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Question: The distances covered by a freely falling body in its first, second, third, nth Seconds of its motio...

The distances covered by a freely falling body in its first, second, third, nth Seconds of its motion
(A) forms an arithmetic progression
(B) forms a geometric progression
(C) do not form any well-defined series
(D) forms a series corresponding to the difference of square root of the successive natural numbers.

Explanation

Solution

Hint
Firstly, we will use the basic formula to find the distance covered by the freely falling body using x=ut+12gt2x = ut + \dfrac{1}{2}g{t^2} the formula. After calculating the value of the distances lastly, we will find the distance covered by the particle in nth time. We will find that our solution consists of the first term, common difference and nth term in sequence. So, our solution will be comparable with Arithmetic progression.

Complete step by step answer
We know that distance is a scalar quantity that refers to the position of an object with respect to a pre-assigned reference point.
Let us assume that x1,x2,x3,....xn{x_1},{x_2},{x_3},....{x_n} are the distance covered by the falling body at the respective time t1,t2,t3,....tn{t_1},{t_2},{t_3},....{t_n}
Initial velocity of that body is zero because it was at rest.
The acceleration of the body will be gravitational acceleration (g) because it is falling freely
For our easier calculation we are taking the value of g is 10ms210\dfrac{m}{{{s^2}}}
Now we can calculate the respective distance covered by the particle using this formula
⇒x=ut+12gt2\Rightarrow x = ut + \dfrac{1}{2}g{t^2}....(i)
⇒x=12gt2\Rightarrow x = \dfrac{1}{2}g{t^2}
⇒x1=0+12g(1)2=5m\Rightarrow {x_1} = 0 + \dfrac{1}{2}g{\left( 1 \right)^2} = 5m
⇒x1+x2=0+12g(2)2=20m\Rightarrow {x_1} + {x_2} = 0 + \dfrac{1}{2}g{\left( 2 \right)^2} = 20m
⇒x2=(20−5)=15m\Rightarrow {x_2} = (20 - 5) = 15m
⇒x1+x2+x3=0+12g(3)2=45m\Rightarrow {x_1} + {x_2} + {x_3} = 0 + \dfrac{1}{2}g{(3)^2} = 45m
⇒x3=25m\Rightarrow {x_3} = 25m
So, the total distance covered by the particle
⇒x1+x2+...+xn=0+12g(n2)=5n2m\Rightarrow {x_1} + {x_2} + ... + {x_n} = 0 + \dfrac{1}{2}g({n^2}) = 5{n^2}m
If we subtract the distance covered by the particle in (n−1)th{(n - 1)^{th}} time from distance covered by the particle in nth time we will get xn{x_n}
⇒xn=5n2−[5(n−1)2]=10n−5=5+10(n−1)....(ii)\Rightarrow {x_n} = 5{n^2} - \left[ {5{{\left( {n - 1} \right)}^2}} \right] = 10n - 5 = 5 + 10\left( {n - 1} \right)....(ii)
⇒x1,x2,....xn\Rightarrow {x_1},{x_2},....{x_n} forming an AP with d=10d = 10
We know that the Arithmetic progression is, an=a1+(n−1)d....(iii){a_n} = {a_1} + (n - 1)d....(iii)
Here, an{a_n} = the nth term in sequence
a1{a_1} = the first term in the sequence
d= the common difference between successive terms
If we observe the value of x1,x2,x3{x_1},{x_2},{x_3} ... we will find they have common difference is (d)=x3−x2=x2−x1=10(d) = {x_3} - {x_2} = {x_2} - {x_1} = 10
So comparing equation (ii) with (iii) we get that the distance covered by a freely falling body formed an Arithmetic progression (Option-A).

Note
Actually, one can use the value of gravity 9.8 ms2\dfrac{m}{{{s^2}}} instead of 10 ms2\dfrac{m}{{{s^2}}} . But here we use 10 ms2\dfrac{m}{{{s^2}}} to minimize calculations and get an answer quickly to check whether it forms an AP series or not. If one uses the value of g=9.8 ms2\dfrac{m}{{{s^2}}} then they have complicated calculations and it consumes more time.