Cindy in FL. Posted September 20, 2012 Share Posted September 20, 2012 I cannot understand this problem at all. Can anyone explain it to me? A plane flies straight for 672.1 km and then turns around and heads back. The plane then lands at an airport that is only 321.9 km away from where the pilot turned around. If the plane's average velocity over the entire trip was 42 m/sec, how much time did the entire trip take? The answer they give uses the displaced distance (350.2 km) to calculate the time. Why doesn't it use the distance travelled? Didn't it take time to do that? Help! Cindy Quote Link to comment Share on other sites More sharing options...
regentrude Posted September 20, 2012 Share Posted September 20, 2012 I cannot understand this problem at all. Can anyone explain it to me?A plane flies straight for 672.1 km and then turns around and heads back. The plane then lands at an airport that is only 321.9 km away from where the pilot turned around. If the plane's average velocity over the entire trip was 42 m/sec, how much time did the entire trip take? The answer they give uses the displaced distance (350.2 km) to calculate the time. Why doesn't it use the distance travelled? Didn't it take time to do that? The issue here is the difference between velocity and speed. The average velocity is defined as the displacement per time. Displacement is final position minus initial position: velocity = (xf-xi)/t Displacement is not the same thing as distance traveled! Average speed would be distance traveled per time, but that is not what the problem asked you to calculate. Quote Link to comment Share on other sites More sharing options...
Cindy in FL. Posted September 20, 2012 Author Share Posted September 20, 2012 The answer in the book has you calculating the displacement and then multliplying that by the speed given and that is the answer. That does not make any sense to me at all. Can anyone explain it to me in very easy to understand words? Cindy Quote Link to comment Share on other sites More sharing options...
KAR120C Posted September 20, 2012 Share Posted September 20, 2012 (edited) The answer in the book has you calculating the displacement and then multliplying that by the speed given and that is the answer. That does not make any sense to me at all. Can anyone explain it to me in very easy to understand words? If you had to get from your house to the store a block away, and it took you an hour to get there, you'd have gone one block per hour. The fact that you took so long because on the way you stopped to chat with a neighbor, took a wrong turn through someone's yard, backtracked to find your keys.. whatever... doesn't change the fact that it took an hour to get to your destination a block away. Now you could have done all those things really quickly, running all over town to eventually get to the store, but your goal was just the one block distance and it took an hour to get there. Basically when you're talking about velocity and displacement as technical terms in physics, you're dealing with two points and and ideal route between them (a straight line). So the average velocity in that one direction refers to just that imaginary straight line and how long it took to get from one end of it to the other, and nevermind what circuitous route was actually taken. The math part.... it shouldn't be displacement times velocity.... Velocity is displacement/ time (like meters/ second), so if you have displacement and velocity, to find time you'd divide the displacement by the velocity, not multiply.... Edited September 20, 2012 by KAR120C accidentally hit post too soon... Quote Link to comment Share on other sites More sharing options...
Cindy in FL. Posted September 20, 2012 Author Share Posted September 20, 2012 Erica, I understand what you are saying. But, in the problem given, it's a straight line there and then you only come part way back. The answer in the book has you only using the distance from the starting point to the point the plane landed on its way back, not the entire trip. They want to know how much time the entire trip took. I don't see how you get the right answer if you are only using the distance from take off to the landing point. You are not using the distance you flew out and back. Cindy Quote Link to comment Share on other sites More sharing options...
Arcadia Posted September 20, 2012 Share Posted September 20, 2012 The answer in the book has you calculating the displacement and then multliplying that by the speed given and that is the answer. The question gave the velocity and not the speed. velocity = displacement/time 42 m/sec = 350.2 km/time time = 350.2 km / 42m/sec = 8,338 sec = 139 minutes = 2 hrs 19 min Quote Link to comment Share on other sites More sharing options...
KAR120C Posted September 20, 2012 Share Posted September 20, 2012 Erica, I understand what you are saying. But, in the problem given, it's a straight line there and then you only come part way back. The answer in the book has you only using the distance from the starting point to the point the plane landed on its way back, not the entire trip. They want to know how much time the entire trip took. I don't see how you get the right answer if you are only using the distance from take off to the landing point. You are not using the distance you flew out and back. Cindy It's really like you were saying the pilot needed to get from Detroit to Cincinnati, but he flew to Nashville by accident and then had to come back. The time it took to get to Cincinnati may have been ridiculously long, but he doesn't get credit for his detour. His intended route was Detroit to Cincinnati, and it took a long time to get there... so his average velocity was low even if his speed (all the way to Nashville and back to Cincinnati) was pretty high. Quote Link to comment Share on other sites More sharing options...
Cindy in FL. Posted September 20, 2012 Author Share Posted September 20, 2012 Thanks for taking the time. I understand now that you just have to use the formula and get the answer. (Even if it seems a little squirrely to me! :lol:) Cindy Quote Link to comment Share on other sites More sharing options...
KAR120C Posted September 20, 2012 Share Posted September 20, 2012 Thanks for taking the time. I understand now that you just have to use the formula and get the answer. (Even if it seems a little squirrely to me! :lol:) Cindy It's not so much that you have to blindly use a formula, just that you have to take the perspective of the people from Cincinnati who are waiting for their plane to get in. All they care about is that it took way too long to get there - not where he went on the way. Quote Link to comment Share on other sites More sharing options...
Cindy in FL. Posted September 20, 2012 Author Share Posted September 20, 2012 I see what you're saying. If they had that perspective in the question, I might have gotten it more easily. I don't think Physics and I are going to be BFFs! Cindy Quote Link to comment Share on other sites More sharing options...
Arcadia Posted September 20, 2012 Share Posted September 20, 2012 Thanks for taking the time. I understand now that you just have to use the formula and get the answer. (Even if it seems a little squirrely to me! :lol:) Sometimes it is the Physics textbook rather than the person. My hubby needs to draw out everything for most physics problem and prefer applied physics textbooks. He learns better when textbooks has lots of diagrams. Just look around at your library or bookstore and see whatever textbook works for you. Quote Link to comment Share on other sites More sharing options...
Cindy in FL. Posted September 20, 2012 Author Share Posted September 20, 2012 I'll do that. Thanks again for the help! Cindy Quote Link to comment Share on other sites More sharing options...
kiana Posted September 20, 2012 Share Posted September 20, 2012 (edited) Thanks for taking the time. I understand now that you just have to use the formula and get the answer. (Even if it seems a little squirrely to me! :lol:) Cindy Well -- you really have to understand what they gave you. When they gave you the average velocity, they gave you displacement/time in the first place. If you wanted to use the entire distance, they'd have had to have given you average SPEED instead of average velocity. ETA: average velocity is actually really tricky to work with and the whole reason they gave you this problem. Edited September 20, 2012 by kiana Quote Link to comment Share on other sites More sharing options...
Cindy in FL. Posted September 20, 2012 Author Share Posted September 20, 2012 OK, that makes sense. I was thinking in terms of speed. Have to get my Physics terms cemented in the old noggin! Cindy Quote Link to comment Share on other sites More sharing options...
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