A very long train is inching slowly along at a constant speed of 1 mile/hour. Charlie starts at the back of the train, running at a constant speed and reaches the front of the train in .07 hours, then for some strange and mysterious reason turns around and runs to the back of the train in .05 hours. How long is the train in feet?
WITH WORK PLEASE!!! :)
2006-09-24 16:59:56 · 3 answers · asked by bluenecklace 2 in Education & Reference ➔ Homework Help
This is a trick question right? There is no way to solve this problem given the information presented. The length of the train is impossible to say because we don't know the speed at which Charlie runs. The standard equation to figure the distance from the front of the train to the back would be distance = time X speed, so the distance would be .05 or .07 times the speed on that running attempt. We have the time it takes, but not the speed at which Charlie travels therefore, no answer to the question. The speed of the train is unimportant.
2006-09-24 17:18:37 · answer #1 · answered by Magic One 6 · 0⤊ 0⤋
this is a problem in relativity
It's been a while, but I'll try to help as much as I can
velocity of train cte: vt = 1 mi/hr
length of train cte: L mi
velocity of Charly running back and forth on the train cte: vc mi/hr
Relative Velocity of Charly running towards the front in 0.07 hr:
Vr1 = L/0.07 = 14.285 L mi/hr
Relative Velocity of Charly running towards the back in 0.05 hr:
Vr2 = L/0.05 = 20 L mi/hr
also we know that
Vr1 = Vt + Vc => Vc = Vr1 - Vt = 14.285L - 1
and
Vr2 = Vt - Vc => Vc = Vt - Vr2= 1 - 20L = 1 - 20L
assuming that Charly runs at a constant velocity Vc = cte
therefore : 20L + 14.285L = 2 => L = 0.0583 miles
I'm not 100% sure cos my brains are a little rusty but it's something like that
2006-09-25 00:44:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋
thats more like physics and science not algebra to me
2006-09-25 00:07:38 · answer #3 · answered by austin_penguin 4 · 0⤊ 0⤋