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Old June 30th, 2008, 03:39 AM
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Unhappy 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?
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Old June 30th, 2008, 04:26 AM
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Originally Posted by amenbreakz View Post
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:
d = 348t
and
1118 - d = 168t

Solve this as a set of simultaneous equations.

-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
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Old 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
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Old June 30th, 2008, 04:56 AM
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hi thanks! Is it possible to see your working for this in full. Would be mega appreciated.

Amenbreakz
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d = 348t
and
1118 - d = 168t
I'll do this much:

The top equation is already solved for d, so put this into the bottom equation:
1118 - (348t) = 168t

516t = 1118

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
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Old June 30th, 2008, 05:04 AM
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Quote:
Originally Posted by amenbreakz View Post
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
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Old June 30th, 2008, 05:13 AM
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thanks people! I hope with practice my brain will start working again!
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Old 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
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Old June 30th, 2008, 06:33 AM
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Quote:
Originally Posted by amenbreakz View Post
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
\frac{1118~m}{516~m/s} = 2.16667~s
is the time it takes for the cars to meet, so

(348~m/s) \left ( \frac{1118~m}{516~m/s} \right ) = (348~m/s)(2.16667~s) = 754~m

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
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Old June 30th, 2008, 06:37 AM
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Thank's again chap!
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Old June 30th, 2008, 05:53 PM
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Quote:
Originally Posted by amenbreakz View Post
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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Old June 30th, 2008, 06:42 PM
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Quote:
Originally Posted by amenbreakz View Post
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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