Question

The velocity of a particle increases from $u$ to $v$ in a time $t$ during which it covers a distance $S$. If the particle has a uniform acceleration, which one of the following equations does not apply to the motion?
(A) $2S = (u + v)t$
(B) $a = \dfrac{{v - u}}{t}$
(C) ${v^2} = {u^2} - 2as$
(D) $S = \left( {u + \dfrac{1}{2}at} \right)t$

Answer 1

Hint
To answer this question, we just need to use the three equations of motion. Then, manipulating these equations according to the relations in the options given, we will get the answer.
The formulae used to solve this question is
$\Rightarrow v = u + at$
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow {v^2} = {u^2} + 2as$
$u = $initial velocity, $v = $final velocity, $s = $displacement, $a = $acceleration, and $t = $time

Complete step by step answer
According to the question, we have initial velocity $ = u$, final velocity $ = v$, distance covered $ = S$, and time elapsed $ = t$.
Let $a$ be the acceleration of the particle.
As the motion is uniformly accelerated, so all the three equations of motion are valid here.
So, we have the distance $S = ut + \dfrac{1}{2}a{t^2}$ (1)
On taking $t$ common, we have
$\Rightarrow S = \left( {u + \dfrac{1}{2}at} \right)t$
So, option (D) is correct.
Now, according to the first equation of motion, we have
$\Rightarrow v = u + at$
$\Rightarrow v - u = at$
Dividing by $t$, we get
$\Rightarrow a = \dfrac{{v - u}}{t}$ (2)
So, option (B) is also correct.
Now, substituting (2) in (1), we get
$\Rightarrow S = ut + \dfrac{1}{2}\left( {\dfrac{{v - u}}{t}} \right){t^2}$
$\Rightarrow S = ut + \dfrac{1}{2}\left( {v - u} \right)t$
Taking $t$ common, we get
$\Rightarrow S = \left[ {u + \dfrac{1}{2}\left( {v - u} \right)} \right]t$
$\Rightarrow S = \left( {u + \dfrac{v}{2} - \dfrac{u}{2}} \right)t$
On simplifying, we get
$\Rightarrow S = \left( {\dfrac{u}{2} + \dfrac{v}{2}} \right)t$
$\Rightarrow S = \left( {\dfrac{{u + v}}{2}} \right)t$
Multiplying by $2$ on both sides
$\Rightarrow 2S = \left( {u + v} \right)t$
So, option (A) is also correct.
Also, according to the second equation of motion, we have
$\Rightarrow {v^2} - {u^2} = 2as$
Adding${u^2}$ both the sides, we get
$\Rightarrow {v^2} = {u^2} + 2as$
So, option (C) is incorrect.
Our answer is option (C).

Note
For a uniformly accelerated motion, the average velocity of a particle is equal to the average of the initial and the final velocities of the particle. That is, ${v_{avg}} = \dfrac{{u + v}}{2}$ . Multiplying this average velocity with the total time taken will give the total distance covered in that much time. From there also, we can prove the first option. This trick should be remembered for a uniformly accelerated motion.