# How fast is the distance between them changing after 5 seconds?

## Homework Statement

A passenger train is approaching a railway crossing from 105m to the east, at 54km/h. Meanwhile a train approaches from 140m to the north, at 72km/h. How fast is the distance between them changing after 5 seconds?

## Homework Equations

Pythagorean Theorem?

## The Attempt at a Solution

It looks to me that there are four constants in this question, namely:
The distance and velocity of the train which are 140m and 72km/h respectively.
The distance and velocity of the car which are 105m and 54km/h respectively.

I drew a right angled triangle and labeled x the car distance on the horizontal and y the train distance on the vertical with z marked on the hypotenuse.

I guess before i go further, i need to know if i am setting up this problem right first, because i am confused as to how to setup all the given data. Help?

The problem seems set up correctly, and Pythagoras' theorem is definitely the way to proceed.

Mark44
Mentor
This is what is called a related rates problem. You need to set up two different sets of relationships: one for the distances, and the other for the rates (velocities in this case).

From Pi-Bond's suggestion you can get a relationship between x and y to get the distance between the train and the car as a function of t. From that equation, differentiate to get an equation that involves the derivatives of the three quantities with respect to time.

Okay, so i tried the following as a start:

When y=140,dx/dt=?
So, with pythagoras' theorem:
x^2 + y^2 = z^2
derivative of above is:
2x(dx/dt) + 2y(dy/dt) = 0 not sure if i should go further because i don't know if i am on the right track?

Why have you differentiated z^2 as zero? It is also changing with time, so you should do the same as you did with x^2 and y^2.

I got confused and could not figure out what z actually represented.
But, is this right for the new derivation?
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)

To get a value for z, i used pythagorean theorem like this, but again i don't know if i am right or not...i just want to show i am trying:

105^2 + 140^2 = z^2
11025 + 19600 = z^2
z^2 = 30625
therefore z = 175

Mark44
Mentor
1irishman said:
I got confused and could not figure out what z actually represented.
But, is this right for the new derivation?
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
z is the distance between the train and the car. Since both are moving toward each other, dx/dt < 0 and dy/dt < 0, so dz/dt is negative as well.

You are given the positions of the car and train at a particular moment. You're asked for dz/dt at a time 5 seconds later.

1irishman said:
To get a value for z, i used pythagorean theorem like this, but again i don't know if i am right or not...i just want to show i am trying:

105^2 + 140^2 = z^2
11025 + 19600 = z^2
z^2 = 30625
therefore z = 175
I didn't check your work, but this isn't z in general, only at the moment when x = 105 and y = 140. One second later x, y, and z will have different values.

okay thank you. I tried plugging in these numbers and then isolating dz/dt at 5 seconds:

2(105)(54) + 2(140)(72) = 2(5)(dz/dt)
31500 = 10dz/dt
Is this right so far?

okay i missed the negative sign for dz/dt? I am confused how to go further.

You substituted the wrong values for x and y; they are at time 0 seconds. You need to figure out their values after 5 seconds, and use those to find z, and hence dz/dt. Use the formula relating speed, distance and time to find x and y.

Okay, the only formula i know that relates speed distance and time is:
v = d/t
i have a time of 5 seconds and velocity of 54km/h for the car
so, the car's distance is d = v*t = (54)*(5) = 270m I don't know i can't figure this one out

Mark44
Mentor
okay thank you. I tried plugging in these numbers and then isolating dz/dt at 5 seconds:

2(105)(54) + 2(140)(72) = 2(5)(dz/dt)
31500 = 10dz/dt
Is this right so far?
No. You are completely ignoring the fact that different units are being used. The distances given are in meters, but the speeds are in km/hr.

Okay, the only formula i know that relates speed distance and time is:
v = d/t
i have a time of 5 seconds and velocity of 54km/h for the car
so, the car's distance is d = v*t = (54)*(5) = 270m I don't know i can't figure this one out
Again, your units are not compatible. Velocity's units are generally distance/time, so when you multiply velocity by time to get distance, the units for time and the units that are part of the velocity have to be the same.

I have obviously done as much as i know how to do. Thank you for your help.

Mark44
Mentor
Converting from one set of units to another isn't all that difficult. For example, you can convert the 54 km/hr to m/sec like so:

54 km/hr = 54 km/hr * 1000 m/km * 1/3600 hr/sec = 54,000/ 3600 km/hr * m/km * hr/sec =15 m/sec.

In this calculation, all of the units except m in the numerator and sec in the denominator cancel, and you're left with units of m/sec.

I have searched high and low for examples of this type of question, so that i could follow the solution process, but have not been able to find anything. So, unless someone is willing to help me more, i guess i'll just have to seek help elsewhere. It is not that difficult to fly an airplane, or drive an 18-speed articulated truck either, but only when you know how.

Mark44
Mentor
Don't give up so easily. We're willing to help you, but are notwilling to do the work for you (by PF policy).

You have most of the work done, so now it's just a matter of following through.

You have an equation that relates the distances at arbitrary times: x2 + y2 = z2, and you have an equation that relates the rates (derivatives) of these quantaties: 2x*dx/dt + 2y*dy/dt = 2z * dz/dt.

Solve this equation for dz/dt, and then substitute the values for the rates and positions at time t = 5 seconds.

Where you got bogged down before was substituting quantities whose units were incompatible.

I don't know i can't figure this one out

You're actually doing very well. On behalf of all tutors everywhere, I apologize to irishman for comments like "this is easy".

Very few people find Calculus to be easy, when learning it. Tutors sometimes forget how much time & effort they put in learning the right methods in the first place. Also, once you've got the principles of related rates understood, you'll be able to do all sorts of related rate questions.
http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx
http://people.hofstra.edu/Stefan_Waner/RealWorld/tutorials/frames4_4.html [Broken]
tutor some related rate questions, but not yet one exactly like yours.

We can convert everything into km or everything into meters.

Did you have any success applying Mark's method to the other speed?
Perhaps, to encourage you and get you back on the right
"track" (pun intended) look at
http://www.wolframalpha.com/input/?i=72 kilometer/hour to meter/second&t=ff3tb01

2x*dx/dt + 2y*dy/dt = 2z * dz/dt.
Solve this equation for dz/dt, and then substitute the values for the rates and positions at time t = 5 seconds.
Will get us there, but it's much simpler to substitute all the constants first, then solve for dz/dt. One key for related rates: don't substitute constants until after completing the derivative.
Please, for your sake keep trying. Learning calculus is never easy, but think of how you'll feel when you get it right.

After you've converted the units & encouraged yourself that you can do this, read the following.
That being said, I learned two principles the hard way in physics.
A. It helps to clarify, by adding words like initial distance, distance at 5 seconds, etc. Then keep initial distance separate from distance at 5 seconds.
B.I could even draw 3 triangles for this question. Label each triangle with a title A:initial distances, etc.
A. initial distances
B. distance at 5 seconds. We'll need this to find z after 5 seconds.
C. one with just the rates. (non essential, leave out if it confuses you.)

Last edited by a moderator:
Firstly, i would like to thank nickalh for showing some compassion for the learner's initial experiences with Calculus and the challenges that it often presents. Second, i have a great deal of respect for the high degree of education and body of knowledge that the contributors to this forum possess. I understand fully and appreciate that the policy of PF is to not do the work for the student; having said that though, showing the student some empathy and providing them with some additional problem solving tools and ways of conceptualizing a given problem in the manner that nickalh has has proven effective results. It only took me a few minutes to solve this problem the rest of the way myself with the aforementioned approach. Mark, thank you for your contributions as well; you too were instrumental in helping me solve this problem effectively.

2(105)(15) + 2(140)(20) = 2(175)-dz/dt
8750/350=350/350-dz/dt
dz/dt = - 25m/s

The distance between the car and the train was changing at a rate of - 25m/s after 5 seconds.

Most importantly, I'm glad you've decided to give us another chance.

At first glance, you've got the right answer to a different question.
Don't the 105 and 140 you've plugged in represent initial distances?
So 25 m/s is the right answer for "What is the rate of change between them at initial time zero?"

This may happen to be the same answer, for"What is the rate of change between them at time =5 seconds?"
(The constraint that all distances must fit in Pythagorean theorem somehow may mean dz/dt is unaffected.) However, on a test I would feel very uncomfortable leaving that as the answer. Plus mastering this concept will make physics that much easier.

2x*dx/dt + 2y*dy/dt = 2z * dz/dt.
substitute the values for the rates and positions at time t = 5 seconds. Solve this equation for dz/dt
(paraphrased)​

I would feel much more confident applying
v=d/t to the horizontal distance & then to the vertical distance
to get us distances at 5 seconds.

This is where the two or 3 triangles I mentioned in my first post comes in. Heck I teach my college students, find colored pencils, draw a large diagram. In our case, we can put both triangles on the same diagram, one very large for the initial distances & one with dimensions half the size representing 5 seconds later. Make sure the diagram is large enough to have clear labels on the smaller triangle. In your labels, be sure to distinguish
A. initial distance from intersection.
B. distance traveled in 5 seconds.
C. distance to intersection
With the new velocities in compatible units, go ahead and solve v=d/t
once for the horizontal, westbound train
and once for the vertical, southbound train.
The distance that equation gives us, is the distance travelled, NOT the distance to the intersection. So you'll need a subtraction, once for the north/south train and again for the east/west train.

Next to last step, whew, almost done.
Apply pythagorean theorem to the new distances to find the hypotenuse or slant distance after 5 seconds.

At this point, we know 5 of the 6 variables.
x, y, z at time = 5 seconds.
dx/dt & dy/dt were given to us in the initial question.
Finally, go back to
2x*dx/dt + 2y*dy/dt = 2z * dz/dt.
substitute the values for the rates and positions at time t = 5 seconds. Solve this equation for dz/dt

Possible trivia:
This problem assumes the westbound & southbound velocities are constant.

I hope you'll find the arithmetic themselves aren't terribly difficult. It's knowing which calculations to do when that seems to be the challenge.
A. Experience,
B. seeing similar problems in other classes,
C. recognizing if I put in distances for t=0, my results will be valid for t=0, but if I put in distance for t=5, my results will be valid for t=5
helps here.

Here's hoping the trains don't cause a wreck.

Hi nickalh!

the distances i got at 5 seconds after converting from km/h to m/s where 100m for the train and 75m for the car; and finally, 125m as the z value(the distance between the car and the train at 5 seconds). It may be a coincidence that i got the same answer(-25m/s) for the initial distance and the distance after 5 seconds, so i am not what to say about that, but i did draw the three triangles you mentioned and i do find it very very helpful for me as the new kid on the block so to speak. No train wrecks here :-)

I had a big long explanation as to why you're close, but the method is not sound.
I also discussed the textbook authors created a poor problem: multiple methods with common mistakes lead to the correct number.
Aughhh, logging in on another tab cost me several paragraphs of work.

Heck I teach my college students, find colored pencils, draw a large diagram. In our case, we can put both triangles on the same diagram, one very large for the initial distances & one with dimensions half the size representing 5 seconds later. Make sure the diagram is large enough to have clear labels on the smaller triangle. In your labels, be sure to distinguish
A. initial distance from intersection.
B. distance traveled in 5 seconds.
C. distance to intersection after 5 seconds.
With the new velocities in compatible units, go ahead and solve v=d/t
once for the horizontal, westbound train
and once for the vertical, southbound train.
The distance that equation gives us, is the distance travelled, NOT the distance to the intersection. So you'll need a subtraction, once for the north/south train and again for the east/west train.
We need
C. values: distance to intersection,
not B. values: distance traveled in 5 seconds.
.
Apply,

(A. initial distance from intersection) = (B. distance traveled in 5 seconds) + (C. position at 5 seconds vs. intersection)

C. represents how much farther from the current position after 5 seconds, to the intersection.
Then substitute the C. values into 2x dx/dt +2... equation.

Thank you nickalh :-) I will have to move on to the next question, but i now see all the good points you have made and appreciate the time and effort you have put into this question to help me. :-)