Question

A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to the 3rd second is:
A. 33%
B. 40%
C. 66%
D. 77%

Answer 1

Hint: When the displacement is asked, and the input is time and the acceleration, the formula to be used is:
$s = ut + \dfrac{1}{2}\left( {a{t^2}} \right)$

Complete step by step solution:
For this problem, the displacement at the nth second formula is to be applied and derived from the equation of motion.
Deriving the equation for displacement at nth second –
Displacement covered at ‘n’ seconds =$s = un + \dfrac{1}{2}\left( {a{n^2}} \right)$
Displacement covered at ‘n-1’ seconds =$s = u\left( {n - 1} \right) + \dfrac{1}{2}\left( {a{{\left( {n - 1} \right)}^2}} \right)$
If we subtract the displacement covered at ‘n-1’ seconds and ‘n’ seconds, we obtain the formula for displacement at the nth second
Displacement at nth second = $s = u + \dfrac{1}{2}a\left( {2n - 1} \right)$
Using the above formula, we should calculate the displacement at 3rd second and 4th second and find the percentage change among them –
Since the particle is starting from rest, u=0
Displacement at 3rd second –
${s_3} = \dfrac{1}{2}a\left( {2\left( 3 \right) - 1} \right)$
Simplifying
$
  {s_3} = \dfrac{1}{2}a\left( {6 - 1} \right) \\
  {s_3} = \dfrac{1}{2}a\left( 5 \right) \\
  {s_3} = \dfrac{{5a}}{2} \\
 $
Displacement at 4th second –
${s_4} = \dfrac{1}{2}a\left( {2\left( 4 \right) - 1} \right)$
Simplifying
$
  {s_4} = \dfrac{1}{2}a\left( {8 - 1} \right) \\
  {s_4} = \dfrac{1}{2}a\left( 7 \right) \\
  {s_4} = \dfrac{{7a}}{2} \\
 $
Percentage change in the displacement –
$\dfrac{{{s_4} - {s_3}}}{{{s_3}}} \times 100$
Substituting the values, we get –
$
  \dfrac{{\dfrac{{7a}}{2} - \dfrac{{5a}}{2}}}{{\dfrac{{5a}}{2}}} \times 100 \\
   = \dfrac{{\dfrac{{2a}}{2}}}{{\dfrac{{5a}}{2}}} \times 100 \\
   = \dfrac{{a \times 2}}{{5a}} \times 100 \\
   = \dfrac{2}{5} \times 100 \\
   = 40\% \\
 $

Hence, the correct option is Option B.

Note:
There is a very quick shortcut method to attempt this problem. If a similar problem like this appears in JEE, you can solve it in a few seconds through this trick.
Whenever you get the nth second, the percentage change is always given by the fraction:
$\dfrac{2}{{2n - 1}}$
For instance, in the Entrance test, there is a similar problem saying to calculate the percentage increase in its displacement during the 8th second compared to the 7th second.
Instead of going around all these steps, you can find the answer in an instant by using the above fraction –
Percentage – $\dfrac{2}{{2n - 1}} = \dfrac{2}{{2\left( 7 \right) - 1}} = \dfrac{2}{{14 - 1}} = \dfrac{2}{{13}} = 0.1538$
So, the answer is 15.3%.
You can save a lot of time by using this shortcut in your JEE entrance test.