A body starts from rest, what is the ratio of the distance travelled by the body during the 4th and 3rd s?

Options

A. `(7)/(5)`
B. `(5)/(7)`
C. `(7)/(3)`
D. `(3)/(7)`

Answer

To solve the problem of finding the ratio of the distance traveled by a body during the 4th and 3rd seconds, we can follow these steps:

### Step 1: Understand the problem
The body starts from rest, which means its initial velocity (U) is 0. We need to find the distance traveled during the 4th second (S4) and the 3rd second (S3).

### Step 2: Use the formula for distance traveled in the nth second
The formula for the distance traveled in the nth second is given by:
\[ S_n = U + \frac{A}{2} (2n - 1) \]
where:
- \( S_n \) is the distance traveled in the nth second,
- \( U \) is the initial velocity,
- \( A \) is the acceleration,
- \( n \) is the second in question.

### Step 3: Calculate S4 (distance traveled in the 4th second)
Substituting \( n = 4 \) into the formula:
\[ S_4 = U + \frac{A}{2} (2 \cdot 4 - 1) \]
Since \( U = 0 \):
\[ S_4 = 0 + \frac{A}{2} (8 - 1) \]
\[ S_4 = \frac{A}{2} \cdot 7 \]
\[ S_4 = \frac{7A}{2} \]

### Step 4: Calculate S3 (distance traveled in the 3rd second)
Now substituting \( n = 3 \) into the formula:
\[ S_3 = U + \frac{A}{2} (2 \cdot 3 - 1) \]
Again, since \( U = 0 \):
\[ S_3 = 0 + \frac{A}{2} (6 - 1) \]
\[ S_3 = \frac{A}{2} \cdot 5 \]
\[ S_3 = \frac{5A}{2} \]

### Step 5: Find the ratio of S4 to S3
Now we need to find the ratio of the distances traveled in the 4th and 3rd seconds:
\[ \text{Ratio} = \frac{S_4}{S_3} = \frac{\frac{7A}{2}}{\frac{5A}{2}} \]

### Step 6: Simplify the ratio
The \( \frac{2}{2} \) cancels out and \( A \) also cancels out:
\[ \text{Ratio} = \frac{7}{5} \]

### Final Answer
The ratio of the distance traveled by the body during the 4th second to the distance traveled during the 3rd second is:
\[ \frac{7}{5} \]

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