Question

A body starts to fall freely under gravity. The distances covered by it in first, second and third seconds are in the ratio:
A. \[1:3:5\]
B. \[1:2:3\]
C. \[1:4:9\]
D. \[1:5:6\]

Answer 1

Hint: First of all, we will find out the distance moved by the body in zero second and after one second, then we will find the difference to find the distance moved for the first second. We will carry on the same procedure for two more cases. Then we will find the ratio.

Complete step by step answer:
In the given question, the supplied data are as follows:

A body was dropped from a point, which falls purely under the action of gravity.
We are asked to find the ratio of the distances covered by it in first, second and third seconds respectively.

To begin with, we will take the help from one of the equations from the laws of motion, which is given below:
\[S = ut + \dfrac{1}{2}g{t^2}\] …… (1)
Where,
\[S\] indicates the distance covered.
\[u\] indicates the initial velocity.
\[g\] indicates the acceleration due to gravity.
\[t\] indicates the time taken.

Now, to find the distance covered in the first second, it can be calculated by the difference in the distances covered at time equals \[0\,{\text{s}}\] and \[1\,{\text{s}}\] .
So, the distance covered at time \[0\,{\text{s}}\] is:
${S_0} = u{t_0} + \dfrac{1}{2}gt_0^2 \\
\implies {S_0} = u \times 0 + \dfrac{1}{2}g \times {0^2} \\
\implies {S_0} = 0 \\$

The distance covered after time \[1\,{\text{s}}\] is:
${S_1} = u{t_1} + \dfrac{1}{2}gt_1^2 \\
\implies {S_1} = u \times 1 + \dfrac{1}{2}g \times {1^2} \\
\implies {S_1} = u + \dfrac{g}{2} \\$
So, the distance covered in the first second is:
$\Delta {S_1} = {S_1} - {S_0} \\
\implies \Delta {S_1} = u + \dfrac{g}{2} - 0 \\
\implies \Delta {S_1} = u + \dfrac{g}{2} \\$

Now, to find the distance covered in the second seconds, it can be calculated by the difference in the distances covered at time equals \[1\,{\text{s}}\] and \[2\,{\text{s}}\] .
So, the distance covered at after time \[1\,{\text{s}}\] is already found as:
\[{S_1} = u + \dfrac{g}{2}\]

The distance covered after time \[2\,{\text{s}}\] is:
${S_2} = u{t_2} + \dfrac{1}{2}gt_2^2 \\
\implies {S_2} = u \times 2 + \dfrac{1}{2}g \times {2^2} \\
\implies {S_2} = 2u + 2g \\$
So, the distance covered in the second seconds is:
$\Delta {S_2} = {S_2} - {S_1} \\
\implies \Delta {S_2} = 2u + 2g - u - \dfrac{g}{2} \\
\implies \Delta {S_2} = u + \dfrac{{3g}}{2} \\$

Now, to find the distance covered in the third seconds, it can be calculated by the difference in the distances covered at time equals \[3\,{\text{s}}\] and \[2\,{\text{s}}\] .
So, the distance covered at after time \[2\,{\text{s}}\] is already found as:
\[{S_2} = 2u + 2g\]

The distance covered after time \[3\,{\text{s}}\] is:
${S_3} = u{t_3} + \dfrac{1}{2}gt_3^2 \\
\implies {S_3} = u \times 3 + \dfrac{1}{2}g \times {3^2} \\
\implies {S_3} = 3u + \dfrac{{9g}}{2} \\$
So, the distance covered in the second seconds is:
$\Delta {S_3} = {S_3} - {S_2} \\
\implies \Delta {S_3} = 3u + \dfrac{{9g}}{2} - 2u - 2g \\
\implies \Delta {S_3} = u + \dfrac{{5g}}{2} \\$

Since, this is the case of free fall, so the initial velocity of the body will be zero i.e. \[u = 0\] .
So, putting this value in all the differences found, we get:
\[\Delta {S_1} = \dfrac{g}{2}\]
\[\Delta {S_2} = \dfrac{{3g}}{2}\]
\[\Delta {S_3} = \dfrac{{5g}}{2}\]

Now, the ratio is calculated as:
$\Delta {S_1}:\Delta {S_2}:\Delta {S_3} \\
= \dfrac{g}{2}:\dfrac{{3g}}{2}:\dfrac{{5g}}{2} \\
= 1:3:5 \\$
Hence, the required ratio is found to be \[1:3:5\] .

So, the correct answer is “Option A”.

Note:
While solving this problem, many students seem to make mistakes by simply calculating the distance for time one second, followed by two seconds and three seconds. However, it is not like that. We have to find out the distance moved in \[n - th\] second.