Question

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?

Answer 1

please help me this one

Answer 2

s1st=s2st
s1+s2=s3
6x(3+t)=10t
18=4t
t=4.5 hr

Answer 3

what is s1st?

Answer 4

Did you follow that with understanding?

Answer 5

not really, please explain

Answer 6

They will catch up when they have traveled the same distance. Agree?

Answer 7

yes

Answer 8

and distance (d) is equal to speed (s) times time (t). Is that clerar?

Answer 9

So lets set up some equations for this scenario.

The slower cyclist starts out 3 hours earlier and his speed is 1/2 of the fast cyclist.

Answer 10

Oop not quite 1/2 but slower by 4 mph lol

Answer 11

The slow cyclist, traveling at 6 mph, has an 18-mile head start over the faster 10 mph biker. The biker approaches the cyclist at 4 mph (10 - 6)mph. The time required for the biker to uvertake the cyclist is (18mi)/(4 mi/h) = 4.5 hours.

Answer 12

ok

Answer 13

so what is your equation for this ?

Answer 14

First cyclist distance will be the product of t+3 hrs and speed is 6mph.
Second cyclist distance will be t times his speed of 10 mph.

Set these two distance equal:
6(t+3) = 10t
6t+18=10t
4t=18
t=4.5 hours

Answer 15

Please review all provided answers and I believe you will see the light.

Answer 16

you wonderful

Answer 17

i understand now dear

Answer 18

Using distance = rate * time, solving for time.

Answer 19

Good luck with your studies.

Answer 20

thank you so much