Could you show me how to solve the following question?
Runner A is initially 4.0 mi west of a flagpole and is running with a constant velocity of 6.0 mi/h due east. Runner B is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when their paths cross?
thanks alot.
2006-08-22 11:43:24 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
the two equations made from the problem both equal x (the poistion relative to the pole as a function of time) set them equal to each other and then solve for the time. next, plug the time into either of the distance equations to get the position.
4 - 6t =x = -3 + 5t
7 = 11t
t = 7/11
x = 4 - 6*7/11 = distance from flag pole
x = roughly 0.1818181818 miles or 2/11ths of a mile
2006-08-22 11:50:13 · answer #1 · answered by promethius9594 6 · 1⤊ 0⤋
welll
lets asume that their paths cross after some T period of time
if their paths cross that means that the sum of lengths they have run must be the initial distance between them
6T+5T=7
T=7/11
now lets see. A is 6*7/11mi east from innitial position and its 42/11mi=3 and 9/11mi
that means A didnt even get to flagpole. he is 2/11 mi west to a flagpole now. and so is B too
2006-08-22 12:02:17 · answer #2 · answered by lika n 1 · 1⤊ 0⤋
Easy enough. How far apart are the runners initially? 7 miles. What is their combined velocity of approach? 11 mph. When will they meet? Do the math. Where will they meet? Do the math.
2006-08-22 11:52:02 · answer #3 · answered by Anonymous · 0⤊ 0⤋
S= d/T
speed = distance/time
use this formula then... search which of those is the same in both cases..make an equation...find what you need
2006-08-22 11:50:20 · answer #4 · answered by freezing school 5 · 0⤊ 0⤋