\(\sqrt{\frac{3}{2}}\)
Motorboat Speed Ratio Calculation
This problem involves calculating the ratio between a motorboat's speed in still water and the river's flow speed, based on changes in travel time.
Let \( v_b \) be the speed of the motorboat in still water, and \( v_s \) be the speed of the river flow. Let \( d \) be the distance between points A and B.
Original Journey Time Calculation
The total time \( T \) for a round trip (A to B and back to A) depends on speeds downstream and upstream.
- Speed downstream (with the current): \( v_b + v_s \)
- Speed upstream (against the current): \( v_b - v_s \)
- Time downstream: \( t_{down} = \frac{d}{v_b + v_s} \)
- Time upstream: \( t_{up} = \frac{d}{v_b - v_s} \)
- Total original time: \( T = t_{down} + t_{up} = \frac{d}{v_b + v_s} + \frac{d}{v_b - v_s} \)
- Combine terms: \( T = d \frac{(v_b - v_s) + (v_b + v_s)}{(v_b + v_s)(v_b - v_s)} = d \frac{2v_b}{v_b^2 - v_s^2} \)
Modified Journey Time Calculation
The motorboat's speed in still water is doubled to \( 2v_b \).
- New speed downstream: \( 2v_b + v_s \)
- New speed upstream: \( 2v_b - v_s \)
- New time downstream: \( t_{down, new} = \frac{d}{2v_b + v_s} \)
- New time upstream: \( t_{up, new} = \frac{d}{2v_b - v_s} \)
- Total new time: \( T_{new} = t_{down, new} + t_{up, new} = \frac{d}{2v_b + v_s} + \frac{d}{2v_b - v_s} \)
- Combine terms: \( T_{new} = d \frac{(2v_b - v_s) + (2v_b + v_s)}{(2v_b + v_s)(2v_b - v_s)} = d \frac{4v_b}{(2v_b)^2 - v_s^2} = d \frac{4v_b}{4v_b^2 - v_s^2} \)
Relating the Times
The problem states the new round trip time \( T_{new} \) is 20% of the original time \( T \).
\( T_{new} = 0.20 \times T = \frac{1}{5} T \)
Substitute the expressions for \( T \) and \( T_{new} \):
\( d \frac{4v_b}{4v_b^2 - v_s^2} = \frac{1}{5} \left( d \frac{2v_b}{v_b^2 - v_s^2} \right) \)Cancel \( d \) and \( v_b \) (since \( v_b \neq 0 \)):
\( \frac{4}{4v_b^2 - v_s^2} = \frac{2}{5(v_b^2 - v_s^2)} \)Solving for the Speed Ratio
Cross-multiply to solve for the relationship between \( v_b \) and \( v_s \):
\( 4 \times 5(v_b^2 - v_s^2) = 2 \times (4v_b^2 - v_s^2) \) \( 20(v_b^2 - v_s^2) = 8v_b^2 - 2v_s^2 \) \( 20v_b^2 - 20v_s^2 = 8v_b^2 - 2v_s^2 \)Rearrange the equation:
\( 20v_b^2 - 8v_b^2 = 20v_s^2 - 2v_s^2 \) \( 12v_b^2 = 18v_s^2 \)Find the ratio \( \frac{v_b^2}{v_s^2} \):
\( \frac{v_b^2}{v_s^2} = \frac{18}{12} = \frac{3}{2} \)Take the square root to find the ratio \( \frac{v_b}{v_s} \):
\( \frac{v_b}{v_s} = \sqrt{\frac{3}{2}} \)Conclusion
The actual speed of the motorboat is \( \sqrt{\frac{3}{2}} \) times higher than the speed of the river flow.
Important Questions from Partial Speed
A train having length 210 metres takes 25 seconds to cross a 540 metres long bridge. How much time will the train take to cross a 630 metres long bridge?
A person X from a place A and another person Y from a place B set out at the same time to walk towards each other. The places are separated by a distance of 15 km. X walks with a uniform speed of 1.5 km / hr and Y walks with a uniform speed of 1 km / hr in the first hour, with a uniform speed of 1.25 km / hr in the second hour and with a uniform speed of 1.5 km / hr in the third hour and so on.
Which of the following is / are correct?
1. They take 5 hours to meet.
2. They meet midway between A and B.
Select the correct answer using the code given below:
The speed of a train is 120 kmph. What is the distance covered by it in 15 minutes?
I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways ?
Karan had covered two third of a certain distance when his car had a breakdown. He parked it and covered the remaining distance on foot. His time of travel on foot was 9 times his time of travel on car. What is the ratio of his walking speed with respect to his car’s speed?
