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# Derivation of the second equation of motion is:$\left( a \right)$ $d\theta = wd2t$$\left( b \right) d\theta = wdt$$\left( c \right)$ $d\theta = wd3t$$\left( d \right)$ $d\theta = wd{t^2}$

Last updated date: 10th Sep 2024
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Hint We know that the product of the velocity and the time will be equal to the displacement of the body. If velocity is not constant that is we can say that velocity keeps on increasing or decreasing. By using the formula for the displacement $s = ut + \dfrac{1}{2}a{t^2}$we will be able to find the relation between them.
Formula used:
Displacement,
$s = ut + \dfrac{1}{2}a{t^2}$
Here,
$s$, will be the displacement
$u$, will be the initial velocity
$a$, will be the acceleration
$t$, will be the temperature

Complete Step By Step Solution So as know displacement will be equal to the
Displacement, $s = ut + \dfrac{1}{2}a{t^2}$
By using the above equation and converting the units as-
$s$ as$\theta$, $u$as ${w_0}$and $a$as $\alpha$
We get the equation as,
$\theta = {w_0}t + \dfrac{1}{2}\alpha {t^2}$
Now differentiating the above equation with respect to time,
We get
$\Rightarrow \dfrac{{d\theta }}{{dt}} = w$
Now by taking the $dt$ right side of the equation,
We get
$\therefore d\theta = wdt$
Therefore the option $B$ will be the correct option.

$\bullet$ ${v^2} - {u^2} = 2as$
$\bullet$ $s = ut + \dfrac{1}{2}a{t^2}$
$\bullet$ $v = u + at$