The Motor Boat

Discussion in 'Physics & Math' started by Motor Daddy, May 31, 2013.

Thread Status:
Not open for further replies.
  1. Motor Daddy Valued Senior Member

    Messages:
    5,425
    Pete restated a problem in another thread and I have a question about that statement. The statement was:


    If a motor boat travels at 8km/hr for 90 minutes, the boat would have traveled 6km each way, for a total of 12 km in 90 minutes, which is a speed of 8km/hr.

    If the boat travels for 5 kilometers each leg of the trip, then the boat would have traveled a total of 10 kilometers in 90 minutes, and that would be a speed of 6.666 km/hr.

    If the boat traveled at 8km/hr in still water for 5 km each leg of the trip, then it traveled 10 kilometers at the speed of 8km/hr, and therefore traveled for 1.25 hours, not 1.5 hours.



    The problem states that the boat traveled at 8km/hr in still water. If it did travel at 8km/hr then it only took 1.25 hours to travel 10 kilometers. If it traveled at 8km/hr for 90 minutes then it would have traveled 12 kilometers round trip.

    Which one is it? Did the boat travel for 90 minutes at 6.666km/hr for a total distance of 10 kilometers, or did it travel at 8km/hr for 75 minutes (1.25 hours) which is a total distance traveled of 10 kilometers?
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. Neddy Bate Valued Senior Member

    Messages:
    2,548
    That speed only holds true for still water. When there is a current in the water, the speed is different. Hence the question, "What is the speed of the current in the water?"


    Correction, that would be an average speed of 6.666 km/hr. The speed upstream can be different than the speed downstream, and yet the average speed can still be 6.666 km/hr. The question still is, "What is the speed of the current in the water?"

    You do understand that "current" means the water is not still, but flowing like a river, correct?
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Motor Daddy Valued Senior Member

    Messages:
    5,425

    I am not talking about the current speed, I am talking about the stated facts of the problem. Pete says the boat travels 5 kilometers each direction for a total of 90 minutes. If so, then the boat in still water could not have traveled at 8km/hr, it only traveled at 6.666km/hr for 90 minutes.




    I am asking a question about the stated fact that the boat travels at 8km/hr in still water. How could that be if the travel time is 90 minutes and the distance is 5km each leg?
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Neddy Bate Valued Senior Member

    Messages:
    2,548
    There is no "still water" in the problem. The water is flowing with a constant CURRENT. The "still water" benchmark is just for reference.
     
  8. Motor Daddy Valued Senior Member

    Messages:
    5,425
    Then there is no 8km/hr still water speed. How was the still water speed of 8km/hr known? By timing how much time it takes to travel 5km upstream and 5km downstream in still water?
     
  9. Neddy Bate Valued Senior Member

    Messages:
    2,548
    8km/hr is what the speed of the boat would be if the current speed was zero. But the current speed is not zero in the problem.
     
  10. eram Sciengineer Valued Senior Member

    Messages:
    1,877
    Motor Daddy, this is the kind of misconception that a 10 year old might have.

    Please Register or Log in to view the hidden image!




    You state that 1 hr 30 mins conflicts with 5km. Nope.

    Why? Both your scenarios are wrong and you have forgotten about The third variable, the speed of the current.


    Here is the correct scenario: if we throw Motor Daddy in the river and he doesn't know how to swim, he will drift at the speed of the current.

    Please Register or Log in to view the hidden image!



    He will see the boat move at 8km/hr. The boat will move off from a starting marker, reach an ending marker, then move back to the starting marker.

    Both markers will be seen drifting at a certain speed. Now we introduce the two variables, the distance between the markers (5km), and the duration (1 hr 30 mins).

    This will let us find the third variable, the speed of the current.


    Motor Daddy, you should be ashamed of yourself.

    Please Register or Log in to view the hidden image!

     
  11. Motor Daddy Valued Senior Member

    Messages:
    5,425
    How do you know the speed of the boat in still water?
     
  12. eram Sciengineer Valued Senior Member

    Messages:
    1,877
    8km/hr was given, you dumbo.

    Please Register or Log in to view the hidden image!

     
  13. Aqueous Id flat Earth skeptic Valued Senior Member

    Messages:
    6,152
    The problem stated that the boat was subjected to a current.

    No, the problem said the total travel time was 90 min. it says that the boat turned around and came back after going upstream at a speed of 5 km/hr.

    The problem said the speed in still water was 8 km/hr. You can't alter that since it was given in the problem.

    Pete may have rounded down to 5 km. But it makes no sense to say the boat traveled at the same speed upstream as downstream.

    Note, there are three boat speeds in this problem, not two:

    (1) 8 km/hr - still water
    (2) 5 km/hr - upstream - the boat travels slower than it did in still water due to working against a current of 3 km/hr
    (3) 11 km/hr - downstream - the boat travels faster than it did in still water due to working with the current of 3 km/hr


    To get the average speed, sum and divide by two: (5 + 11) / 2 = (16 / 2) = 8 km/ hr. in other words, on average you lose no time driving into the current because you recover the lost time on the return leg, with the help of the same current that slowed you down in the first place.

    Maybe that was rounded down. The actual distance is 5.15625 km.

    (a) 5 km/hr x t[sub]u[/sub] = 11 km/hr x t[sub]d[/sub] --> t[sub]d[/sub] = (5/11) x t[sub]u[/sub]

    (b) t[sub]u[/sub] + t[sub]d[/sub] = 1.5 hr --> t[sub]u[/sub] = 1.5 hr - t[sub]d[/sub]

    substitute

    t[sub]u[/sub] = 1.5 hr - (5/11) x t[sub]u[/sub]

    t[sub]u[/sub] (16/11) = 1.5 hr = 90 min

    t[sub]u[/sub] = 1.03125 hr (61.875 min) --> t[sub]d[/sub] = 1.5 - 1.03125 = 0.46875 hr (28.125 min)

    5 km/hr * 1.03125 hr = 5.15625 km (upstream), 11 km/hr * 0.46875 hr = 5.15625 km (downstream)

    61.875 min + 28.125 min = 90 min total

    5.15625 km + 5.15625 km = 10.3125 km total
     
  14. Motor Daddy Valued Senior Member

    Messages:
    5,425
    I'll make it a little more clear.

    I want to measure the speed of the boat in still water to get a baseline as to how much time it takes for the boat to travel 10 km in still water. I have a 5km long test bed to travel up and back in still water and record the time. How much time does it take for the boat to travel the 5km up and 5km back in still water?

    According to Pete, the boat travels at 8km/hr in still water, so it takes 1.25 hours to travel the 10 km in still water, right?
     
  15. Motor Daddy Valued Senior Member

    Messages:
    5,425
    Funny how everyone talks crap while they think they are right, but when they learn they are wrong they are dead silent.

    The crickets have arrived!! (Again!)
     
  16. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    Call the speed of the boat in still water v1, and the speed of the current is v2, both measured relative to the bank of the river. When going upstream, the boat's speed relative to the bank is therefore v1 - v2, and when going downstream it is v1 + v2.

    From the data in the question we have

    5 km = (v1 - v2)t_up ... (1)
    5 km = (v1 + v2)t_down ... (2)
    t_up + t_down = 1.5 hours ... (3)

    where t_up and t_down are the time spent travelling upstream and downstream, respectively.

    From (1) and (2) we find

    t_up = 5 km / (v1 - v2)
    t_down = 5 km / (v1 + v2)

    And (3) then gives

    [5 km / (v1 - v2)] + [5 km / (v1 + v2)] = 1.5 hours

    or

    1/(v1 - v2) + 1/(v1 + v2) = (1.5/5) (hours/km).

    A little algebra...

    [v1 + v2 + v1 - v2]/(v1^2 - v2^2) = (1.5/5) (hours/km)

    or

    [v1^2 - v2^2] / 2 v1 = (5/1.5) (km/hr)

    [v1^2 - v2^2] = (10/1.5) v1

    v2^2 = v1^2 - (10/1.5)v1

    v2 = sqrt(v1^2 - (10/1.5)v1)

    But we are told that v1 = 8 km/hr. Therefore

    v2 = sqrt(8^2 - (80/1.5)) = sqrt(10.67) = 3.27 km/hr.

    The speed of the current is 3.27 km/hr.
     
  17. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    Motor Daddy:

    I see you still haven't come to grips with frames of reference. (How many years has it been now?)

    See my solution above. The boat never travelled at 8 km/hr relative to the bank of the river, because you have to add or subtract the speed of the water the boat is moving in to get the boat's speed relative to the bank. Relative to the water, on the other hand, the boat travelled at 8 km/hr for the whole trip.

    In terms of distance, it travelled 12 km relative to the water, but only 10 km relative to the bank.

    You take your boat to some still water, crank up the engine and time how long it takes the boat to cover a known distance.

    Right.

    So, all clear now?
     
  18. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    Yes, that is quite obvious.
     
  19. Motor Daddy Valued Senior Member

    Messages:
    5,425
    No, I'm not clear now, and I'm not trying to be difficult.

    James, You bought a boat and measured the speed in still water to be 8km/hr. How did you determine the boat traveled 8km/hr in still water? Did you measure the time it took for the boat to travel a distance, or did you just claim that the boat travels 8km/hr in still water without having measured anything? How do you know the boat travels 8km/hr in still water?
     
  20. Motor Daddy Valued Senior Member

    Messages:
    5,425
    You're digging yourself in a deep hole, James. The speed of the boat relative to the bank in still water is 8km/hr, period. You already measured it and Pete took your word for it. The boat traveled 10 kilometers relative to the bank in 1.25 hours.
     
  21. Motor Daddy Valued Senior Member

    Messages:
    5,425
    Right. So where does the 90 minutes come into play? If the boat traveled 10 km for 90 minutes instead of the 75 minutes, then the speed is 6.666km/hr, NOT 8km/hr.

    It's IMPOSSIBLE for a boat to travel 10 kilometers in 90 minutes at the speed of 8km/hr, IMPOSSIBLE!
     
  22. Motor Daddy Valued Senior Member

    Messages:
    5,425
    You call me up and tell me you bought a boat! Being a reasonable person you want to find out how fast it travels in still water at wide open throttle. Lucky for you I have a test facility that is at rest, just like the embankment in Einstein's chapter 9, but instead of the train, it's a boat, your boat, and we are going to time you to see how much time it takes for the boat to travel 10 km along the boat embankment so we know the top speed of the boat in still water. You fire it up and go WOT as the clock starts. You cross the finish line in 1.25 hours, so the speed of the boat is 8km/hr along the embankment. The time is 1.25 hours for you to travel 10 km, NOT 1.5 hours. If it took you 1.5 hours then your boat is slower at 6.666km/hr.

    Which is it, James, does your boat travel 10km along the embankment in 1.25 hours at the speed of 8km/hr, or does it take 90 minutes for the boat to travel 10 km along the bank at the speed of 6.666km/hr? You tell me?
     
  23. ash64449 Registered Senior Member

    Messages:
    795
    Motor Daddy, I don't understand why are you so confused. Leave it.. It is just a question that can be answered by simple algebra.

    Don't you agree with the algebra provided by James R? He is right. And everyone got the same answer.
     
Thread Status:
Not open for further replies.

Share This Page