a reply to:
theatreboy
Oh, this should be fun.
The trick is to get the units consistent. You have feet combined with miles per hour, and that simply will not do. To convert from miles per hour to
feet per second, we need to multiply by 5280 (feet in a mile), then divide by 3600 (seconds in an hour). So we get the following (I am rounding to two
digits):
4 mph = 5.87 fps
35 mph = 51.33 fps
Now we have everything in consistent units, let's look at the problem again:
At the start point (t0), we do not know where the northbound pack of cars is. We do know where the southbound pack of cars is: the first car is beside
our pedestrian. That means the last car will be past him at some future time. Now, how long is this pack of cars? There are seven cars, each separated
by a car length, so that's 7+6=13 car lengths. Each car length is 8 feet, so that's 104 feet long.
(I'm not going to mention that these are extremely short cars spaced way, way, way too close for safety!)
Now, at some future time, we do know where the northbound cars are. That happens when the sixth southbound car is directly beside the pedestrian; the
first northbound car is exactly beside the pedestrian as well. When does this happen? Well, between the first and sixth southbound cars (center to
center) is 10 car lengths (five cars and five spaces), so that's 80 feet. So our second time point happens when the cars have traveled 80 feet,
right?
Wrong! Relativity matters here. The southbound cars are traveling at 51.33 fps,
while the pedestrian is traveling at 5.87 fps in the opposite
direction. We need the speed of the cars
relative to the pedestrian. Opposite directions, so the speeds add. The second point in time occurs
when the cars have traveled 80 feet at the relative speed to the pedestrian, which is 51.33 + 5.87 = 57.2 fps. 80 Ć· 57.2 = 1.4 s (seconds). So this
second point in time is at t1 = 1.4 s. Since we will later want to know how far our pedestrian has traveled at this point, we can go ahead and
calculate that he has traveled 8.22 feet (5.87 fps x 1.4 s) at t1 = 1.4 s.
Now we use a little logic: the southbound cars are almost past him, but the northbound cars are just getting to him. So he will probably be able to
cross as soon as the northbound cars pass him. When will that happen?
We know the total length of the northbound cars is 104 feet. We also know that the first northbound car is beside him, so we deduct half a car length,
or 4 feet, for a total of 100 feet. How long will it take the northbound cars to move 100 feet relative to the pedestrian? In this case, since both
the cars and the pedestrian are traveling in the same direction, their speeds subtract. 51.33 - 5.87 = 45.46 fps. 100 ft Ć· 45.46 fps = 2.2 s.
So we now know the road will be clear at 2.2 s past our t1 point of 1.4 s, which is a total of t2 = 3.6 s.
But we're not done yet! The southbound lanes next to the pedestrian will be clear before the northbound lanes will clear. Our pedestrian can start
crossing before that point, as long as he is only halfway across when the northbound lanes clear. So how long will it take him to cross two lanes of
traffic? 16 ft Ć· 5.87 fps = 2.73 s. According to that, he can start crossing at 3.6 - 2.73 = 0.87 s.
The problem is that at 1.4 seconds, the first two lanes are not yet clear! So he can't start crossing at 0.87 s; he will have to wait until the first
two lanes are clear before he begins to cross. Since we know now that the northbound lanes will be clear by the time he gets to them, we don't have to
worry about them. When will the southbound lanes be clear?
We know that from the start point, the southbound cars will clear after they have traveled 100 ft relative to the pedestrian, which is at a relative
speed of 57.2 fps. So 100 ft Ć· 57.2 fps = 1.75 s. The first lane will clear sooner, though, when the cars have moved 100 - 16 = 84 feet relative to
the pedestrian. That is 84 ft Ć· 57.2 fps = 1.47 s.
To double check, the time our pedestrian will take to cross one lane is 8 ft Ć· 5.87 fps = 1.36 s. The last car will have covered the needed 16 feet
to clear the inner lane in 16 ft Ć· 51.33 fps (the pedestrian is moving sideways so relative and actual speeds are the same) = 0.31 s. So the last car
will have easily cleared the inner lane long before the pedestrian gets there.
So, to sum all this up:
- The pedestrian can start to cross as soon as the outer lane southbound is clear, at 1.47 s.
- He will reach the inner southbound lane at 1.47 + 1.36 = 2.83 seconds. That lane will have cleared at 1.75 s.
- He will reach the middle of the road at 1.47 + 1.36 + 1.36 = 4.19 s. The last car northbound will have easily cleared both northbound lanes by that
time.
So the answer to number 1, at what point can the pedestrian safely cross, is beginning at 1.47 s.
Question 2: How far will he have walked before he can start to cross? 5.87 fps x 1.47 s = 8.63 ft.
Question 3: The answer is yes, there will be a pattern, and requires no actual calculation. The pedestrian's speed is cyclical; he walks, crosses,
walks, crosses back, and repeats. The traffic is cyclical; the cars come in evenly spaced increments of identical length and speed. Therefore the
result will be a cyclical pattern. The crossing back may be at a different relative time than the first crossing, as the traffic is mow moving in the
opposite direction to the first crossing, but the timing will be identical for each direction of crossing.
I was right; that
was fun!
TheRedneck