Math ?? non poker

[Scorpion]

Can someone help?

Charlie was crossing a narrow bridge. When he was halfway across he saw a car on the opposite side 200 yards away and heading toward him. He turned back and ran and barely made it before the car hit him. If Charlie had tried to cross the bridge the truck would have hit him 3 yards before he reached the end. How long is the bridge?

As near as I can figure there are 5 variables- the rate of the truck and Charlie and the time it takes for both of them to get whereever they are going and then the length of the bridge.

[SossMan]

wouldn't you just need the speed of charlie and the truck?

[fnord_too]

[ QUOTE ]
Can someone help?

Charlie was crossing a narrow bridge. When he was halfway across he saw a car on the opposite side 200 yards away and heading toward him. He turned back and ran and barely made it before the car hit him. If Charlie had tried to cross the bridge the truck would have hit him 3 yards before he reached the end. How long is the bridge?

As near as I can figure there are 5 variables- the rate of the truck and Charlie and the time it takes for both of them to get whereever they are going and then the length of the bridge.

[/ QUOTE ]

Ok, I am reading this as 200 yards from Charlie, not 200 yards from the other side of the bridge.

So, lets call half the bridge's length L, Charlie's speed R1 and the Car's speed R2.

Now, the distance between Charlie and where he will get hit if he goes forwars is L - 3. The relative rate of Charlie and the car here is R1+R2.

Charlie and the car meet up L away from Charlies position the other way, and their relative rate is R2-R1.

I think we need times too, but these should cancel somewhere (I'm working this out as I go). Let's call the running toward the car time T1, and the running away from the car time T2.

So now we know (R1+R2)T1 = L - 3 and (R2-R1)T2 = L.

That sais that R1T1 + R2T1 + 3 = L and R2T2 - R1T2 = L so....

T1 = (L-3)/(R1+R2) and T2 = L/(R2-R1)

We know that there is some multiple A such that T2=A*T1.

A turns out to be (LR2-LR1-3R2+3R1)/(LR1+LR2). Now we can get rid of T2.

Now we get (if I haven't screwed up the algebra)
T1*(R1+R2+3/T1) = A*T1*(R2-R1) ==> R1 + R2 + 3/T1 = A*(R2-R1)

This algebra is just getting hairy. I believe you have all the information you need to solve this but a simple way to solve it escapes me at the moment. The general attack is to find the relationship between R1 and R2, then substitue in for one of them and find L. I could be wrong, you may need another bit of info, but it feels like you have enough.

[Lost Wages]

I don't think that there is enough information.

If we let:
L = length of bridge in yards
V(c) = Velocity of Charlie
V(t) = Velocity of car/truck(?)

From
When he was halfway across he saw a car on the opposite side 200 yards away and heading toward him. He turned back and ran and barely made it before the car hit him.
We can set the time it took Charlie to run halway across equal to the time it took the truck to cross + 200 yards.

Since time = distance/velocity, then:

Distance(c)/V(c) = Distance(t)/V(t), or

1) .5L/V(c) = (L + 200)/V(t)

From
If Charlie had tried to cross the bridge the truck would have hit him 3 yards before he reached the end.
we can do the same:

2) (.5L - 3)/V(c) = (200 - .5L + 3)/V(t)

Now we have used all of the information but have 2 equations and 3 unknowns, so unless I am missing something....

Lost Wages

Edit: made an oopsie

[jason1990]

You are right, there are five variables:

x = half the length of the bridge
r_C = Charlie's speed
r_T = Truck's speed
t = time for Charlie to run back and survive
t' = time for Charlie to run forward and get hit

We get four equations:

x = (r_C)t
x + 200 = (r_T)t
x - 3 = (r_C)t'
200 - (x - 3) = (r_T)t'

If we eliminate t and t', we get

x + 200 = (r_T/r_C)x
200 - (x - 3) = (r_T/r_C)(x - 3)

Now we see that we really only had four variables, because the answer only depends on the ratio of the truck's speed to Charlie's speed. So let r = r_T/r_C to get

x + 200 = rx
200 - (x - 3) = r(x - 3)

Multiplying the second equation by x gives

200x - x^2 + 3x = (x + 200)(x - 3)
0 = 2x^2 - 6x - 600
x^2 - 3x - 300 = 0

Since x > 0, we must take the positive root, which makes the bridge about 37.8 yards long.

[BruceZ]

You are all making this much too complicated. Here is the equation where H is half the length of the bridge:

H / (H + 200) = (H - 3) / (200 - H + 3)

Each side is the ratio of the distance Charlie traveled to the distance the car traveled in the same amount of time. These are equal since the ratio of their speeds are equal. This gives the quadratic equation:

H^2 - 3H - 300 = 0

H = [ 3 + sqrt(9 + 1200) ] / 2

Length of bridge = 2H =~ 37.8 yards.