Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
2007-08-22 06:25:48 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
4.5 hours
the first person is 18 miles ahead when the second person starts (6mph x 3 hours)
the second person is gaining at a rate of 4mph (10-6) so how long will it take the person to gain 18 miles at a rate of 4 miles per hour of gain? 18/4=4.5 hours
2007-08-22 06:29:27 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Let the time the cyclists meet be t
Distance travelled by first cyclist = (6*t) + (3*6)
Distance travelled by 2nd cyclist = 10*t
Since the two distances are the same: 10*t = (6*t) + (3*6)
4*t = 18, t = 4.5hrs
2007-08-22 13:38:14 · answer #2 · answered by tj is cool 5 · 0⤊ 0⤋
t1 = t2 + 3
v2 = 10, v1 = 6
x1 = 6(t + 3)
x2 = 10t
Solve for equal x:
6(t + 3) = 10t
6t + 18 = 10t
4t = 18
t = 9/2
2007-08-22 13:34:14 · answer #3 · answered by gebobs 6 · 0⤊ 0⤋
Well..........how much time.....5 hours +3hours= 8hours
i think
2007-08-22 13:39:59 · answer #4 · answered by starwars f 2 · 0⤊ 1⤋
a long time..
2007-08-22 13:31:26 · answer #5 · answered by Anonymous · 0⤊ 0⤋