Question

Starting from rest, when a body moves with uniform acceleration, then distances covered after 1st, 2nd, 3rd, ... seconds are in the ratio
A. $1:2:3:4.....$
B. $1:4:9:16.....$
C. $1:3:5:7.....$
D. $2:3:5:7.....$

Answer 1

Hint: In order to denote the position of an object we define a coordinate system and an origin as a reference point. If a body moves from one position to another position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity. Rate of change of velocity gives acceleration.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$

Complete answer:
Rate of change of displacement with respect to time gives us the velocity and the rate of change of velocity with respect to time gives us the acceleration. Now if we are given with acceleration as a function of time and we were asked to find out the displacement then first we should get the velocity function by integrating the acceleration with in the proper limits given and then later after getting the velocity function, integrate it again with respect to time to get the displacement.
$\eqalign{
  & \dfrac{{dv}}{{dt}} = a \cr
  & \Rightarrow \int {(a)} dt = v \cr
  & \therefore \int {(v)} dt = s \cr} $
Where ‘v’ is the velocity and ‘a’ is the acceleration and ‘t’ is the time.
We have the relation between velocity and displacement and uniform acceleration as
$s = ut + \dfrac{1}{2}a{t^2}$
As the body is starting from rest initial velocity(u=0).
Distance travelled in 1 second is
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow {s_1} = 0 + \dfrac{1}{2}a{(1)^2} \cr
  & \therefore {s_1} = \dfrac{1}{2}a{(1)^2} \cr} $
Distance travelled in 2 seconds is
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow {s_2} = 0 + \dfrac{1}{2}a{(2)^2} \cr
  & \therefore {s_2} = \dfrac{1}{2}a(4) \cr} $
Distance travelled in 3 seconds is
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow {s_3} = 0 + \dfrac{1}{2}a{(3)^2} \cr
  & \therefore {s_3} = \dfrac{1}{2}a(9) \cr} $
Distance travelled in 4 seconds is
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow {s_4} = 0 + \dfrac{1}{2}a{(4)^2} \cr
  & \therefore {s_4} = \dfrac{1}{2}a(16) \cr} $
So finally the ratio will be
$\eqalign{
  & {s_1}:{s_2}:{s_3}:{s_4} = \dfrac{1}{2}a(1):\dfrac{1}{2}a(4):\dfrac{1}{2}a(9):\dfrac{1}{2}a(16) \cr
  & \therefore {s_1}:{s_2}:{s_3}:{s_4}...... = 1:4:9:16..... \cr} $

Hence option B is the correct answer.

Note:
The formula which we had used is only valid if the acceleration is uniform. If it is not uniform then that will be a function of time and we have to integrate acceleration too with respect to time and we get entirely different results. If there is some initial velocity then we can’t get the above ratio.