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?
2006-11-20 04:31:23 · 5 answers · asked by Serrato 1 in Education & Reference ➔ Homework Help
Bicyclist 1 will travel 6 miles in 1 hour
Bicyclist 2 will travel 10 miles in 1 hour
Bicyclist 1 will start out 18 (3 hours * 6 miles/hour) miles ahead of bicyclist 2.
Bicyclist 1 will catch up by 4 (10 - 6) miles/hour.
So, we take
18 miles / 4 miles/hour and get
4.5 hours
It will take the second cyclist 4.5 hours to catch up with the first.
2006-11-20 05:28:56 · answer #1 · answered by Math Helper 3 · 0⤊ 0⤋
distance = rate x time
Since they will meet, the distance travelled by both is the same
Using notation 1 for the first cyclist and 2 for the 2nd...
rate1 x time1 = rate2 x time2
rate1 of cyclist 1 = 6 mph
rate2 of cyclist 2 = 10 mph
time1 of cyclist 1 = t
time 2 of cyclist 2 = t -3 (started 3 hours later)
so 6mph(t) = 10 mph (t-3)
temporarily suspending units
6t = 10 (t-3)
multiply out
6t = 10t -30
combine terms
4t = 30
t = 7.5 hours
t-3 - 4.5 hours for second cyclist
2006-11-20 12:38:34 · answer #2 · answered by Grover 3 · 0⤊ 0⤋
It's an equation
On the left, the first biker, who has covered 18 miles.
18 + 6x = 10x
x = 4,5
4 hours and a half will pass in order to catch up the first biker.
2006-11-20 12:37:06 · answer #3 · answered by Anonymous · 0⤊ 0⤋
10x = 6(x + 3)
10x = 6x + 18
4x = 18
x = 4.5 hours
2006-11-20 12:33:50 · answer #4 · answered by dmb 5 · 0⤊ 0⤋
I believe he answer is 5hrs.
2006-11-20 12:38:07 · answer #5 · answered by Anonymous · 0⤊ 0⤋