Question

A body falling for 2 sec. covers a distance equal to that covered in the next second taking g value as 10m/s2. Then ${\text{s}}$ equals
A. $30{\text{m}}$
B. $10{\text{m}}$
C. $60{\text{m}}$
D. $20{\text{m}}$

Answer 1

In the given question the value of acceleration is constant that is $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ . Therefore we can use the equations of motion given by Sir Newton. Constant acceleration means there is no external force applied other than $g$.

Complete answer:
Given: body falling for $2{\text{ sec}}$,
$g = 10{\text{m}}{{\text{s}}^{ - 2}}$
Now, by using the equation of motion
Distance travelled by the body in $2{\text{ sec}}$ is ${\text{s = ut + }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2}{\text{ }}.........{\text{(i)}}$
Here, in the above equation ${\text{s}}$ is distance covered, $u$ is initial velocity of the object, ${\text{a}}$is acceleration, ${\text{t}}$ is time.
From equation $({\text{i}})$ , ${\text{s = u}} \times {\text{2 + }}\dfrac{1}{2}{\text{a}} \times {({\text{2)}}^2}$
$\Rightarrow {\text{s = 2(u + g)}}$
$\Rightarrow {\text{s = 2(u + 10) }}.........{\text{(ii)}}$
Now the distance covered in next second is given by ${\text{s = u + (2n - 1)}}\dfrac{{\text{g}}}{2}{\text{ }}.........{\text{(iii)}}$
Here the value of n is $3$ as we want to find the distance covered in $3rd$ second.
From equation $({\text{iii}})$, ${\text{s = u + (2}} \times {\text{3 - 1)}}\dfrac{{\text{g}}}{2}$
$\Rightarrow {\text{s = u + (5)}} \times \dfrac{{10}}{2}$
$\Rightarrow {\text{s = u + 25}}\,\,\,\,\,.......{\text{(iv)}}$
Now from $({\text{ii}})$ and $({\text{iv}})$
${\text{u = 5m}}{{\text{s}}^{ - 1}}$ and ${\text{s = 30m}}$
So, distance covered in two seconds $({\text{s}}){\text{ = 30m}}$
Therefore, option (A) is the correct option.

Note:
Here in this question a formula for the distance covered for the ${{\text{n}}^{{\text{th}}}}$ second is used that is ${\text{s = u + (2n - 1)}}\dfrac{{\text{g}}}{2}$ and the equations of motion are only valid for the body experiencing constant acceleration. Here in this question we used the second equation of motion.