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 of finding the ratio of distances covered by a particle during the fourth and third seconds of its motion under uniform acceleration, we can follow these steps:

### Step 1: Understand the Problem
The particle starts from rest, which means its initial velocity \( u = 0 \). It moves with a uniform acceleration \( a \). We need to find the distances covered during the 4th second and the 3rd second.

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

### Step 3: Calculate Distance for the 4th Second
For the 4th second (\( n = 4 \)):
\[ S_4 = \frac{1}{2} a (2 \cdot 4 - 1) = \frac{1}{2} a (8 - 1) = \frac{1}{2} a \cdot 7 = \frac{7a}{2} \]

### Step 4: Calculate Distance for the 3rd Second
For the 3rd second (\( n = 3 \)):
\[ S_3 = \frac{1}{2} a (2 \cdot 3 - 1) = \frac{1}{2} a (6 - 1) = \frac{1}{2} a \cdot 5 = \frac{5a}{2} \]

### Step 5: Find the Ratio of Distances
Now, we can find the ratio of the distances covered during the 4th and 3rd seconds:
\[ \text{Ratio} = \frac{S_4}{S_3} = \frac{\frac{7a}{2}}{\frac{5a}{2}} = \frac{7}{5} \]

### Conclusion
Thus, the ratio of the distances covered during the 4th and 3rd seconds is:
\[ \frac{7}{5} \]