# How many times does the grasshopper jump from one car to another before collision

1. Jul 20, 2009

1. The problem statement, all variables and given/known data

Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 m/s relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?

2. Relevant equations

$$\Delta$$x =vt

3. The attempt at a solution

Answer: $$\Delta$$X (grasshopper) = 15t and 100 = 10t then t = 10s so the answer is $$\Delta$$X (grasshopper) = 15*10 =150 m.

But I have another question. How many times does the grasshopper jump from one car to another before the collision? ( I don’t know how to solve it, please help me what to do).

2. Jul 20, 2009

### Lok

If the answer is not infinity then I've no ideea.

3. Jul 20, 2009

I'd like to know why the answer is infinity?

4. Jul 20, 2009

### Lok

Is it ... nice.

Well i've computed about 5 jumps before I got the ideea that you can go on and on. The ideea is that as the two cars approach collision the distance gets smaller.

So the time spend in each jump is only a fraction of what is left til the collision.

So it's like halving a distance, you can do that for an infinite number of times.

5. Jul 20, 2009

### Lok

In this case the first jump takes 8 s the second 1.6 s the 50th = 49th/5 etc...

6. Jul 20, 2009

Well this is really nice, thank you very much.

7. Jul 20, 2009

### Mentallic

Yes, weirdly enough the answer will be infinite because of the criteria you set it. i.e. the grasshopper basically bounces off the cars. Maybe you would like to place some restrictions on the problem. Maybe counting the number of jumps the grasshopper makes until the cars get close enough that they are a grasshopper's body length apart.

8. Jul 20, 2009

### minger

Last edited: Jul 20, 2009
9. Jul 20, 2009

### HallsofIvy

Staff Emeritus
I just gave a complete answer and then realized this is "homework". Sorry.

The two cars are coming together at 20 m/s each. How long will it take them to collide, in seconds?

The grasshopper is jumping at a constant 15 m/s. How much distance will it cover in that time?

It's as simple as that!

No, the answer is NOT "infinity", in fact, it will be much smaller than you think it ought to be!

Last edited: Jul 20, 2009
10. Jul 20, 2009

### Staff: Mentor

It is not different from the other question - about bouncing ball. First bonuce takes 1 sec. Each next bounce takes half time of the previous one. How many times will the ball bounce before it stops? How long will it take?

11. Jul 20, 2009

### HallsofIvy

Staff Emeritus
Completely true but I think looking at it as a series just makes it harder. In fact, this problem is the basis of an old joke: someone goes up to Erdos (or whatever mathematician you like) and proposes this problem to him (I think it normally involves a bird rather than a grasshopper). Erdos gives the correct answer almost immediately and the person chuckles and says "you know, most people try to do that by summing an infinite series". Erdos looks puzzled and says "But I did do it by summing an infinite series"!

12. Jul 20, 2009

### Staff: Mentor

Agreed

Junior told me that at some exam they were given a question that was - at the end - basically the same thing. Some people used the simple approach, some people did it the hard way...

13. Jul 20, 2009

### Mentallic

I believe otherwise. If you re-read the scenario the OP posted, you'll see there are virtually no restrictions placed on the grasshopper (which of course makes it impossible). Basically:

1) the grasshopper spends 0 time on each car so it is like a perfect bounce
2) the grasshopper is always faster than the cars are travelling since it jumps at 15 m/s while the cars travel at 10 m/s (it doesn't matter how much faster, only that it is faster)
3) Most importantly, the problem never states when the counting should stop, thus, when the cars are so close to each other (closer than the bodylength of a grasshopper), the grasshopper will still be jumping.

The count is only infinite since the count number will follow a hyperbolic path. As distance between the cars, $$d \rightarrow 0$$ the count, $$c \rightarrow \infty$$

I'm using logic to answer this though.

Yep. Finite distance leading to an infinite count.

14. Jul 20, 2009

### diazona

I think HallsofIvy was talking about the distance traveled (which is not infinite), rather than the number of jumps (which apparently is).

15. Jul 20, 2009

### Mentallic

Ahh my mistake. I just assumed since the OP answered the distance question already and all the discussion was about the number of jumps between the cars.
But you have to admit, Hallsofivy's post was quite misleading. There is mention of infinite so I think we both were on the wrong page

16. Jul 20, 2009

### maverick_starstrider

I love this classic question. It's a perfect example of the 'poet's' path vs. the 'peasant's' path. if you take a minute to take a step back and think about it it's incredibly easy. If you don't it's basically impossible.

17. Jul 21, 2009

### HallsofIvy

Staff Emeritus
Okay, I just noticed that this thread was titled "How many times does the grasshopper jump". But the question asked was "What total distance does the grasshopper travel before the cars hit?". That is certainly misleading but the question asked was the question I answered and I don't see how my response was misleading. And the only mention of "infinite" was to assert that the total distance the grasshopper covered was NOT "infinite".

1) If two cars are 200 m apart and coming toward each other at 10 m/s each, how long is it until they collide?
2) What distance will the grasshopper, jumping at 15 m/s, jump in that time?

If you really want to do it the hard way,

When the grasshopper first jumps, the cars are 200 ft apart. The grasshopper is going 15m/s and the car to which it is jumping is going at 10 m/s so they will meet after 200/(10+ 15)= 200/25 s= 8 seconds. In that time the grasshopper has gone 15(8)= 120 m and the car 10(8)= 80 m. 120+ 80= 200 m. At the time the grasshopper arrives at the second car, both cars have gone 80 feet so they are 200- 80- 80= 40 m apart. Do the same thing. The grasshopper jumps back to the first car in time t= 40/(10+15)= 40/25= 8/5 seconds. In that time both cars have gone 10(8/5)= 16 feet so they are now 40- 16- 16= 8 m apart. The grasshopper jumps back to the second car in time t= 8/(10+ 15)= 8/25 s. That should be enough to see the pattern. The total time for the first three jumps is 8+ 8/5+ 8/25 second and the distance is 120+ 120/5+ 120/25 meters.

The total distance is the sum of the geometric series,
$$\sum_{n=0}^\infty 120\left(\frac{1}{5}\right)^n$$

Theoretically, then, there are an "infinite" number of jumps. But this was posted in the physics section, not mathematics! There is no way to calculate the "actual" number of jumps but the total distance is easy.

Last edited: Jul 21, 2009
18. Jul 21, 2009

### Mentallic

Well everyone beforehand was mentioning infinite because they were answering the number of jumps, not the distance. Your response was only misleading to me because I skimmed through it and didn't fully process the part "How much distance will it cover in that time?" and only focused on the last line "No, the answer is NOT "infinity", in fact, it will be much smaller than you think it ought to be!", so yes, it was my fault. Sorry to cause any frustration.

Why wouldn't the geometric series instead be

$$200\sum_{n=0}^\infty \left(\frac{1}{5}\right)^n$$

??

Edit: I found where my thoughts were flawed. Never mind, hallsofivy's sequence is correct.

Last edited: Jul 21, 2009
19. Jul 22, 2009

### HallsofIvy

Staff Emeritus
Actually, I finally went back and read the original post completely. I now see that the original post posed the question "What is the total distance" and answered it completely, using the method I later suggested. Then asked "How many times".

I would however, disagree with an "infinite" number of times. Mathematically, the geometric sum assumes an infinite number of terms but because this was posted under "physics" rather than "mathematics" I would say the answer is not "an infinite number of times" but "it is impossible to determine how many times" without some additional information such as how close the cars must be (> 0) before we consider them to have collided.

Here's a related problem. A man, boy, and dog go for a hike, starting from the same point at the same time. The man walks at 5 mph, the boy at 3 mph and the dog runs at 8 mph, always running between the two. After one hour, the man is exactly 5 miles from the starting point and the boy is 3 miles from it. Where is the dog?

That is impossible to answer because of the ambiguity about where the dog is in first "infinitesmal" time. To see that it is impossible, start with the man 5 miles from the starting point, the boy 3 miles, put the dog anywhere between them and run it backwards: no matter where the dog is between them, they all arrive back at the starting point.

If you start the man and boy, say, 1 m apart, or any non-zero, non-"infinitesmal", distance apart the problem becomes tedious but doable.

20. Jul 22, 2009

### Mentallic

Skimming posts is an addictive habit that needs to be treated. Don't try fight it cold-turkey. I can't...

As I similarly suggested in post #7?

Yes this is true.
But my instincts are telling me that no matter what the dog is doing in the first instant in time (whether going forwards or backwards), it will run the same distance at the end of the hour. So the dog has the chance to end up in 2 places? Each position equidistant from the boy and man.
But again, as you proposed, if we rewind the situation, it doesn't matter where the dog is standing at the end of the hour. It will always end up back in the zero-point of course. Does this mean my instinct is incorrect?

You mean to say if we place restrictions on this scenario? Well that just strips away all the fun then