, ${v}^{2} = {u}^{2} + 2 a s$

Again, $\frac{\mathrm{dx}}{\mathrm{dt}} = u + a t$ (as, velocity = rate of change in displacement)

So, $\mathrm{dx} = u \mathrm{dt} + a t \mathrm{dt}$

Integrating from, $x = 0 \to x$ and, $t = 0 \to t$,we get,

$x = u t + \frac{1}{2} a {t}^{2}$