The Motor Boat

Motor Daddy, I seriously hope you're trolling because otherwise this thread is one of the worst demonstrations of reading comprehension I've ever sen.

 

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My starting equation was 1.5 = 5/(8-S) + 5/(8+S) where S is the current speed. Then multiplying both sides by (8-S)(8+S), you get:

1.5(8-S)(8+S) = 5(8+S) + 5(8-S) or 1.5(64-S^2) = 80 or 96 - 80 = 1.5S^2 = 16 or S = 4/(sq rt 1.5) which is not same as James got nor correct, (prior to red correction).

Proof of error:
Going up stream 5 miles at speed relative to shore of 8-4 = 4 mph takes 1.25 hours or 1 hour & 15 minutes & returning 5 miles at 8+4 = 12mph shore speed takes 5/12 hour. (Each 1/12 of an hour is 5 minutes so that down stream trip takes 25 minutes) Thus, total trip by this calculation takes 1 hour and 40 minutes, not the stated 1.5 hours.

Where is my error?

A few minutes later by edit:
I always was sloppy at algebra. Error is the failure to include the red 1.5 now above. Any way, I think this approach is shorter and more clear than James´ long one.

 

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You have NO IDEA how much water the boat traveled through. All you know is that it takes 1.25hrs for the boat to travel 10km through still water. Where do you get the idea that you know how far compared to the embankment the water traveled when it was current water? Where is there ANY information about the water or how far it traveled with no boat?

 

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Sorry, but I know exactly. It's just algebra. Look at the maths. I independently came up with the same method as Billy T (look at his working to see what I did) and got the identical result.

The proof is that, as you see if you read my post, it gives an answer that completely accounts for the times, distances and speeds, and, as it happens, accords with my personal experience of this phenomenon. So there's no doubt that 3.27km/hr is the right answer.

If you know the speed of the boat through the water, the distance along the bank it went and the time it took to complete the return trip, you have enough information to work out the rest, believe me. That's the power of mathematics.

 

You're wrong! Engines operate at different torque at different RPM. The current water is placing more torque on the crank of the boat motor than it does in still water, causing it to operate at a lower RPM than it operated at when it was in still water. The force on the boat is dependent on the frontal area of the boat and the force of the water, ie, the current water is placing more DRAG on the boat than the still water, so the time is reduced upstream and increased downstream. You have no clue WHAT THE FORCES ARE! You have no clue what force the current is placing on the boat in order to reduce the boat's speed!

 

Out of morbid curiosity, if the motor boat travels at max speed of 8km/hr through still water, and the boat starts at full throttle by a bouy anchored to the bottom of the river and the boat is try to go upstream in a river that has a current of 8 km/hr how far from the bouy will it be after 1 hour? If that is unanswerable in motordaddy world, then how far from the bouy would the boat be +/- 1 km? I have to ask also do you understand how treadmills work?

 

You're barking up the wrong tree. The problem doesn't say that.

Wrong tree again. The problem doesn't say that. In fact the original problem said it slowed to 5 km/hr. As JamesR shows, Pete apparently changed the problem data, so this now calculates to to 4.73 km/hr.

No, you alone are saying that. The problem is telling you the "road" is moving at 3.27 km/hr (0r 3 km/hr in the original problem.)

No, your statement has nothing to do with the problem. Current has everything to do with the problem. For some reason you are in denial of the current.

No, you are mixing still water with moving water and calling both still water.

You really don't know? Then you should have cut to the chase and asked this first.

Do you? And do you know how to apply your answer in order to find the speed of the current?

...at a slower speed (4.73 km/hr) upstream and a faster speed (11.27 km/hr) downstream

(taking into account the change from the original problem in which upstream current was 5 km/hr)

 

The ENGINE is doing work/time (power), do you understand that?

 

No, the torque is the same, just as the torque would be the same driving across a moving aircraft carrier as it would be driving on a road.

 

The work doesn't change, the speed does.

 

The work changes because the current changes!

 

NO! That is like saying the work is the same if your car is traveling 100 MPH vs 10 MPH! Do you not understand that the faster you go the more HP it takes to go 1 MPH faster?

 

Wrong! It takes more HP to increase from 150-151 MPH than it does to increase from 1-2MPH!

 

The engine power never changes. The boat is applying a constant torque to the prop throughout this problem. That's why some folks suggested you consider that the throttle is wide open.

 

The boat can only muster 8 km/hr at full power in still water. So can you answer my question?

 

Not the work done by the motor, which is at constant throttle. The current is doing work, however, but it's constant too (the current doesn't change; the boat merely turns around, and the two work sources now add instead of subtract).

 

That's relative to the road surface which is the same whether on a road or the deck of an aircraft carrier.

 

The current remains constant. The boat changes direction and the sign changes:

Work upstream = Work of the boat - Work of the water
Work downstream = Work of the boat + Work of the water

 

NO it is not! Upstream the engine is operating at a lower RPM then it is when it travels downstream! There is more load on the crank due to the current. The engine is under a greater load, but it was already at wide open throttle, so it has to reduce RPM. You can't give it more throttle!

 

We are not measuring the current, we are measuring the BOAT!