Question

A body starts from rest, what is the ratio of the distance traveled by the body during the $ 4th $ and $ 3rd $ second?
(A) $ \dfrac{7}{5} $
(B) $ \dfrac{5}{7} $
(C) $ \dfrac{7}{3} $
(D) $ \dfrac{3}{7} $

Answer 1

Hint: When an object changes its position concerning some reference point, we call it to be in motion. To study and find different parameters we have equations of motion that are always valid for constant acceleration or uniform motion. we have to find the distance at $ 4th $ and $ 3rd $ second.
Distance in $ nth $ second of a journey is given by,
 $ {S_n} = \dfrac{{a(2n - 1)}}{2} $ .

Complete answer:
To find the distance traveled at different times we use the following formula.
 $ {S_n} = \dfrac{{a(2n - 1)}}{2} $
Here, $ S $ is the distance traveled, $ n $ is the time in seconds and a is the acceleration of the body.
Now, we have to find the distance at $ 4th $ and $ 3rd $ second, so once we keep $ n = 4 $ and $ n = 3 $ in the above formula.
When, $ n = 4 $
 $ {S_4} = \dfrac{{a(2 \times 4 - 1)}}{2} = \dfrac{{7a}}{2} $ …….. (1)
When, $ n = 3 $
 $ {S_3} = \dfrac{{a(2 \times 3 - 1)}}{2} = \dfrac{{5a}}{2} $ ………..(2)
Now, to find the ratio let us divide the equation (1) by (2).
 $ \dfrac{{{S_4}}}{{{S_3}}} = \dfrac{7}{5} $
Hence, option (A) $ \dfrac{7}{5} $ is the correct option.

Note:
We can derive this formula using the following equation of motion.
 $ S = ut + \dfrac{1}{2}a{t^2} $
Here, $ S $ is the displacement, $ u $ is initial velocity, $ t $ is the time, and $ a $ is acceleration.
Now, here we have $ t = n $ (second).
Now we can write,
Distance traveled in $ nth $ seconds = distance traveled in $ n $ second – distance traveled in $ \left( {n - 1} \right) $ second,
Let us use the above equation of motion.
 $ {S_n} = 0 \times n + \dfrac{{a{n^2}}}{2} - 0 \times \left( {n - 1} \right) - \dfrac{{a{{\left( {n - 1} \right)}^2}}}{2} $
 $ {S_n} = \dfrac{{a{n^2}}}{2} - \dfrac{{a{{(n - 1)}^2}}}{2} = \dfrac{{a{n^2} - a{n^2} - a + 2an}}{2} = \dfrac{{a(2n - 1)}}{2} $
Hence, the general formula to calculate the distance traveled at nth second is given by the following formula.
 $ S(nth{\text{ second}}) = \dfrac{{a(2n - 1)}}{2} $
The body starts from rest so we took the initial velocity ( $ u $ ) to be zero.
Similarly, if the body will stop, we will take its final velocity zero.