Question

Equation for uniform accelerated motion for the displacement covered in its nth second of its motion is?
A. ${{S}_{n}}=u+a\left( n-\dfrac{1}{3} \right)$
B. ${{S}_{n}}=u+a\left( n-\dfrac{1}{2} \right)$
C. ${{S}_{n}}=2u+a\left( n-\dfrac{1}{2} \right)$
D. ${{S}_{n}}=\dfrac{u}{2}+a\left( n-\dfrac{1}{2} \right)$

Answer 1

Hint: This problem can be solved by finding out the general equation for the displacement covered for uniform accelerated motion in $n$ seconds using the equations of motion and then extending the same for $n-1$ seconds. By subtracting the displacement covered in $n-1$ seconds from the one covered in $n$ seconds, we can get the expression for the displacement covered in the nth second.

Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ --(equation for uniformly accelerated motion)
Where $s$ is the displacement covered, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time period for which uniformly accelerated motion has taken place.
${{S}_{n}}=S\left( n \right)-S\left( n-1 \right)$
where ${{S}_{n}}$, $S\left( n \right)$, $S\left( n-1 \right)$ are the displacements covered in the nth second, $n$ seconds and $n-1$ seconds respectively.

Complete Step-by-Step solution:
To solve this problem, first we have to find out the displacement covered for uniform accelerated motion in $n$ seconds using the equations of motion and then extend it for $n-1$ seconds. Then, since the displacement covered in the nth second is nothing but the difference of the displacement covered in $n$ seconds and that covered in $n-1$ seconds, we will subtract the displacement covered in $n-1$ seconds from the one covered in $n$ seconds.
For finding the displacement covered in $n$ seconds, we will use the equation of motion,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ --(equation for uniformly accelerated motion)--(1)
Where $s$ is the displacement covered, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time period for which uniformly accelerated motion has taken place.
Putting $t=n$ in (1), we get,
$S\left( n \right)=un+\dfrac{1}{2}a{{n}^{2}}$ -------------------(2)
Where, $S\left( n \right)$is the displacements covered in $n$ seconds.
Now, similarly the displacement covered in $n+1$ will be obtained by putting $t=n-1$ in (1). Hence, we get,
$S\left( n+1 \right)=u\left( n-1 \right)+\dfrac{1}{2}a{{\left( n-1 \right)}^{2}}$ ----------(3)
Where, $S\left( n-1 \right)$ is the displacements covered in $n-1$ seconds.
Now, for finding displacement covered in the nth second we will subtract the displacement covered in $n-1$ seconds from the one covered in $n$ seconds.
${{S}_{n}}=S\left( n \right)-S\left( n-1 \right)$ ---------------(4)
Where ${{S}_{n}}$, $S\left( n \right)$, $S\left( n-1 \right)$ are the displacements covered in the nth second, $n$ seconds and $n-1$ seconds respectively.
Hence, putting (2) and (3) in (4), we get,
${{S}_{n}}=un+\dfrac{1}{2}a{{n}^{2}}-\left[ u\left( n-1 \right)+\dfrac{1}{2}a{{\left( n-1 \right)}^{2}} \right]=un+\dfrac{1}{2}a{{n}^{2}}-un+u-\dfrac{1}{2}a{{\left( n-1 \right)}^{2}}$
$\therefore {{S}_{n}}=u+\dfrac{1}{2}a\left[ {{\left( n \right)}^{2}}-{{\left( n-1 \right)}^{2}} \right]$
$\therefore {{S}_{n}}=u+\dfrac{1}{2}a\left[ \left( n+n-1 \right)\left( n-n+1 \right) \right]=u+\dfrac{1}{2}a\left( 2n-1 \right)\left( 1 \right)$
$\therefore {{S}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right)$
$\therefore {{S}_{n}}=u+a\left( n-\dfrac{1}{2} \right)$
Hence, the displacement covered in the nth second is given by ${{S}_{n}}=u+a\left( n-\dfrac{1}{2} \right)$.
Hence, the correct option is B) ${{S}_{n}}=u+a\left( n-\dfrac{1}{2} \right)$.

Note: A very common error in this question is that students write the displacement covered in the nth second as the displacement covered in $n+1$ seconds minus the displacement covered in $n$ seconds. That is,
${{S}_{n}}=S\left( n+1 \right)-S\left( n \right)$
which is absolutely wrong.
Most students go ahead with this thought process and arrive at a completely wrong answer. To avoid this one can think like this that the displacement covered in the first second must be the displacement covered in one second, S(1), which can only be arrived at by subtracting S(0) (which will obviously be zero) from S(1).
In competitive exams, where one needs to save time, a trick can be used which is to replace n by small numbers (preferably 1), so as to get an equation that is known by the students. For example, the displacement in 1 second can be found by putting $t=1$ in (1) and this is also the displacement in the 1st second $\left( {{S}_{1}} \right)$ as explained above. Now substituting $n=1$ in the options and comparing with the result of ${{S}_{1}}$ that we got, we can eliminate some options.
However, this does not always help and is a short-cut method.