A particle at rest starts moving in a horizontal straight line with a uniform acceleration. The ratio of the distances covered during the fourth and third seconds is

Options

A. `4//3`
B. `26//9`
C. `7//5`
D. `2`

Answer

To solve the problem, we need to find the ratio of the distances covered by a particle during the third and fourth seconds of its motion under uniform acceleration, starting from rest.

### Step-by-Step Solution:

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

2. **Distance Covered in nth Second**:
- The formula for the distance covered by a particle in the nth second is given by:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
- Since the initial velocity \( u = 0 \), the formula simplifies to:
\[ s_n = \frac{1}{2} a (2n - 1) \]

3. **Calculating Distance for Third Second**:
- For \( n = 3 \):
\[ s_3 = \frac{1}{2} a (2 \cdot 3 - 1) = \frac{1}{2} a (6 - 1) = \frac{1}{2} a \cdot 5 = \frac{5a}{2} \]

4. **Calculating Distance for Fourth Second**:
- For \( n = 4 \):
\[ s_4 = \frac{1}{2} a (2 \cdot 4 - 1) = \frac{1}{2} a (8 - 1) = \frac{1}{2} a \cdot 7 = \frac{7a}{2} \]

5. **Finding the Ratio of Distances**:
- Now, we need to find the ratio of the distances covered in the fourth second to the third second:
\[ \text{Ratio} = \frac{s_4}{s_3} = \frac{\frac{7a}{2}}{\frac{5a}{2}} = \frac{7a}{2} \cdot \frac{2}{5a} = \frac{7}{5} \]

6. **Final Answer**:
- The ratio of the distances covered during the fourth and third seconds is:
\[ \frac{s_4}{s_3} = \frac{7}{5} \]