# Velocity question

• Jun 30th 2008, 01:39 AM
amenbreakz
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?
• Jun 30th 2008, 02:26 AM
topsquark
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?

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:
$\displaystyle d = 348t$
and
$\displaystyle 1118 - d = 168t$

Solve this as a set of simultaneous equations.

-Dan
• Jun 30th 2008, 02:37 AM
amenbreakz
hi thanks! Is it possible to see your working for this in full. Would be mega appreciated.

Amenbreakz
• Jun 30th 2008, 02:56 AM
topsquark
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
$\displaystyle d = 348t$
and
$\displaystyle 1118 - d = 168t$

I'll do this much:

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

$\displaystyle 516t = 1118$

Solve for t and put that into either one of the original equations to get d.

-Dan
• Jun 30th 2008, 03:04 AM
nikhil
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
• Jun 30th 2008, 03:13 AM
amenbreakz
thanks people! I hope with practice my brain will start working again!
• Jun 30th 2008, 03:24 AM
amenbreakz
1118/516=348*1118/516 =754m

sorry am not used to the symbols what is / and * in relation to sum
• Jun 30th 2008, 04:33 AM
topsquark
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
$\displaystyle \frac{1118~m}{516~m/s} = 2.16667~s$
is the time it takes for the cars to meet, so

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

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
• Jun 30th 2008, 04:37 AM
amenbreakz
Thank's again chap!(Rock)
• Jun 30th 2008, 03:53 PM
ticbol
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.
• Jun 30th 2008, 04:42 PM
mr fantastic
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.