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A body starts from rest, what is the rat...

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

A

`(7)/(5)`

B

`(5)/(7)`

C

`(7)/(3)`

D

`(3)/(7)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the ratio of the distance traveled by a body during the 4th second to the distance traveled during the 3rd second, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Problem**: The body starts from rest, which means its initial velocity \( U = 0 \). We need to find the distances traveled during the 4th and 3rd seconds. 2. **Use the Formula for Distance in nth Second**: The distance traveled during the nth second can be calculated using the formula: \[ S_n = U + \frac{a}{2} (2n - 1) \] where \( S_n \) is the distance traveled in the nth second, \( U \) is the initial velocity, \( a \) is the acceleration, and \( n \) is the second. 3. **Calculate the Distance for the 4th Second**: - For \( n = 4 \): \[ S_4 = 0 + \frac{a}{2} (2 \cdot 4 - 1) = \frac{a}{2} (8 - 1) = \frac{a}{2} \cdot 7 = \frac{7a}{2} \] 4. **Calculate the Distance for the 3rd Second**: - For \( n = 3 \): \[ S_3 = 0 + \frac{a}{2} (2 \cdot 3 - 1) = \frac{a}{2} (6 - 1) = \frac{a}{2} \cdot 5 = \frac{5a}{2} \] 5. **Find the Ratio of Distances**: - Now we need to find the ratio of the distance traveled in the 4th second to the distance traveled in the 3rd 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 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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