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June 30th, 2008, 03:39 AM
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| | Velocity question Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A? | 
June 30th, 2008, 04:26 AM
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Originally Posted by amenbreakz Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A? | Let A travel a distance of d miles. Then you know that B has traveled 1118 - d miles. You also know that they traveled the same amount of time to get there. Call this time t.
So using v = dt: 
and
Solve this as a set of simultaneous equations.
-Dan
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June 30th, 2008, 04:37 AM
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| | hi thanks! Is it possible to see your working for this in full. Would be mega appreciated.
Amenbreakz | 
June 30th, 2008, 04:56 AM
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| | Quote:
Originally Posted by amenbreakz hi thanks! Is it possible to see your working for this in full. Would be mega appreciated.
Amenbreakz | Quote:
Originally Posted by topsquark 
and  | I'll do this much:
The top equation is already solved for d, so put this into the bottom equation:
Solve for t and put that into either one of the original equations to get d.
-Dan
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"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert | 
June 30th, 2008, 05:04 AM
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| | Check it out Quote:
Originally Posted by amenbreakz Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A? | Velocity is a vector quantitity.
Suppose A is going east and B is going west. now subtract a velocity vector whose direction is from west to east from A and B and magnitude is 168mph.so the velocity of B will become 168-168=0mph and velocity of A will become
348-(-168)=516mph
now for us B is at rest(not actually its at rest) and A is approaching B at a speed 516mph. So time taken by A to reach B=1118/516. Since time taken by A to reach also mean time taken by A to reach meeting point.so distance of meeting point from point A=distance travelled by A at time 1118/516=348*1118/516 =754m
same can also be done by using concept of ratio but I think vector method is easiest.
But you may ask ratio method if you want | 
June 30th, 2008, 05:13 AM
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| | thanks people! I hope with practice my brain will start working again! | 
June 30th, 2008, 05:24 AM
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| | 1118/516=348*1118/516 =754m
sorry am not used to the symbols what is / and * in relation to sum | 
June 30th, 2008, 06:33 AM
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| | Quote:
Originally Posted by amenbreakz 1118/516=348*1118/516 =754m
sorry am not used to the symbols what is / and * in relation to sum | / means division and * means multiply. Unfortunately this expression makes no sense. It should be written with more detail and on more than one line.
1118/516=348*1118/516 =754m
Should be 
is the time it takes for the cars to meet, so
amenbreakz: Something like the line "1118/516=348*1118/516 =754m"
appears every now and again on student papers. It's a common mistake to make (I've seen it a lot) but it is Mathematically incorrect, so I'd advise that in the future you omit the first equal sign. You may lose points for writing such an expression on an exam.
-Dan
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"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert | 
June 30th, 2008, 06:37 AM
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| | Thank's again chap! | 
June 30th, 2008, 05:53 PM
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| | Quote:
Originally Posted by amenbreakz Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A? | Some people see many ways to the solution to the same problem at once. Espescially in exams, it's good to use the fastest way to save time.
Here is how I would have solved your problem in an exam.
Since A and B are traveling toward each other in constant velocities, then their traveled distance as time goes on add up until the sum reaches 1118 miles. So the distance traveled by A or B is proportional to the speeds they travel.
So, from where A started, A and B meets at [(348)/(348 +168)] *1118 = 754 miles.
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June 30th, 2008, 06:42 PM
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| | Quote:
Originally Posted by amenbreakz Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A? | Since we're looking at alternative approaches .....
1. Find the time it takes A and B to meet by pretending that B is stationary and A is moving towards B at 348 + 168 = 516 mph.
Then the time it takes A to reach B is 1118/516 = 13/6 hours.
2. So in real life A travels (348)(13/6) = 754 miles before meeting B.
3. So A and B meet 754 miles from where A started.
4. Check: B travels (168)(13/6) = 364 miles. 754 + 364 = 1118. Edit: Just read nikhil's reply more carefully and it says the same thing. Nikhil, try putting more line spaces etc. in your replies to improve their readability.
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