Question

The ratio of the distance travelled by a freely falling body in the 1^st, 2^nd, 3^rd and 4^th second

• A. 1:2:3:4
• B. 1:4:9:16
• C. 1:3:5:7
• D. 1:1:1:1

Answer 1

The Correct Option is C

The distance \( s \) traveled by a freely falling body in time \( t \) can be calculated using the equation for uniformly accelerated motion: \( s = \frac{1}{2}gt^2 \), where \( g \) is the acceleration due to gravity. For this problem, we focus on the distance traveled during each specific second, rather than the total distance.

To find the distance fallen in each second: 

• First Second: The distance traveled, \( s_1 = \frac{1}{2}g(1^2) = \frac{1}{2}g \).
• Second Second: Total distance at 2 seconds, \( s_2 = \frac{1}{2}g(2^2) \). Calculate distance for the second second: \( \Delta s_2 = s_2 - s_1 = \frac{1}{2}g(4) - \frac{1}{2}g = \frac{3}{2}g \).
• Third Second: Total distance at 3 seconds, \( s_3 = \frac{1}{2}g(3^2) \). Calculate distance for the third second: \( \Delta s_3 = s_3 - s_2 = \frac{1}{2}g(9) - \frac{1}{2}g(4) = \frac{5}{2}g \).
• Fourth Second: Total distance at 4 seconds, \( s_4 = \frac{1}{2}g(4^2) \). Calculate distance for the fourth second: \( \Delta s_4 = s_4 - s_3 = \frac{1}{2}g(16) - \frac{1}{2}g(9) = \frac{7}{2}g \).

Now, the distance traveled each second is \( \frac{1}{2}g, \frac{3}{2}g, \frac{5}{2}g, \frac{7}{2}g \).

The ratio of these distances is: \( 1:3:5:7 \).

Therefore, the ratio of the distances traveled by the body in the 1^st, 2^nd, 3^rd, and 4^th seconds is 1:3:5:7.