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Relative speed; Boats and Streams; Difficulty: Hard

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by regor60 » Mon Jul 18, 2016 6:36 am
Recognize that the ship's speed relative to the water reflects adding the speed of the water against ground and the speed of the ship against ground since both are working against the same distance, 300 meters.

Recognize that the the free floating boat is just a point on the water reflecting the water speed

Let Rate of Ship against ground = Rs
" Boat = Rb = water speed against ground
" Rsb = Rate of Ship relative to Boat > Still water speed> as if water not moving

Distance covered = 300 meters = Rsb x Time

Time = 30 seconds = 1/120 hr

Solve for Rsb = 300m/(1/120)hr = 36000meters/hr = 36kmh
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by Matt@VeritasPrep » Wed Jul 20, 2016 11:01 pm
The wording here is quite poor ("crosses a boat" is unintelligible), but I think the idea is something like
A ship of length 300 meters is traveling upstream at a constant rate on the Suez Canal, which has its own constant rate of 10 km/h. The ship encounters a boat, which is floating freely down the Canal. (Assume the boat has zero weight, and is traveling only under the power of the stream.) If the time from when the ship meets the boat to when the ship passes the boat is 30 seconds, what is the speed of the ship in still water
Let's call the ship's rate in still water s. We know that

(Ship's Rate + Boat's Rate) * 30 seconds = 300 meters

since the ship and the boat are working together to traverse the full 300 meters. (The ship is doing one part of the distance, and the boat is doing the rest, so we can add their rates together.)

We also know that the ship's rate against the current = (s - 10), and the boat's rate with the current is 10, so we have

(s - 10 + 10) * 30 seconds = 300 meters

s = 300 meters/30 seconds = 10 meters/second

To convert this to kilometers per hour, we multiply the top by 60*60 and the bottom by 100, which gives us 36 kilometers/hour.
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by Matt@VeritasPrep » Wed Jul 20, 2016 11:05 pm
That said, I'm not sure my interpretation is correct ... but it does get me to the OA, and seems like the intent of the question!
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