Vector question, please help me

[reign226]

Two trains are travelling on railway lines which cross at right angle at O. Train A is travelling at 50kmh northwards while train B is travelling at 70kmh eastwards and both trains are approaching O. Initially, both trains are 100km from O.

find: 1) Velocity of A relative to B, 2) the shortest distance possible between two of them

answer for 1, 86.02, 2 is 23.26 KM


I can do 1, but I cannot even start to do 2. All my friends I've consulted also can't provide an answer. My teacher went on about finding the difference in angle between aVb and their initial distance, but I don't really get him. Anybody care to help?

[littlebigone]

since the trains are traveling at right angles, the distances between them can easily be given by pythagoras's theorem.

but since they're traveling we need to set the distances to be a function of time.

D = sqroot[(100-70t)^2 + (100-50t)^2]

What we want to do then is differentiate the function and set it equal to zero to find the minimum. I'd rather work without the square root, so i squared the function.

D^2 = S = (100-70t)^2 + (100-50t)^2

dS/dt = 9800t - 14000 + 5000t - 10000 = 0
14800t = 24000
t = 1.621621......

plug this back into the function for D and you go the shortest distance.

I get 23.25 though as my answer not 23.26. I think the way to solve it should be right.

[reign226]

Yes, I tried your method on another similar question and it worked flawlessly. The answer I got is also off by a few fraction, like, 0.0001 meters, but I believe calculus over whatever the heck the book is presenting. Thanks a lot. Now what happens if they are not at right angles to each other? Do I then resolve their velocities such that they are at right angles?

[reign226]

for example:

Car A: North 30 degrees to east at speed 20m/s
Car B: North 30 degrees to west at speed 30 m/s

Both are approaching a common point O. The initial distance from point O for A and B are 800m and 1km respectively. Answer is 130.something.

[littlebigone]

I don't have time now but I think that there is some trigonometric law that allows you to calculate the sides for triangles the are not right triangles. Either sine law or cosine law. I think it's cosine law.

but anyways, it would pretty much follow the same method, except that since it isn't a right angle anymore, you would need to check for when each car passes O so that you can adjust the angle to suit the current situation.

For example, in your case,

A takes 40 secs to get to O
B takes 33.3333 secs to get to O

Initially, the angle between them is 60 degrees

then at 33.3333 it is a straight line

for 33.33333 < t < 40 the angle is 120 degrees

then at 40 secs it's another straight line

anything more than 40 secs is 60 degrees.

fine the minimum by differentiating the different functions that you come up with. make sure that the minimum t is in the domain for the function.

Again, I'm not too sure and I don't have time to verify this method.

I don't quite understand how you would do your method. But please do post your solutions online. it sounds interesting.

[reign226]

Something stange I discovered. I resolved both distance and speed of car B (initially) to be 90 degrees to car A's path by finding the product of both distance and speed with cos 30 (or sin 60, take your pick). I kept the distance and speed of car A the same. Then I did the maths the same way using Pythagoras and found the answer to be about 1.6 off the one given.

Then I reversed and resolved Car A's speed and distance and kept Car B's constant. The answer given however is off by 0.8 seconds to the one I found using the previous method. Aren't they supposed to give the same thing? Straange.

One more thing, this is a vectorial question. What happens in the exam if we solved this question without applying vector principles? I think the point of the question is to test the candidate's talent on a specific maths skill (vector in this case) but if we use another method, we just went out of the 'answer scheme' and thus I think they won't accept the answer. Will they?

[doodle]

If train A is stationary and only train B is moving, it becomes quite easy to find the nearest distance between the two trains.

We will convert the present problem to the simpler problem (of only one moving train) by calculating the relative velocity between the two trains. Suppose the velocity of train A is the reference velocity. Then the velocity of train B with respect to the reference velocity is Vba = Vb - Va = [70,0] - [0,50] = [70,-50]. Now, position train A and train B at coordinates (0,-100) and (-100,0) respectively. Train A is stationary and train B is moving relative to train A at [70,-50]; see figure below.

-deleted-

A bit of geometry will tell you that theta = arctan(50/70) and d = 40 * sin(theta) = 23.25km.

[kelynn]

Two trains are travelling on railway lines which cross at right angle at O. Train A is travelling at 50kmh northwards while train B is travelling at 70kmh eastwards and both trains are approaching O. Initially, both trains are 100km from O.

find: 1) Velocity of A relative to B, 2) the shortest distance possible between two of them

answer for 1, 86.02, 2 is 23.26 KM


I can do 1, but I cannot even start to do 2. All my friends I've consulted also can't provide an answer. My teacher went on about finding the difference in angle between aVb and their initial distance, but I don't really get him. Anybody care to help?

after u have find the 1st part,u have to draw the displacement diagram of train A relative to B, and then find the initial displacement of the train A and trian B,that is square root of (100000+100000).then u transfer the line of the velocity of train A relative to B to the initial point of B.then u find the angle between them.that all.try ur best.