Understanding Free Fall and Distance Travelled

When a body falls freely from rest, it means it starts with an initial velocity of zero and moves under the influence of gravity alone. The acceleration acting on the body is the constant acceleration due to gravity, denoted by \(g\). To find the distance travelled by a body falling freely from rest, we can use the kinematic equation:

\(s = ut + \frac{1}{2}at^2\)

Where:

• \(s\) is the distance travelled
• \(u\) is the initial velocity
• \(t\) is the time taken
• \(a\) is the acceleration

For a body falling freely from rest, \(u = 0\) and \(a = g\). So the equation simplifies to:

\(s = 0 \cdot t + \frac{1}{2}gt^2 = \frac{1}{2}gt^2\)

Calculating Distances in Successive Seconds

We need to find the distance travelled in the first second, the second second, and the third second. This is different from the distance travelled in the first 1 second, first 2 seconds, and first 3 seconds.

Let's calculate the total distance travelled from the start (rest) at the end of 1 second, 2 seconds, and 3 seconds:

• Distance travelled in the first 1 second (\(S_1\)):
  Using \(t = 1\) s, \(S_1 = \frac{1}{2}g(1)^2 = \frac{1}{2}g\). This is the distance covered in the first second.
• Distance travelled in the first 2 seconds (\(S_2\)):
  Using \(t = 2\) s, \(S_2 = \frac{1}{2}g(2)^2 = \frac{1}{2}g(4) = 2g\).
• Distance travelled in the first 3 seconds (\(S_3\)):
  Using \(t = 3\) s, \(S_3 = \frac{1}{2}g(3)^2 = \frac{1}{2}g(9) = \frac{9}{2}g\).

Now, let's find the distance travelled *during* each specific second:

• Distance travelled in the first second (\(d_1\)):
  This is the total distance covered at \(t=1\) s, as the body started from rest at \(t=0\) s. So, \(d_1 = S_1 = \frac{1}{2}g\).
• Distance travelled in the second second (\(d_2\)):
  This is the distance covered between \(t=1\) s and \(t=2\) s. It is the total distance covered in the first 2 seconds minus the total distance covered in the first 1 second. So, \(d_2 = S_2 - S_1 = 2g - \frac{1}{2}g = \frac{4}{2}g - \frac{1}{2}g = \frac{3}{2}g\).
• Distance travelled in the third second (\(d_3\)):
  This is the distance covered between \(t=2\) s and \(t=3\) s. It is the total distance covered in the first 3 seconds minus the total distance covered in the first 2 seconds. So, \(d_3 = S_3 - S_2 = \frac{9}{2}g - 2g = \frac{9}{2}g - \frac{4}{2}g = \frac{5}{2}g\).

We can summarize the distances travelled in each successive second in a table:

Finding the Ratio of Distances in Free Fall

The question asks for the ratio of the distances travelled in the first, second, and third second. This ratio is \(d_1 : d_2 : d_3\).

Ratio \( = \frac{1}{2}g : \frac{3}{2}g : \frac{5}{2}g\)

To simplify the ratio, we can divide each term by the common factor \(\frac{1}{2}g\). Since \(g\) is the acceleration due to gravity, it is non-zero.

Ratio \( = \frac{\frac{1}{2}g}{\frac{1}{2}g} : \frac{\frac{3}{2}g}{\frac{1}{2}g} : \frac{\frac{5}{2}g}{\frac{1}{2}g}\)

Ratio \( = 1 : 3 : 5\)

This shows that the distances travelled by a body falling freely from rest in successive equal intervals of time (like 1 second, 2nd second, 3rd second, etc.) are in the ratio of odd numbers \(1:3:5:7...\). This is a well-known result related to uniformly accelerated motion starting from rest.

Thus, the distances travelled in the first, second and third second are in the ratio 1 : 3 : 5.