Starting from rest, when a body moves with uniform acceleration, then distances covered after 1st, 2nd, 3rd, ... seconds are in the ratio

Options

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

Explanation

To find the distances covered by a body starting from rest with uniform acceleration after the 1st, 2nd, and 3rd seconds, we can use the equations of motion. Here’s the step-by-step solution:

### Step 1: Understand the equation of motion
The equation for distance covered under uniform acceleration is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) is the distance covered,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.

### Step 2: Set initial conditions
Since the body starts from rest, the initial velocity \( u = 0 \). Therefore, the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]

### Step 3: Calculate distances for each second
1. **Distance covered in the 1st second (\( S_1 \))**:
- For \( t = 1 \) second:
\[ S_1 = \frac{1}{2} a (1^2) = \frac{1}{2} a \]

2. **Distance covered in the 2nd second (\( S_2 \))**:
- For \( t = 2 \) seconds, we need to find the total distance covered in 2 seconds and subtract the distance covered in the first second:
\[ S_2 = \text{Total distance in 2 seconds} - \text{Distance in 1 second} \]
\[ S_2 = \left(\frac{1}{2} a (2^2)\right) - S_1 \]
\[ S_2 = \left(\frac{1}{2} a \cdot 4\right) - \left(\frac{1}{2} a\right) = 2a - \frac{1}{2} a = \frac{3}{2} a \]

3. **Distance covered in the 3rd second (\( S_3 \))**:
- For \( t = 3 \) seconds, we find the total distance covered in 3 seconds and subtract the distance covered in the first two seconds:
\[ S_3 = \text{Total distance in 3 seconds} - \text{Distance in 2 seconds} \]
\[ S_3 = \left(\frac{1}{2} a (3^2)\right) - (S_1 + S_2) \]
\[ S_3 = \left(\frac{1}{2} a \cdot 9\right) - \left(\frac{1}{2} a + \frac{3}{2} a\right) \]
\[ S_3 = \frac{9}{2} a - 2a = \frac{9}{2} a - \frac{4}{2} a = \frac{5}{2} a \]

### Step 4: Find the ratio of distances
Now we have:
- \( S_1 = \frac{1}{2} a \)
- \( S_2 = \frac{3}{2} a \)
- \( S_3 = \frac{5}{2} a \)

To find the ratio \( S_1 : S_2 : S_3 \):
\[ S_1 : S_2 : S_3 = \frac{1}{2} a : \frac{3}{2} a : \frac{5}{2} a \]
Cancelling \( \frac{1}{2} a \) from each term gives:
\[ 1 : 3 : 5 \]

### Final Answer
Thus, the distances covered after the 1st, 2nd, and 3rd seconds are in the ratio \( 1 : 3 : 5 \).

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