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Math Quiz Thread

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steve_bank

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An object is on a circular path radius 200 at 1 meter per second.

An object is traveling along a path x^2/4.

How fast dies the second object have to travel to intercept the object on the circular path in the first quadrant.

At t0 the object on the circular path is at (0,200) and the object on the second path (0,0).

Adding

The object on the circular path at t0 starts at (0,200) moving counter clock wise.

The object on the parabolic path at t0 starts at (0,0) moving in the first quadrant.
 
Last edited:
There are infinitely many solutions. Or to put it differently, there is no solution -- there is no speed at which the second object has to travel.
 
Bomb#20 said:
There are infinitely many solutions. Or to put it differently, there is no solution -- there is no speed at which the second object has to travel.
Click to expand...
There is only one solution. A first step might be skethching the curves.
 
Bomb#20 said:
There are infinitely many solutions. Or to put it differently, there is no solution -- there is no speed at which the second object has to travel.
Click to expand...
No, there clearly must be a solution.

We have a circle and a parabola, as this is restricted to the first quadrant there is exactly one intercept between them.

We know x^2 + y^2 = 40,000. We also know that y = x^2/4. Thus x^2 + (x^2/4)^2 = 40,000. Thus x^2 + x^4/16 = 40,000. I'm not going to take it further because I do not remember the equations for distance along a circle nor for distance along a parabola.
 
The line integral calculates distance off a curve.
 
Loren Pechtel said:
Bomb#20 said:
There are infinitely many solutions. Or to put it differently, there is no solution -- there is no speed at which the second object has to travel.
Click to expand...
No, there clearly must be a solution.
Click to expand...
No, what there clearly must be a solution to is the problem of finding some speed that will result in interception. But that's not what the problem asks for. It asks how fast does "the second object --> HAVE TO <-- travel". Since the problem of finding a speed that works does not have a --> UNIQUE <-- solution, whatever solution you come up with is not a speed the second object --> HAS TO <-- travel at. The second object has its choice of going at any of the infinite number of intercepting speeds.

We have a circle and a parabola, as this is restricted to the first quadrant there is exactly one intercept between them.

We know x^2 + y^2 = 40,000. We also know that y = x^2/4. Thus x^2 + (x^2/4)^2 = 40,000. Thus x^2 + x^4/16 = 40,000. I'm not going to take it further because I do not remember the equations for distance along a circle nor for distance along a parabola.
Click to expand...
Hint: You are treating this as a problem in space. It's a problem in time. Your calculation tells us WHERE the two objects meet, not WHEN.

steve_bank said:
The line integral calculates distance off a curve.
Click to expand...
"An object is on a circular path radius 200 at 1 meter per second." The path is a circle. The speed is constant. The intersection point isn't going away. The object will arrive at the intersection point over and over, about every 21 minutes, infinitely many times.

steve_bank said:
Bomb#20 said:
There are infinitely many solutions. Or to put it differently, there is no solution -- there is no speed at which the second object has to travel.
Click to expand...
There is only one solution. A first step might be skethching the curves.
Click to expand...
It's not an issue of sketching the curves; it's an issue of reading the problem statement.
 
If I posted the problem as two straight lines it would have been obvious.

Will two ships tackling non parallel collide knowing the velocities?

You know the distance along the circle at which the parabola intersects the circle. Straighten out the two curves and place side by side. At what velocity will the parabolic object reach the end of the line at the same time as the object on the circle.
 
steve_bank said:
If I posted the problem as two straight lines it would have been obvious.
Click to expand...
If you posted the problem as two straight lines it would have been a different problem.

Will two ships tackling non parallel collide knowing the velocities?

You know the distance along the circle at which the parabola intersects the circle. Straighten out the two curves and place side by side. At what velocity will the parabolic object reach the end of the line at the same time as the object on the circle.
Click to expand...
It's time to stop arguing and calculate. It's very simple.

If the second object's speed is 0.13085261754 m/s, where will the two objects be after 1542.55940744 sec.?

If the second object's speed is 0.70595369339 m/s, where will the two objects be after 285.922346 sec.?
 
Same problem. You just need to find the path distance of two non linear paths.

I'll work out the path integrals and post an answer.

A circle radius 100 meters. Two objects traveling on the circle.

object 1 at (0,200) at 1m/s
object 2 at (0,-200) at 2 m/s

How long before they collide?

Consider this a test for the Star Fleet space navigator certification program.
 
bilby said:
Never mind where the objects are, where can I buy a clock that's accurate to 0.00000001 seconds, or a device that can measure speeds to the nearest 0.000000000001 m.s-1??
Click to expand...
Acme. (For some reason they keep them in the frictionless surface department. Ask somebody for directions -- I presume you speak Spherical Cow?)
 
steve_bank said:
A circle radius 100 meters. Two objects traveling on the circle.

object 1 at (0,200) at 1m/s
object 2 at (0,-200) at 2 m/s
Click to expand...
The objects are 400 meters apart. They can't both be traveling on the same 100 meter radius circle. Did you mean a 200 meter radius circle, or did you mean for the objects to be at (0,100) and (0, -100)?
 
steve_bank said:
Same problem. You just need to find the path distance of two non linear paths.

I'll work out the path integrals and post an answer.
Click to expand...
In case you want to save some work, you can get the path distances along the circle with simple arithmetic, from those you can get the arrival point with trig, and from that you can get the path distance along the parabola by googling an on-line parabolic arc length calculator, which will compute the path integral for you. The formula is some godawful thing with hyperbolic cosines.
 
Bomb#20 said:
steve_bank said:
Same problem. You just need to find the path distance of two non linear paths.

I'll work out the path integrals and post an answer.
Click to expand...
In case you want to save some work, you can get the path distances along the circle with simple arithmetic, from those you can get the arrival point with trig, and from that you can get the path distance along the parabola by googling an on-line parabolic arc length calculator, which will compute the path integral for you. The formula is some godawful thing with hyperbolic cosines.
Click to expand...
True.
Bomb#20 said:
steve_bank said:
A circle radius 100 meters. Two objects traveling on the circle.

object 1 at (0,200) at 1m/s
object 2 at (0,-200) at 2 m/s
Click to expand...
The objects are 400 meters apart. They can't both be traveling on the same 100 meter radius circle. Did you mean a 200 meter radius circle, or did you mean for the objects to be at (0,100) and (0, -100)?
Click to expand...
Ya got me pard. (0,-100) (0,100)

I'll add.

The objects are spheres with a .1 meter diameter. A collision is when the centers of the objects are within 1 diameter of each other.
 
steve_bank said:
Bomb#20 said:
steve_bank said:
A circle radius 100 meters. Two objects traveling on the circle.

object 1 at (0,200) at 1m/s
object 2 at (0,-200) at 2 m/s
Click to expand...
The objects are 400 meters apart. They can't both be traveling on the same 100 meter radius circle. Did you mean a 200 meter radius circle, or did you mean for the objects to be at (0,100) and (0, -100)?
Click to expand...
Ya got me pard. (0,-100) (0,100)

I'll add.

The objects are spheres with a .1 meter diameter. A collision is when the centers of the objects are within 1 diameter of each other.
Click to expand...
That last bit makes it a good deal harder, if you want the answer precise. It's the starting path length minus the final path length divided by the relative speed, so about (100 pi - .1) / 1 or (100 pi - .1) / 3, depending on whether the two objects are going along the circle in the same direction or opposite directions. Is that the answer you were aiming for? Or did you mean the final path length to be negligibly longer than .1 meter, the length of a circular arc with radius 100 and chord length .1? I make the correction for the curvature of a .1 meter arc at about 4 nm.
 
For an iterative solution the bounds are required.

Work it either way.


It took me a bit to work out the solution. I'll post it after a while. There is an exact solution as first defined.
 
v1 velocity objects
v2
sdif Distance offset at t0. Half circle.

sdif = PI*r
v1 * t = sdif + v2 * t
t(v1 – v2) = sdif
t = sdif/(v1 - v2)

Iterated solution

Code: 
Sub Main

Dim Doc As Object
Dim Sheet As Object
Doc = ThisComponent
Sheet = Doc.Sheets (0)
Cell = Sheet.getCellByPosition(0,0)
cell.string = "DIST 1"
Cell = Sheet.getCellByPosition(1,0)
cell.string = "DIST 2"
Cell = Sheet.getCellByPosition(2,0)
cell.string =" TIME"
Cell = Sheet.getCellByPosition(0,1)
Cell.Value = 0  
Cell = Sheet.getCellByPosition(1,1)
Cell.Value = 0  
Cell = Sheet.getCellByPosition(2,1)
Cell.Value = 0
Cell = Sheet.getCellByPosition(0,3)
cell.string = "RUNNING"

dim s1,s2,t,dt,sdif,v1,v2,r,tmax as  double
dim Nmax,i  as  long
r = 100.
sdif = pi * r 'inital spearation
s1 = 0.
s2 = sdif
v1 = 2.
v2 = 1.
t = 0.
dt =  .0001
tmax = 800  'seconds
Nmax = tmax/dt
for i = 1 to Nmax
     s1 = s1 + (v1*dt)
     s2 = s2 + (v2*dt)
     if(s1 > s2) OR (s1 = s2) then
       exit for
     end if    
      t = t + dt
 next i
 
 Cell = Sheet.getCellByPosition(0,3)
cell.string = "DONE"
Cell = Sheet.getCellByPosition(0,1)
Cell.Value  =s1  
Cell = Sheet.getCellByPosition(1,1)
Cell.Value = s2  
Cell = Sheet.getCellByPosition(2,1)
Cell.Value = t

End Sub


For r = 100m v1 - 2, v2 = 1

Calculated time 314.159265 seconds
Iterated time .0001 second resolution 314.159300 seconds
 
After thinking about it some more I realize Bomb is right.

We were coming at it from a standpoint of the two objects moving to the intercept point. However, suppose the object on the parabola is moving very slowly--instead of hitting the orbiting object on the first orbit it hits on the second orbit. Even slower, on the third orbit. There are infinitely many such solutions.
 
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