The fly, the truck, and the Ferrari

[BMAD]

I am pretty sure bonanova asked a variation of this question but I can't seem to locate it... Anyways:

Suppose there is a Ferrari and a truck on a collision course. They will crash in two hours.The truck is going 60mph and the Ferrari is going 80mph. There is also a magical fly on the Ferrari. When the two hour time began the fly instantly achieved and maintained its maximum speed. The fly flew to the truck and back and to the truck and back. Upon the returning from the second trip, the fly, the truck, and the Ferrari all collide. How fast was the fly going?

[BASKARAN R]

140 mph

[bonanova]

16 hours ago, BMAD said:

I am pretty sure bonanova asked a variation of this question but I can't seem to locate it... Anyways:

Suppose there is a Ferrari and a truck on a collision course. They will crash in two hours.The truck is going 60mph and the Ferrari is going 80mph. There is also a magical fly on the Ferrari. When the two hour time began the fly instantly achieved and maintained its maximum speed. The fly flew to the truck and back and to the truck and back. Upon the returning from the second trip, the fly, the truck, and the Ferrari all collide. How fast was the fly going?

I don't recall that I posted it. The way I first heard it, tho, the question gives the speed of the fly and asks how far it flew. This form hasn't been posted that I can tell. What intrigues me about this version is that ...

Spoiler

The fly cannot collide with both vehicles after a finite number of trips.

When the vehicles collide, the fly must have made an infinite number of trips.

Its speed must be greater than the speeds of both vehicles in order to fly back and forth between them. So at each head-on collision with the approaching vehicle, the fly is moving away from the trailing vehicle. So it can't collide with both vehicles after the second trip. Or after any finite number of trips.

It's very much like Zeno's paradox of moving halfway toward a target. You'll get there eventually, but not after a finite number of steps.

 

[BMAD]

22 hours ago, bonanova said:

I don't recall that I posted it. The way I first heard it, tho, the question gives the speed of the fly and asks how far it flew. This form hasn't been posted that I can tell. What intrigues me about this version is that ...

  Hide contents

The fly cannot collide with both vehicles after a finite number of trips.

When the vehicles collide, the fly must have made an infinite number of trips.

Its speed must be greater than the speeds of both vehicles in order to fly back and forth between them. So at each head-on collision with the approaching vehicle, the fly is moving away from the trailing vehicle. So it can't collide with both vehicles after the second trip. Or after any finite number of trips.

It's very much like Zeno's paradox of moving halfway toward a target. You'll get there eventually, but not after a finite number of steps.

 

There is a solution but I find it unsatisfactory.  Using kinematics we can set up a system and find two solutions though one doesn't make sense.

[bonanova]

The problem is a paradox of sorts. At all times the fly is flying *away* from one of the vehicles. So the vehicles *can't* collide with each other during any part of the fly's journey! Yet of course the vehicles *do* collide in finite time. And they *do* collide, at the same time, with the fly. That's why the fly must make an infinite number of trips. After any finite number of fly-trips, the vehicles are a calculable distance apart.

[bonanova]

Here's a simulation without actually writing the program:

Spoiler

OP says: Fly begins on the Ferrari, bounces off truck, bounces off Ferrari, bounces off truck then hits Ferrari at the same time (2 hours from start) that the  truck does. Ferrari speed is 80, truck is 60.  OP asks: what is fly speed?

Let the vehicles collide at x=0 by having the Ferrari begin at x=-160 and the truck at x=+120.
There are three fly-speed regimes of interest:

1. Fly < 80.
   In this case the fly gets behind the Ferrari and never catches up.
   The fly therefore cannot collide with the truck when the Ferrari does.
    
2. Fly = 80.
   In this case the fly accompanies the Ferrari to the truck.
   The three collide after two hours at x=0.
   But it happens on the fly's first collision, not its not fourth.
    
3. Fly > 80.
   Since fly is faster than Ferrari it will get to x=0 before the Ferrari, and will hit the truck at some x>0 and turn around
   Since fly is faster than truck it will get back to x=0 before the truck, and will hit the Ferrari at some x<0 and turn around
   Since fly is faster than Ferrari it will get back to x=0 before Ferrari and will hit truck at some x>0 and turn around.
   Since fly is faster than truck it will get to the x=0 collision point before the truck does (and before the Ferrari does.)
   The fly's fourth collision will involve only the Ferrari, will not happen at x=0 and will happen before 2 hours have elapsed.

Conclusion: There is no fly speed that satisfies the OP.

 

[BMAD]

5 hours ago, bonanova said:

Here's a simulation without actually writing the program:

  Hide contents

OP says: Fly begins on the Ferrari, bounces off truck, bounces off Ferrari, bounces off truck then hits Ferrari at the same time (2 hours from start) that the  truck does. Ferrari speed is 80, truck is 60.  OP asks: what is fly speed?

Let the vehicles collide at x=0 by having the Ferrari begin at x=-160 and the truck at x=+120.
There are three fly-speed regimes of interest:

1. Fly < 80.
   In this case the fly gets behind the Ferrari and never catches up.
   The fly therefore cannot collide with the truck when the Ferrari does.
    
2. Fly = 80.
   In this case the fly accompanies the Ferrari to the truck.
   The three collide after two hours at x=0.
   But it happens on the fly's first collision, not its not fourth.
    
3. Fly > 80.
   Since fly is faster than Ferrari it will get to x=0 before the Ferrari, and will hit the truck at some x>0 and turn around
   Since fly is faster than truck it will get back to x=0 before the truck, and will hit the Ferrari at some x<0 and turn around
   Since fly is faster than Ferrari it will get back to x=0 before Ferrari and will hit truck at some x>0 and turn around.
   Since fly is faster than truck it will get to the x=0 collision point before the truck does (and before the Ferrari does.)
   The fly's fourth collision will involve only the Ferrari, will not happen at x=0 and will happen before 2 hours have elapsed.

Conclusion: There is no fly speed that satisfies the OP.

what about this approach:

Spoiler

Total distance: 280 miles at start, Say Ferrari is at location zero with fly and truck is at location 280.  Ferrari's speed is 80mph, Truck's speed is 60mph, and fly's speed is x.

Trip 1: The fly would meet the truck at Time1 = 280/(60+x), The fly and truck would be at location x*time1 or 280-60*time1, Ferrari would be at location 80*time1

Trip 2: The fly would meet the Ferrari at Time2 = (280-60*time1 - 80*time1)/(80+x), the fly and Ferrari would be at location 80*time1+ 80*time2, truck would be at location 280-60*time1-60*time2.

Trip 3: The fly would meet the truck at time3 = ((280-60*time1-60*time2) - (80*time1+ 80*time2))/(60+x), The fly and truck would be at location 280-60*time1-60*time2-60*time3, Ferrari would be at location 80*time3+80*time2+80*time1

Trip 4:The fly, the Ferrari, and The truck meet at time4 = ((280-60*time1-60*time2-60*time3)-(80*time1+80*time2+80*time3))/(80+x)

given the constraints that

time1+time2+time3+time4 = 2 and

80(time1+time2+time3+time4)= 280 - 60(time1+time2+time3+time4)

we can derive two answers, though I admit the answers are unsatisfactory fulfilling.

 

 

 

Quote

 

 

Edited April 11, 2016 by BMAD

[bonanova]

I think those equations describe the conditions described in the OP. But I don't think there is a solution that satisfies the OP. Maybe that is just what you are saying. But I don't think it's the fault of the equations; it's the claim in the OP that the three all collide at the conclusion of the second round trip of the fly. That just can't happen if all three have constant speeds.

Consider the situation at time1 + time2 + time3, when the fly has hit the truck for the second time. The fly completes his trip by returning to the Ferrari. But it doing so it must perform an impossible task. (1) It must fly back to the Ferrari at a speed that is greater than the speed of the truck (it's at least the speed of the Ferrari), but (2) it must arrive at the same time that the truck arrives. That is not possible. The fly will arrive at the Ferrari before the truck does.

[BMAD]

4 hours ago, bonanova said:

I think those equations describe the conditions described in the OP. But I don't think there is a solution that satisfies the OP. Maybe that is just what you are saying. But I don't think it's the fault of the equations; it's the claim in the OP that the three all collide at the conclusion of the second round trip of the fly. That just can't happen if all three have constant speeds.

Consider the situation at time1 + time2 + time3, when the fly has hit the truck for the second time. The fly completes his trip by returning to the Ferrari. But it doing so it must perform an impossible task. (1) It must fly back to the Ferrari at a speed that is greater than the speed of the truck (it's at least the speed of the Ferrari), but (2) it must arrive at the same time that the truck arrives. That is not possible. The fly will arrive at the Ferrari before the truck does.

What you say is true if and only if time 2,3 & 4 is greater than zero.  at zero there are two solutions, 80mph,  60mph for the fly meaning that the fly never left either car until the collision. But since it is given that the fly starts with the car, the fly must be going 80mph.

Edited April 12, 2016 by BMAD

[bonanova]

In the case of 80 mph the fly, Ferrari and truck all meet on the fly's first and only collision. The OP asks the fly to move to the truck, back to the Ferrari, back to the truck and finally back to the Ferrari. Then, "upon returning from the second trip," the three collide. The 80 mph case has them colliding upon reaching the truck for the first and only time.

Spoiler

If we try to fix this by saying there are multiple fly-collisions separated by 0 minutes, in order to have a "fourth" fly collision, we still have to say that the three of them all meet on the first of them, and that happens *before* the fourth one -- exactly 0 minutes before. In any event, the OP does not mention the three of them colliding on the first fly-collision, and even if we say the first one happens at the same time as the fourth, the OP does not allow for the three of them to meet at the first (or second or third) fly collision.