Don't need a mathematician, either. Or not much of one. 60 mph = 1 mile/minute. 30 mph = 1/2 mile per minute. 10 miles + 5 miles..... There is a much more complex way to do it with algebra, but most people set it up wrong, and can't solve it anyway. And of course this answer assumes infinitely high acceleration from 0 to 60 or 30 and absolutely constant speed thereafter, an absolutely precise clock, etc. Or other mostly irrational boundary conditions. Best engineering answer: "about 15"
No algebra needed. I got an answer that matches Anon @3:41PM through a different order of calculations. The approach speed is 60 + 30 = 90mph = 1-1/2 mile per minute. Times 10 is 15 miles.
Like he said, this is an idealized problem that ignores real world factors - acceleration time, variation in speed, and uncertainties in every measurement.
15 miles apart
ReplyDeleteDon't need a mathematician, either. Or not much of one. 60 mph = 1 mile/minute. 30 mph = 1/2 mile per minute. 10 miles + 5 miles..... There is a much more complex way to do it with algebra, but most people set it up wrong, and can't solve it anyway. And of course this answer assumes infinitely high acceleration from 0 to 60 or 30 and absolutely constant speed thereafter, an absolutely precise clock, etc. Or other mostly irrational boundary conditions. Best engineering answer: "about 15"
ReplyDeleteIs the answer 15miles apart?
ReplyDeleteNo algebra needed. I got an answer that matches Anon @3:41PM through a different order of calculations. The approach speed is 60 + 30 = 90mph = 1-1/2 mile per minute. Times 10 is 15 miles.
ReplyDeleteLike he said, this is an idealized problem that ignores real world factors - acceleration time, variation in speed, and uncertainties in every measurement.