Question

Derive the equation of motion by the calculus method or the graphical method.

Answer 1

Hint
The three equations of motion form the basics of classical mechanics. The equations establish relations between the physical quantities that define the characteristics of motion of a body, such as the acceleration of the body, the displacement and the velocity of the body.
$a=\dfrac{dv}{dt}$ , $v=\dfrac{ds}{dt}$

Complete step by step answer
We know that the acceleration of a boy is the rate of change of its velocity. Mathematically, we can say that
$\begin{align}
  & a=\dfrac{dv}{dt} \\
 & \Rightarrow adt=dv \\
\end{align}$
Integrating both sides of the equation, we get
$\begin{align}
  & \int\limits_{0}^{t}{adt=\int\limits_{u}^{v}{dv}} \\
 & a\int\limits_{0}^{t}{dt=\int\limits_{u}^{v}{dv}} \\
 & \Rightarrow a\left[ t \right]_{0}^{t}=\left[ v \right]_{u}^{v} \\
 & \Rightarrow at=v-u \\
 & \Rightarrow v=u+at \\
\end{align}$
This is the first equation of motion where $v$ is the final velocity of the body, $u$ is the initial velocity of the body and $t$ is the time of motion of the body. To derive the second equation of motion, we will use the fact that the velocity of the body is the rate of change of its displacement, that is
$\begin{align}
  & v=\dfrac{ds}{dt} \\
 & \Rightarrow ds=vdt \\
\end{align}$
We will substitute the value of the final velocity of the body from the first equation of motion
$\begin{align}
  & ds=(u+at)dt \\
 & \Rightarrow ds=udt+atdt \\
\end{align}$
Integrating both sides of the equation, we get
$\begin{align}
  & \int\limits_{0}^{s}{ds}=\int\limits_{0}^{t}{udt}+\int\limits_{0}^{t}{atdt} \\
 & \Rightarrow \left[ s \right]_{0}^{s}=u\left[ t \right]_{0}^{t}+a\left[ \dfrac{{{t}^{2}}}{2} \right]_{0}^{t} \\
 & \Rightarrow s=ut+\dfrac{1}{2}a{{t}^{2}} \\
\end{align}$
This is the second equation of motion.
To derive the third equation of motion, we will use both the expression used for the first and the second laws, that is $a=\dfrac{dv}{dt}$ and $v=\dfrac{ds}{dt}$
We can now say that $a\dfrac{ds}{dt}=v\dfrac{dv}{dt}$
Integrating both sides of the expression, we have
$\begin{align}
  & \int\limits_{0}^{s}{ads=\int\limits_{u}^{v}{vdv}} \\
 & \Rightarrow a\left[ s \right]_{0}^{s}=\dfrac{1}{2}\left[ {{v}^{2}} \right]_{u}^{v} \\
 & \Rightarrow as=\dfrac{{{v}^{2}}-{{u}^{2}}}{2} \\
 & \Rightarrow {{v}^{2}}={{u}^{2}}+2as \\
\end{align}$
This is the third equation of motion.

Note
The equations of motion can also be derived using graphical and algebraic methods. Note that we have not used constants of integration as the limits of integration are already known to us. The calculus method of deriving quantities and equations is most feasible because the description of motion through graphs is not always possible.