Orion VIII Posted October 26, 2006 Report Share Posted October 26, 2006 Does this make sense to any of you? A fishing boat travels with the tide in the morning. It took 20 minutes to cover 6km. The return trip was against the tide and took 36 minutes. What was the still water speed of the boat and the speed of the tide? Thanks for any help! Link to comment Share on other sites More sharing options...
Board Admin A. Wong Posted October 27, 2006 Board Admin Report Share Posted October 27, 2006 No, it doesn't make any sense, but I assume it has to do with vectors going against each other. Link to comment Share on other sites More sharing options...
Orion VIII Posted October 27, 2006 Author Report Share Posted October 27, 2006 No, it doesn't make any sense, but I assume it has to do with vectors going against each other. Thanks We are doing intersections of lines and relationship of them to word problems, if it helps anyone... Link to comment Share on other sites More sharing options...
Promagstyle Posted October 27, 2006 Report Share Posted October 27, 2006 ^d(Distance) = 6Km = 6000m ^t (time) = 20minutes / 60 minutes(hour) = 0.33 secs V (velocity) = ? ^ = Delta V = ^d X ^t = 6000m X 0.33secs = 1980 m/s = Meters per second 1980meters / 1000km = 1.98 Kilometers s = 60 hours 1.98km X 60 minutes (hour) = 118.8 Km/h Answer..... 118.8 km/h..... um not sure if this is correct. I'm using OAC Physics Formulas. Link to comment Share on other sites More sharing options...
Matt Demers Posted October 27, 2006 Report Share Posted October 27, 2006 If the boat was being carried by the tide, why would it have it's own speed in this problem? Link to comment Share on other sites More sharing options...
Buurin Posted October 27, 2006 Report Share Posted October 27, 2006 With tide, boat covers 6000m in 20 minutes = 1200s. Velocity is 6000/1200 = 5m/s. Against tide, velocity is 6000/(36*60) = 25/9 m/s = ~2.77777... m/s Assuming tide and boat velocity are both constant (only the direction differs), tide's velocity is, take an average of the velocity difference here, 1/2*(5-(25/9)) = 10/9 m/s = ~ 1.1111... m/s Boat's still water velocity is thus 5 - 10/9 = 35/9 m/s = 3.888888..... m/s. Maybe 12U Physics (Disclaimer: When I took it it's still OAC ), but probably no need to go there. Link to comment Share on other sites More sharing options...
Orion VIII Posted October 27, 2006 Author Report Share Posted October 27, 2006 I figured out that the boat travels 10 km/h and the tide 4 km/h. I will report bakc whether this is right or not when i get it backed marked. Link to comment Share on other sites More sharing options...
Enviro 500 Posted October 27, 2006 Report Share Posted October 27, 2006 Does this make sense to any of you? A fishing boat travels with the tide in the morning. It took 20 minutes to cover 6km. The return trip was against the tide and took 36 minutes. What was the still water speed of the boat and the speed of the tide? Thanks for any help! Let's say a = speed of boat (in still water) b = speed of tide 1/3a + b = 6 3/5a- b = 6 a + 3b = 18 3a-5b = 30 3a + 9b = 54 -14b = -24 b = 1.71 a + 3 x 1.71 = 18 a + 5.13 = 18 a = 12.87 CHECK (Standard procedure for me ) 1/3 x 12.87 + 1.71 = 6 3/5 x 12.87 - 1.71 = 6.012 See if that works for you or not...........if you want fractions, a = 90/7, b = 12/7 Dave Link to comment Share on other sites More sharing options...
Mike Posted October 27, 2006 Report Share Posted October 27, 2006 I figured out that the boat travels 10 km/h and the tide 4 km/h. I will report bakc whether this is right or not when i get it backed marked. Here is the simple answer to this problem: 20 min for 6 km is 18 km/h 36 min for 6 km is 10 km/h let x be the speed of the boat and y be the speed of the tide so x + y = 18 and x-y = 10 this implies that x = 14 and y = 4, so the speed of the boat is 14 km/h and speed of the current is 4 km/h. Link to comment Share on other sites More sharing options...
MaT Posted October 27, 2006 Report Share Posted October 27, 2006 That's the closest one IMO. Link to comment Share on other sites More sharing options...
Enviro 500 Posted October 27, 2006 Report Share Posted October 27, 2006 Here is the simple answer to this problem: 20 min for 6 km is 18 km/h 36 min for 6 km is 10 km/h let x be the speed of the boat and y be the speed of the tide so x + y = 18 and x-y = 10 this implies that x = 14 and y = 4, so the speed of the boat is 14 km/h and speed of the current is 4 km/h. This doesn't seems to work out if I reverse your equation: 14 / 3 + 4 would give you 8.67.........since the speed of the tide is a constant speed, so it should not be divided by 3.........let's say the tide travels at 2km/h in 20 minutes, then it would continue to travel at 2km/h after 1/2/3 hours. If I go 14/3 - 4, that would give 0.67. Dave Link to comment Share on other sites More sharing options...
Orion V Posted October 27, 2006 Report Share Posted October 27, 2006 ^d(Distance) = 6Km = 6000m ^t (time) = 20minutes / 60 minutes(hour) = 0.33 secs V (velocity) = ? ^ = Delta V = ^d X ^t = 6000m X 0.33secs = 1980 m/s = Meters per second 1980meters / 1000km = 1.98 Kilometers s = 60 hours 1.98km X 60 minutes (hour) = 118.8 Km/h Answer..... 118.8 km/h..... um not sure if this is correct. I'm using OAC Physics Formulas. Sometimes in physics, you have to think about the numbers for a second; would it make sense for a boat (on earth in this world) to travel at 1980m/s =7128km/h. Now I don't know about you but I've never heard of any car that can go at that speed. LOL. More like a spaceship. Just something for you to think about. Link to comment Share on other sites More sharing options...
Mike Posted October 27, 2006 Report Share Posted October 27, 2006 This doesn't seems to work out if I reverse your equation: 14 / 3 + 4 would give you 8.67.........since the speed of the tide is a constant speed, so it should not be divided by 3.........let's say the tide travels at 2km/h in 20 minutes, then it would continue to travel at 2km/h after 1/2/3 hours. If I go 14/3 - 4, that would give 0.67. Dave It works out perfectly. When the boat is traveling with the current the combined speed is 18 (14+4) km'h so 6 km is covered in 1/3 hour or 20 min. When the boat is travelling against the current, the combined speed is 10 (14-4) km/h so the 6 km is covered in 6/10 hours or 36 min. Link to comment Share on other sites More sharing options...
Enviro 500 Posted October 27, 2006 Report Share Posted October 27, 2006 It works out perfectly. When the boat is traveling with the current the combined speed is 18 (14+4) km'h so 6 km is covered in 1/3 hour or 20 min. When the boat is travelling against the current, the combined speed is 10 (14-4) km/h so the 6 km is covered in 6/10 hours or 36 min. if it's asking for 1 hour, that works perfectly, yes you are right on that. But, you can't take 18 and divide that by 3 since it's only the velocity of the boat that's being divided by 3, the speed of the current stays the same as "4" What you're doing is this: 14 + 4 = 18 14/3 + 4/3 = 18/3 = 6 But what is supposed to be done is this: 14 + 4 = 18 14/3 + 4 = 9 Same method applied to the second equation Dave Link to comment Share on other sites More sharing options...
Mike Posted October 27, 2006 Report Share Posted October 27, 2006 if it's asking for 1 hour, that works perfectly, yes you are right on that. But, you can't take 18 and divide that by 3 since it's only the velocity of the boat that's being divided by 3, the speed of the current stays the same as "4" What you're doing is this: 14 + 4 = 18 14/3 + 4/3 = 18/3 = 6 But what is supposed to be done is this: 14 + 4 = 18 14/3 + 4 = 9 Same method applied to the second equation Dave Listen, your equations are wrong! Its as simple as that. So substituting anything into them is equally wrong. The reason they are wrong is that the water isn't still it is also travelling at a speed so while the boat covers a/3 in 20 minutes the water covers b/3 so you get a/3+b/3=6 or a+b=18. Similarly with the second equation Link to comment Share on other sites More sharing options...
Orion VIII Posted October 27, 2006 Author Report Share Posted October 27, 2006 Listen, your equations are wrong! Its as simple as that. So substituting anything into them is equally wrong. The reason they are wrong is that the water isn't still it is also travelling at a speed so while the boat covers a/3 in 20 minutes the water covers b/3 so you get a/3+b/3=6 or a+b=18. Similarly with the second equation Your equations are, IMO, correct. Let x be boat stillwater speed. Let y tide speed. Let s be total speed. Assume boat travels one constant speed and the tide speed is constant. In AM: x+y=1 In PM: x-y=10 Eventually I got x=10 y=4 Hopefully what I (we) did is right. I hope to find out on Monday. Thanks everyone! Link to comment Share on other sites More sharing options...
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