Question

A body starts from rest and travels with uniform acceleration. If the distance covered in first 2 seconds is 'x' and next 2 seconds is 'y', then

• A. y = x
• B. y = 2x
• C. y = 3x
• D. y = 4x

Hint:

According to Galileo's law of odd numbers, the distances traversed during equal intervals of time by a body falling from rest stand to one another in the same ratio as the odd numbers beginning with unity (1:3:5:7...). Since the time intervals are equal (2s each), the ratio of distances \(x:y\) will be 1:3, which means \(y=3x\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given that a body starts from rest with constant acceleration. We need to find the relationship between the distance covered in the first 2 seconds and the distance covered in the subsequent 2 seconds.

Step 2: Key Formula or Approach:
We use the equation of motion for displacement: \(s = ut + \frac{1}{2}at^2\).
Since the body starts from rest, the initial velocity \(u=0\). The formula simplifies to \(s = \frac{1}{2}at^2\).

Step 3: Detailed Explanation:
Let the uniform acceleration be \(a\).
The distance covered in the first 2 seconds (\(t_1 = 2\) s) is \(x\).
\[ x = \frac{1}{2}a(t_1)^2 = \frac{1}{2}a(2)^2 = \frac{1}{2}a(4) = 2a \]
The distance covered in the "next 2 seconds" means the distance traveled between \(t=2\) s and \(t=4\) s. This can be found by calculating the total distance in 4 seconds and subtracting the distance covered in the first 2 seconds.
Total time for both intervals is \(t_2 = 4\) s.
Total distance covered in 4 seconds is \(s_{total}\).
\[ s_{total} = \frac{1}{2}a(t_2)^2 = \frac{1}{2}a(4)^2 = \frac{1}{2}a(16) = 8a \]
The distance covered in the next 2 seconds, \(y\), is:
\[ y = s_{total} - x = 8a - 2a = 6a \]
Now, we find the relationship between \(y\) and \(x\):
We have \(x = 2a\) and \(y = 6a\).
\[ y = 6a = 3 \times (2a) = 3x \]

Step 4: Final Answer:
The relationship between y and x is \(y = 3x\).