Alan_Au
1721
Well, you see, when two people are attracted to each other, and one of them is a male born on a Tuesday…
I have twin boys, both born on a Tuesday.
That is all.
John Many Jewsday, born on a Tuesday
How does the old rhyme go?
Monday’s child is fair of face
Tuesday’s child is full of shit
Something like that.
13.44748/284.384743875832982.
Each of them obviously had a 50/50 chance of being a twin, so combined it was 100%.
99%
There’s a 1% chance the firstborn are born just before midnight, and the other after.
(Just for fun, read this: “What are the odds of twins being born in different years?”)
So Tanya Khovanova decided to necro this whole issue by publishing “Martin Gardner’s Mistake” in the latest issue of The College Mathematics Journal. Short version: Gardner first answered the question posed in the original post as 1/3, but then retracted his answer, arguing that the question is ambiguous.
Boo yah!
When I asked my personal statistician this question, and he ripped one of my eyeballs out of its socket, I figured out the answer pretty GB quick, tell you what, right now.
A man steps onto a moving walkway at an airport. The man’s velocity in feet/sec is given by f(t) = 2 + t, so his initial velocity is 2 fps and it increases at 1 fps per second. The walkway’s velocity in feet/sec is given by g(t) = 4 - 2t, so its initial velocity is 4 fps in the same direction as the man’s, but its velocity is dropping (and soon becomes negative).
How far does the man get from the walkway’s entrance, and how long does it take for him to be brought back (against his will) to the entrance?
(This question came up in the algebra class I teach, and I immediately assigned it for extra credit.)
I don’t see how you can answer this problem without knowing the rate at which the velocity of the slidewalk drops/reverses, and what the maximum negative speed is. At which point it is actually a calculus problem, isn’t it?
Rejected on grounds of no man or machine has infinitely accelerating velocity, they need top speeds in either direction.
Ignoring that, it’s a rephrased “how high does the accelerating rocket get before gravity pulls it back to the ground?” From a potential energy change perspective it’s just double the time it takes to get to the peak, right?
h(t) = f(t) + g(t) = 2 + t + 4 - 2t = 6 - t
h(t) = 0? 6 = t.
So 12 seconds to get there and get back.
Going out:
Start velocity = 6.
Final velocity = 0.
Average velocity = 3.
Time = 6
Distance = average velocity * time = 3 * 6 = 18.
Coming back:
Start velocity = 0
Final velocity = ?
Distance = (start + final) / 2 * time = 18 = final / 2 * time = 18
final * time = 36
We know final velocity is just time, so substitute:
time * time = 36
time = 6.
My rusty memory of physics led to computations that were far less elegant than Jason’s, but yielded the same answers. The man gets a maximum of 18 feet from the entrance before he starts returning against his will. He gets to the 18 feet mark at 6 seconds, & takes 6 more seconds to return to the entrance, for a total of 12 seconds.
Enidigm
1738
At 1 second, he advances 5 feet
At second 2, he advances 4 feet (9 total)
At second 3, he advances 3 feet (12 total)
At second 4, he advances 2 feet (14 total)
At second 5, he advances 1 foot (15 total)
At second 6, he advances 0 feet (15 total)
At second 7, he retreats 1 foot (14 total)
Ect.
He doesn’t move in jolts of stops & starts; he moves continuously.
So at the end of 1 second, he’s moved a total of 5.5 feet, not 5.
Or in general clunky hazy calculus memory:
total distance(t) = 6t - .5t^2,
which is the integral of the man’s velocity, which is
v(t) = 6 - t, as Jason pointed out
Three of my students approached me after class with answers that looked like Endigm’s. (They are algebra students, not calculus students.)
