A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th second to that covered in 5 seconds is

Options

A. `(1)/(25)`
B. `(3)/(25)`
C. `(9)/(25)`
D. `(1)/(25)`

Answer

To solve the problem of finding the ratio of the distance covered by a body during the 5th second to that covered in 5 seconds, we can follow these steps:

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

### Step 2: Use the Formula for Distance Covered in the nth Second
The distance covered during the nth second can be calculated using the formula:
\[ S_n = u + \frac{a}{2} (2n - 1) \]
Since \( u = 0 \), the formula simplifies to:
\[ S_n = \frac{a}{2} (2n - 1) \]

### Step 3: Calculate the Distance Covered in the 5th Second
Substituting \( n = 5 \) into the formula:
\[ S_5 = \frac{a}{2} (2 \cdot 5 - 1) = \frac{a}{2} (10 - 1) = \frac{a}{2} \cdot 9 = \frac{9a}{2} \]

### Step 4: Use the Formula for Total Distance Covered in t Seconds
The total distance covered in \( t \) seconds is given by:
\[ S = ut + \frac{1}{2} a t^2 \]
Again, since \( u = 0 \), this simplifies to:
\[ S = \frac{1}{2} a t^2 \]
Substituting \( t = 5 \):
\[ S_5 = \frac{1}{2} a (5^2) = \frac{1}{2} a \cdot 25 = \frac{25a}{2} \]

### Step 5: Find the Ratio
Now, we need to find the ratio of the distance covered in the 5th second to the total distance covered in 5 seconds:
\[ \text{Ratio} = \frac{S_5}{S} = \frac{\frac{9a}{2}}{\frac{25a}{2}} = \frac{9a}{25a} = \frac{9}{25} \]

### Conclusion
The ratio of the distance covered by the body during the 5th second to that covered in 5 seconds is:
\[ \frac{9}{25} \]