Question

$A \: body \: starting \: from \: rest \: has \: \\ an \: acceleration \: of \: 5m/ {s}^{2} . \\ calculate \: the \: distance \: travelled \: \\ by \: it \: in \: 4th \: second.$

#Answer the question with steps and also explain that how did you reached to your answer#

Answer 1

distance travelled in nth second =U+a/2(2n-1)
=0+5/2(2×4-1)
=2.5×7=17.5m
derivation of the implemented formula has been elaborated in the above answer XD

Answer 2

Let us consider the general case first.

Suppose a body starts with velocity u  and moves with a uniform acceleration $a$

Suppose we have to find the distance travelled in the $n^{th}$ second.  

The logic we are going to use is:

Distance covered in $n^{th}$ second = (Total Distance covered in n seconds) - (Total Distance covered in n-1 seconds)


That is, to put it simply, suppose we want to find the distance travelled in the 5th second. Then we first find the total distance covered in 5 seconds, and from it, we subtract the total distance covered in 4 seconds. 

The total distance covered in n seconds can be obtained as follows:


$s=ut+\frac{1}{2}at^2 \\ \\ \implies s_n = un + \frac{1}{2}an^2$


So, now we can write:

Total Distance Covered in n seconds = $s_n$
Total Distance Covered in (n-1) seconds = $s_{n-1}$


Thus, if we represent the distance travelled in $n^{th}$ second as $d_n$ we have:



$d_n = s_n - s_{n-1} \\ \\ \implies d_n = (un + \frac{1}{2}an^2) - (u(n-1)+\frac{1}{2}a(n-1)^2) \\ \\ \implies d_n = \cancel{un} + \frac{an^2}{2} - \cancel{un} + u - \frac{a(n-1)^2}{2} \\ \\ \implies d_n = u + \frac{a}{2} \times (n^2-(n-1)^2) \\ \\ \\ \left[\text{Using } x^2-y^2=(x+y)(x-y)\right] \\ \\ \\ \implies d_n = u + \frac{a}{2} (n+(n-1))(n-(n-1)) \\ \\ \implies d_n = u + \frac{a}{2} (2n-1)(1) \\ \\ \\ \implies \boxed{\bold{d_n = u + \frac{a}{2}(2n-1)}}$


Thus, we have obtained the formula for distance travelled in $n^{th}$ second.
_____________________________


Coming to the question. The body starts from rest and has a uniform acceleration of 5 $m/s^2$. We need the distance travelled in the 4th second. So our data is:

$u = 0 \, \, m/s \\ a = 5 \, \, m/s^2 \\ n = 4$

Our answer would be:

$d_n = u + \frac{a}{2}(2n-1) \\ \\ \implies d_4 = 0 + \frac{5}{2}(2(4)-1) \\ \\ \implies d_4 = \frac{5}{2} (8-1) \\ \\ \implies d_4 = \frac{5 \times 7}{2} \\ \\ \implies d_4 = \frac{35}{2} \\ \\ \\ \implies \boxed{d_4 = 17.5 \, \, m }$


Thus, The Distance Travelled in the 4th second is 17.5 metres.