Motor Daddy
Valued Senior Member
Are you saying that if your car travels down the road for 90 minutes at the speed of 8km/hr that you will have traveled 10km? Is that what you are saying?
The Motor Boat
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Are you saying that if your car travels down the road for 90 minutes at the speed of 8km/hr that you will have traveled 10km? Is that what you are saying?
How come you have arrived at that sort of conclusion? Do simple algebra for Speed.
If my car(i don't have!!) traveled at the 8km/h at the constant speed and there is no opposing force or any other forces involved to make my speed decrease,
Then Distance that i have traveled should be 12 km by:
S=D/T.
D=S*T.
Time is 3/2 hours.(90 minutes.)
SO D=8*(3/2)
Which 12 km. And you must understand that i have arrived at the answer through algebra.
Motor boat question also through algebra. But there is current that flows against the boat and also in way-back trip it helps to increase the speed. so you need to include them to for total time.
What is that you are confusing? It is simple. I have done this problem Almost 2 years ago. When i was in 9 std.
Motor Daddy
Valued Senior Member
If that is what you and everyone is saying then I have a fast car, and it travels 100 miles in one hour at a speed of 82.456 MPH! It doesn't travel 100 MPH, it travels 82.456 MPH for 1 hour and is 100 miles down the road from where it started!
I get great gas mileage with your new math! (rolls eyes)
Motor Daddy
Valued Senior Member
Not according to Pete. Pete says that the distance is 10 km, the speed is 8km/hr, and the time is 90 minutes. Where do you get 12km when Pete says it's 10km?
How is that If you travel at 82.456 MPH for 1 hour and reach at a distance of 100 miles and i am still surprised how did you got that from my simple math!!!!
S=D/T
IF speed is 82.456 MPH and it travels at the time period for 1 hour,then
S*T=D
SO,
82.456*1=82.456 miles.
How did you get the answer that it travels 100 miles in 1 hour?
Motor Daddy
Valued Senior Member
The same way you get 8km/hr for 90 minutes is 10 km. You have funny math so I can use it too, no?
Elementary Math. what is the answer for 8*(3/2)?????
10 km or 12 km??? And too i got that from simple algebra for calculating speed....
You are confusing something for sure...
Who got 10 km??? I got 12 km...
Explain elaborately how did you get 100 miles through equations so that we can correct you.
Motor Daddy
Valued Senior Member
Pete. Pete says that if I test my boat in STILL WATER that it travels 8km/hr for 90 minutes and travels 10 kilometers.
Ok. I think i understood your confusion.
Boat has strong tendency to travel 8km/h. But due to opposing force,it travels 8km/h-3.27km/h.
So boat does not travel 8km/h but 8km/h-3.27km/h.
When returning backwards,It travels at the speed of 8km/h+3.27km/h and not 8km/h. Got that?
EDIT: If the water is still(a case),it doesn't have any speed. So Time taken to reach back for boat(round-trip)in two cases in different.
Boat has strong tendency to travel 8km/h. But due to opposing force,it travels 8km/h-3.27km/h.
So boat does not travel 8km/h but 8km/h-3.27km/h.
When returning backwards,It travels at the speed of 8km/h+3.27km/h and not 8km/h. Got that?
EDIT: If the water is still(a case),it doesn't have any speed. So Time taken to reach back for boat(round-trip)in two cases in different.
Leave it. Simple algebra from me in terms of speed. Leave rest.
Motor Daddy
Valued Senior Member
Boat doesn't have tendency to do squat! We measured the boat in order to find the still water speed. It traveled 10km in 1.25 hours! That is 8km/hr in still water.
Motor Daddy
Valued Senior Member
Simple algebra says 8km/hr for 90 minutes is 12km, not 10 km!
Motor Daddy
Valued Senior Member
The current has NOTHING to do with the time it takes for the boat to travel 10 km in still water!
Motor Daddy:
Obviously you didn't understand my previous posts in this thread. I thought they were fairly clear.
I told you before. At least do me the courtesy of reading what I write. Here it is again: you take the boat. You put it in still water. You time how long it takes the boat to go a certain distance. You divide the distance by the time taken, and that's your 8 km/hr.
Got any problem with that?
I can't see any hole in that.
Who said anything about 1.5 hours?
The only time 1.5 hours came up was in a problem that didn't involve still water. In fact, it explicitly involved a flowing current. Did you miss that part of the problem?
Go back and read the problem. Go back and read my comprehensive solution to that problem - the one with the flowing river, not the still water.
Is the boat moving in still water, or is the river flowing? You tell me.
Is the road still, or is it flowing like a.... like a river perhaps?
Pete was talking about a flowing current. You're talking about still water. Two different situations can give two different answers.
I know you struggle with this kind of thing, but the entire solution to this problem depends on you appreciating that a boat's speed relative to a flowing river is not the same as its speed relative to a stationary river bank.
If you ever get yourself over this difficult conceptual hurdle, then you might just, one day, be ready to start studying Einstein's relativity.
Where does he say that? Link me to the post.
Right!
Right!
Now, how about solving that problem, then? You've got to first base. Now do the rest.
Obviously you didn't understand my previous posts in this thread. I thought they were fairly clear.
I told you before. At least do me the courtesy of reading what I write. Here it is again: you take the boat. You put it in still water. You time how long it takes the boat to go a certain distance. You divide the distance by the time taken, and that's your 8 km/hr.
Got any problem with that?
I can't see any hole in that.
Who said anything about 1.5 hours?
The only time 1.5 hours came up was in a problem that didn't involve still water. In fact, it explicitly involved a flowing current. Did you miss that part of the problem?
Go back and read the problem. Go back and read my comprehensive solution to that problem - the one with the flowing river, not the still water.
Is the boat moving in still water, or is the river flowing? You tell me.
Is the road still, or is it flowing like a.... like a river perhaps?
Pete was talking about a flowing current. You're talking about still water. Two different situations can give two different answers.
I know you struggle with this kind of thing, but the entire solution to this problem depends on you appreciating that a boat's speed relative to a flowing river is not the same as its speed relative to a stationary river bank.
If you ever get yourself over this difficult conceptual hurdle, then you might just, one day, be ready to start studying Einstein's relativity.
Where does he say that? Link me to the post.
Right!
Right!
Now, how about solving that problem, then? You've got to first base. Now do the rest.
Motor Daady: I´m sure james has it correct in post 12, but it is longer than need be proof.
Going up stream 5 miles against a current of S mph will take 5/(8-S) hours will it not? And the return trip going down stream will take 5/(8+S) will it not? We are told the total trip took 1.5 hours.
I.e. solve this equation: 1.5 = 5/(8-S) + 5/(8+S) for S.
Use trial and error if your not up to that algebraically. Here is a hint for the algebraic approach. Get rid of the fractions by multiplying both sides of equation by:
(8-S)(8+S). Then equation becomes: 1.5(8-S)(8+S) = 5(8+S) + 5(8-S).
The left side will produce a quadratic term (one with S^2) so after you have it reduced to standard form: aS^2 + bS + c = 0 apply the "quadratic formuale" to get S.
I´m just guessing but nearly sure your answer is the one with the + sign. (the solution with the negative sign is going to be a negative time (must arrive back at the start before you start the trip, if that were possible, I bet.)
Going up stream 5 miles against a current of S mph will take 5/(8-S) hours will it not? And the return trip going down stream will take 5/(8+S) will it not? We are told the total trip took 1.5 hours.
I.e. solve this equation: 1.5 = 5/(8-S) + 5/(8+S) for S.
Use trial and error if your not up to that algebraically. Here is a hint for the algebraic approach. Get rid of the fractions by multiplying both sides of equation by:
(8-S)(8+S). Then equation becomes: 1.5(8-S)(8+S) = 5(8+S) + 5(8-S).
The left side will produce a quadratic term (one with S^2) so after you have it reduced to standard form: aS^2 + bS + c = 0 apply the "quadratic formuale" to get S.
I´m just guessing but nearly sure your answer is the one with the + sign. (the solution with the negative sign is going to be a negative time (must arrive back at the start before you start the trip, if that were possible, I bet.)
Motor Daddy
Valued Senior Member
I understand perfectly clear what you are saying, but I am telling you that you are mixing apples and oranges and calling it oranges. It's not oranges, James, it's a mixture of apples and oranges.
I read what you wrote and I understand what you are saying, do you understand what I am saying? That is the question.
What distance did you travel in the boat relative to the embankment??
What time, the "I started the stop watch at the starting point on the embankment, and I stopped the stopwatch when the boat was 10km from that point" time?
I have no problem with you measuring off 10 kilometers on the embankment, starting the boat at one end of the distance, starting a stop watch and traveling at a constant speed until you travel a distance of 10km and then you stop the stop watch. I have absolutely no problem with that (assuming it is an Einstein embankment (which I can prove is at a zero velocity)).
What I do have a problem with is you pretending that the velocity of the boat is then relative to the water, when you did ALL the measurements according to the embankment. I have serious problems with that.
Right, who did? It takes 1.25 hours for the boat to travel 10 km along the embankment 5km each way, assuming zero acceleration at the turn around, unless of course you want to say there is acceleration at the turn around, and in that case I can do that too, so you're in luck!!
No, I understood it perfectly clear. If the boat traveled 10km in 1.25 hours then that is the time we are testing, 1.25 hours. If your boat in current does not complete the task in 1.25 hours then you're SOL!
I read it.
The only test of the boat you've done is the boat traveling in still water. I want to remind you, that little boat motor of yours was giving it everything it had against the still water. What do you think will happen when the boat motor gives it everything it has against a current? You've NOT performed that test.
Flowing like a river compared to what? I know, do you? James, if you remember, I showed you the embankment and the train in motion, measured using light. You remember that, right, that I put the train AND the embankment in motion and showed you the numbers? I hope you didn't already forget that, because each the water, and the boat are in motion in the preferred frame. You know that, right?
The boat's speed was measured compared to the embankment, remember? The boat's speed was NOT measured relative to the water. Do you understand what you are saying??
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Motor Daddy
Valued Senior Member
I wouldn't bet the farm on it.
The boat traveled 10km in 1.25 hours in still water, so how far did it travel in current in 1.5 hours?
exchemist
Valued Senior Member
12km, of course, Motor Daddy. But the point is, it travelled through 12km of water.
Try looking at it this way:-
It travelled upstream at a speed, relative to the bank of 8-3.27 = 4.73 km/hr. So it took 5/4.73 = 1.06hrs to go up. During this time it had to travel through 1.06 x 8 = 8.48km of water.
It then turned round and went downstream. This time it travelled, relative to the bank, at 8+3.27 = 11.27 km/hr. So it took 5/11.27 = 0.44hrs to return. It travelled through 8 x 0.44 = 3.52km of water during this time.
Total time was thus 1.06 + 0.44 = 1.5hrs
And the total distance through the water it travelled was 8.48 + 3.52 = 12km.
Hey presto.
To be fair, if you had sculled on the Thames for 40 years as I have, I think you would be able to visualise this more easily. The stronger the current, the longer the time required for a return trip, because the time spent going against the current is always longer than the time spent going with it. (In the limiting case, when the current is the same as the boat speed it takes an infinitely long time to get upstream, because you are sculling on the spot, relative to the bank! It's a horrible feeling.)
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