A body starting from rest moves with constant acceleration. What is the ratio of distance covered by the body during the fifth second of time to that covered in the first 5.00 s ?

Answer

To solve the problem, we need to find the ratio of the distance covered by a body during the fifth second of time to the distance covered in the first five seconds, given that the body starts from rest and moves with constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The body starts from rest, which means the initial velocity \( u = 0 \). It moves with a constant acceleration \( a \).

2. **Distance Covered in the First 5 Seconds**:
We can use the formula for the distance covered under constant acceleration:
\[ S = ut + \frac{1}{2} a t^2 \]
Since \( u = 0 \), the formula simplifies to:
\[ S_1 = \frac{1}{2} a t^2 \]
For the first 5 seconds (where \( t = 5 \)):
\[ S_1 = \frac{1}{2} a (5^2) = \frac{1}{2} a (25) = \frac{25a}{2} \]

3. **Distance Covered During the Fifth Second**:
The distance covered during the \( n \)-th second can be calculated using the formula:
\[ S_n = u + \frac{a}{2} (2n - 1) \]
Again, since \( u = 0 \):
\[ S_5 = 0 + \frac{a}{2} (2 \cdot 5 - 1) = \frac{a}{2} (10 - 1) = \frac{a}{2} (9) = \frac{9a}{2} \]

4. **Finding the Ratio**:
Now, we need to find the ratio of the distance covered during the fifth second \( S_5 \) to the distance covered in the first 5 seconds \( S_1 \):
\[ \text{Ratio} = \frac{S_5}{S_1} = \frac{\frac{9a}{2}}{\frac{25a}{2}} \]
The \( \frac{a}{2} \) cancels out:
\[ \text{Ratio} = \frac{9}{25} \]

5. **Final Result**:
Thus, the ratio of the distance covered during the fifth second to the distance covered in the first five seconds is:
\[ \frac{9}{25} \quad \text{or} \quad 0.36 \]