Question

Write three equations of uniformly accelerated motion relating the initial velocity $\left( u \right)$, final velocity $\left( v \right)$, acceleration $\left( a \right)$ and displacement $\left( S \right)$. Derive the following equations for a uniformly accelerated motion:
A. $v = u + at$
B. $s = ut + \dfrac{1}{2}a{t^2}$
C. ${v^2} = {u^2} + 2as$

Answer 1

Hint: All the three equations for the uniformly accelerated motion can be derived by using the expressions for velocity, displacement, and acceleration and integrating them. The first equation is just the formula for acceleration rearranged. And the other two can be derived from it.

Complete step by step answer:
It is given in the question, that for a body possessing a uniformly accelerated motion,
The initial velocity is represented by, $u$
The final velocity is represented by, $v$
Acceleration is represented by, $a$
And its displacement is represented by $S$.
Then the three equations of motion can be derived as follows-
Deriving $v = u + at$
We know that the acceleration of a body is given by the amount of change in its velocity at a given time interval-
$a = \dfrac{{dv}}{{dt}}$
Rearranging this,
$ \Rightarrow adt = dv$
If an object changes its velocity from $u$ to $v$ in a given time interval $t$. Then, integrating this equation-
$\int_0^t a dt = \int_u^v {dv} $
$ \Rightarrow a\left[ t \right]_0^t = \left[ v \right]_u^v$
Solving the limits,
$a\left[ {t - 0} \right] = \left[ {v - u} \right]$
$ \Rightarrow at = v - u$
Rearranging,
$v = u + at$……….$(1)$
(B) Deriving $s = ut + \dfrac{1}{2}a{t^2}$
To derive the second equation, we need the help of the first equation,
We know that,
$v = \dfrac{{ds}}{{dt}}$
(Velocity is the derivative of displacement of the body)
Thus we can write,
$ds = vdt$
From equation, $(1)$ we get the expression for the velocity as-
$v = u + at$
Substituting this value, we have-
$ds = \left( {u + at} \right)dt$
It can be written in the integral form as-
$\int {ds = \int {udt} + } \int {atdt} $
Putting the limits and integrating,
$\int_0^s {ds} = \int_0^t {udt} + \int_0^t {atdt} $
$ \Rightarrow \left[ s \right]_0^s = \left[ {ut} \right]_0^t + \left[ {\dfrac{{a{t^2}}}{2}} \right]_0^t$ $\left\{ {\int {ndx = \dfrac{{{n^2}}}{2}} } \right\}$
$ \Rightarrow \left( {s - 0} \right) = u\left( {t - 0} \right) + \dfrac{1}{2}a\left( {{t^2} - 0} \right)$
Simplifying,
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$……….$(2)$
(C) Deriving ${v^2} = {u^2} + 2as$
This equation of motion does not contain the terms of time.
To derive this we use the formula of velocity and acceleration.
Velocity of a body can be given by the rate of change of its displacement with respect to time-
$v = \dfrac{{ds}}{{dt}}$
Rearranging,
$dt = \dfrac{1}{v}ds$
We know that acceleration is defined as rate of change of velocity with respect to time,
$a = \dfrac{{dv}}{{dt}}$
Rearranging,
$dt = \dfrac{1}{a}dv$
Equating both of these equations,
$dt = \dfrac{1}{v}ds = \dfrac{1}{a}dv$
Cross multiplying the terms,
$ \Rightarrow ads = vdv$
Writing this equation in integral form,
$\int_0^s {ads = \int_u^v {vdv} } $
(The velocity ranges from $u$ to $v$ )
Integrating and writing the limits,
$a\left[ s \right]_0^s = \left[ {\dfrac{{{v^2}}}{2}} \right]_u^v$
$ \Rightarrow a\left[ {s - 0} \right] = \dfrac{1}{2}\left[ {{v^2} - {u^2}} \right]$
Simplifying this,
$as = \dfrac{1}{2}\left( {{v^2} - {u^2}} \right)$
$ \Rightarrow {v^2} - {u^2} = 2as$
$ \Rightarrow {v^2} = {u^2} + 2as$……….$(3)$
All three equations of motion are derived.

Note: This derivation was based on the calculus method to derive the equations of motion for a uniformly accelerating body. Two other approaches that are the graphical and algebraic methods can also be used to derive these equations.