A train leaves town A for town B traveling at 35 mi/h. At the same time, a second train leaves town B for town A at 45mi/h. If the two towns are 320 mi apart how long will it take for the two trains to meet??
2006-11-12 04:02:44 · 5 answers · asked by sugalums 1 in Science & Mathematics ➔ Geography
vA = 35 m/h
vB = 45 m/h
dA + dB = 320 m (dA, dB are the distances traveled by each train)
tA = tB = t (the 2 trains will travel for the same amount of time before they meet)
vA (t) = dA
vB (t) = dB = 320 - dA
35(t) = dA
45(t) = 320 - dA
dA = 35t = 320 - 45t
80t = 320
t = 4 hours
2006-11-12 04:11:19 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The two trains will meet in four hours f(t)=d/s.
Let t= the time it takes to ravel the distance between the towns
Let s= the relative speed of both trains
Let d= the distance between the towns
Let a & b = the speed of each train respectively
The relative speed of both trains (s) is 80 mph, s = a + b. the distance is given there for d = 320; therefore t = 320/80. Train a will have traveled 140 miles and train be 180 miles, relative distance traveled is 320 miles at a relative speed of 80 mph.
2006-11-12 13:10:09 · answer #2 · answered by ikeman32 6 · 0⤊ 0⤋
four hours.
Every hour, the distance between them is reduced by 80 miles (35 mph + 45 mph).
80 miles divided into 320 is 4 hours.
2006-11-12 13:48:57 · answer #3 · answered by kent_shakespear 7 · 0⤊ 0⤋
4 hours
2006-11-12 12:12:46 · answer #4 · answered by _Hithere_ 1 · 0⤊ 0⤋
first turn trains speed to m/sec.
use the formula,
v=u+at or s=ut+1/2at2 or V2=U2+2as
2006-11-12 12:06:31 · answer #5 · answered by dog b 1 · 0⤊ 0⤋