A body starts from rest. What is the retio of the distance traveled by the body during the `4^(th)` and `3^(rd)` seconds

Options

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

Answer

To find the ratio of the distance traveled by a body during the 4th second to the distance traveled during the 3rd second, we can use the formula for the distance traveled in the nth second of motion. The formula is given by:

\[ s_n = u + \frac{a}{2} \cdot (2n - 1) \]

where:
- \( s_n \) is the distance traveled in the nth second,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( n \) is the nth second.

Since the body starts from rest, the initial velocity \( u = 0 \).

### Step 1: Calculate the distance traveled during the 4th second (n = 4)

Using the formula for the 4th second:

\[ s_4 = 0 + \frac{a}{2} \cdot (2 \cdot 4 - 1) \]
\[ s_4 = \frac{a}{2} \cdot (8 - 1) \]
\[ s_4 = \frac{a}{2} \cdot 7 \]
\[ s_4 = \frac{7a}{2} \]

### Step 2: Calculate the distance traveled during the 3rd second (n = 3)

Using the formula for the 3rd second:

\[ s_3 = 0 + \frac{a}{2} \cdot (2 \cdot 3 - 1) \]
\[ s_3 = \frac{a}{2} \cdot (6 - 1) \]
\[ s_3 = \frac{a}{2} \cdot 5 \]
\[ s_3 = \frac{5a}{2} \]

### Step 3: Calculate the ratio of distances

Now, we can find the ratio of the distance traveled during the 4th second to the distance traveled during the 3rd second:

\[ \text{Ratio} = \frac{s_4}{s_3} = \frac{\frac{7a}{2}}{\frac{5a}{2}} \]

### Step 4: Simplify the ratio

The \( \frac{a}{2} \) 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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