Question

2. 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?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

Answer 1

4 1/2 hours

Answer 2

how?

Answer 3

can u draw it 4 me?

Answer 4

the 1st cyclist travels @ 6mph and started 3 hours earlier than 2nd cyclist
hence already travelled 18 miles ..

Hours Miles traveled
3 18 miles (1st) 0 miles (2nd) -- 2nd starts @ 10mph after 3 hrs
4 24 10
5 30 20
6 36 30
7 42 40
7.5 45 45 -- hence 7.5 hrs - 3 hrs = 4.5 hrs

Answer 5

why did u subtract -3 at the end?

Answer 6

?

Answer 7

since it was from the time the second started biking

Answer 8

the question was from the time the second one started cycling so he did it 3 hrs later to 1st cyclist

Answer 9

but we gave the first biker a head start by 3 and the second biker at 0

Answer 10

Here is one way to think about it.
the first biker goes 3 hr * 6 m/hr = 18 m in 3 hours

now we start the clock. the first biker will be 18+6t away
where t is the time in hours
meanwhile the second biker has started, and he goes 10t in t hours

after a certain amount of time we expect
10t = 18+6t (they have traveled exactly the same distance. that means the 2nd biker has caught up to the 1st biker)

solve for t

Answer 11

1st cyclist:

head start - 3hrs so travel advantage of 18 miles as he was travelling @ 6mph
so
after 1st hr with reference to the 2nd cyclist
1st cyclist traveled 24 miles ( 18 miles for head start + 6mph -his normal speed )
2nd cyclist @ 10 mph - 10 miles
so after 4.5 hrs both traveled 45 miles ( 1st cyclist = 18 miles head start + 6 miles for 1st hr + 6 miles for 2nd hr + 6 miles for 3 hr + 6 miles for 4th hr + 3 miles for half an hour )
and also 45 miles traveled by 2nd cyclist ( as he travels @ 10 mph )