# Homework Help: Math Methods problem

1. May 3, 2005

### gordda

I got this question in a text book and i don't really know what it is asking
Here is the question:

A submarine is travelling due east and heading straight for a point P. A battleship is travelling due south and heading for the same point P. Both ships are travelling at a velocity of 30km/hr. Initially, their distances from P are 210km for the submarine and 150km for the battleship. How close will the two vessels come to each other????

Is the question asking intial distance or do i need to find a rule that relates them then find the distance. i don't know what it is meant. any insight to this question would be greatly appreciated.

Thanx :)

2. May 3, 2005

### TenaliRaman

gorda,
Note that the two vehicles are moving towards the same point with same speed but from different starting distances. It means that one of them is likely to reach P before the other. Now as they move towards P, the distance between these two vehicles will decrease. The distance will continue to decrease till they reach some point, after which their distance will increase again. So u are supposed to find out how close they get before the distance between them increases again.

-- AI

3. May 3, 2005

### gordda

if that was the case, then wouldn't that mean that the closest distance is when the submarine reaches Point P and the battle ship is at 60km away from Point P. cause that is when they are the closest

4. May 3, 2005

### jdavel

gordda,

".... cause that is when they are the closest"

No it's not. Try figuring out how close they are a little later.

5. May 3, 2005

### StatusX

The rigorous way to do this problem would be to use calculus. Find the distance between the ships as a function of time and determine when this reaches a minimum.

But here's a trick, if you're interested. It might be a little advanced, but I think you'll find it's pretty intuitive. Since the two ships have the same speed, you can take advantage of symmetry to make this a lot easier. What symmetry? Well, you can flip the setup about the northeast-southwest line through P, switch which ship is the "submarine" and which is the "battleship", and all you have done is shift to a new point in time. Draw it and you'll see what I mean. You can only do this because the two ships have the same velocity, so which one is the "submarine" and which is the "battleship" is completely arbitrary.

One thing to notice about this transformation is that it preserves the distance between the ships. This means that whatever the seperation is now, at some other point in time, that speration will come up again. However, the ships are moving with constant velocities, so after they've gotten as close as they'll get, they'll only get further and further apart. What does this mean? At the closest point, the transformation will do nothing. The transformation is a shift in time, but at this point that shift has to be 0, because there can only be one point in time when the ships have this minimum seperation. So just find where the ships have to be for this transformation to do nothing.

Maybe this isn't easier, but it gives the answer with no equations at all, so I thought it was kinda cool.

Last edited: May 3, 2005
6. May 3, 2005

### Werg22

Ok there is how you do it:

$$d^2=(210-vt)^2 + (150-vt)^2$$
$$d^2=210^2 - 420vt + vt^2 + 150^2 - 300vt + vt^2$$
$$d^2=2vt^2 - 720vt + 66600$$
$$d^2=2(30)t^2 - 720(30)t + 66600$$
$$d^2=1800t^2 - 21600t + 66600$$
$$d^2=1800(t^2 - 12t + 36-36) + 66600$$
$$d^2=1800((t-6)^2 -36) + 66600$$
$$d^2=1800(t-6)^2 - 64800 + 66600$$
$$d^2=1800(t-6)^2 + 1800$$

Just find what value for t makes the distance the shorter possible ;)

Last edited: May 3, 2005
7. May 3, 2005

### Werg22

Thanks Status X I figured out!