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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.
Additional information
The equations of motion are simple equations that describe the state of motion of a point object, provided the acceleration is constant throughout the motion. The capacities are characterized in Euclidean space in old-style mechanics, yet are supplanted by bent spaces in relativity.
There are two descriptions of motion: Kinematics and Dynamics. Kinematics deals with motion where the force is not taken into account. Dynamics considers force and energy.
Now, coming to Kinematics, the equations of motion are:
$ \bullet $ ${v^2} - {u^2} = 2as$
$ \bullet $ $s = ut + \dfrac{1}{2}a{t^2}$
$ \bullet $ $v = u + at$
Note Motion is a movement with velocity and acceleration. In material science, movement is an adjustment in the position of an article over the long haul. Movement is depicted regarding relocation, separation, speed, quickening, time, and speed.
Speed, being a scalar, is the rate at which an article covers separation concerning time. The normal speed is the separation as for time (a scalar amount) proportion. Speed is oblivious of bearing.
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.
Additional information
The equations of motion are simple equations that describe the state of motion of a point object, provided the acceleration is constant throughout the motion. The capacities are characterized in Euclidean space in old-style mechanics, yet are supplanted by bent spaces in relativity.
There are two descriptions of motion: Kinematics and Dynamics. Kinematics deals with motion where the force is not taken into account. Dynamics considers force and energy.
Now, coming to Kinematics, the equations of motion are:
$ \bullet $ ${v^2} - {u^2} = 2as$
$ \bullet $ $s = ut + \dfrac{1}{2}a{t^2}$
$ \bullet $ $v = u + at$
Note Motion is a movement with velocity and acceleration. In material science, movement is an adjustment in the position of an article over the long haul. Movement is depicted regarding relocation, separation, speed, quickening, time, and speed.
Speed, being a scalar, is the rate at which an article covers separation concerning time. The normal speed is the separation as for time (a scalar amount) proportion. Speed is oblivious of bearing.
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