distance formula....algebra help please???

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?

i need to know the formula used to find the answer to this problem. if you know the formula i'd really appreciate your help. thanks.

Answer 1

There is no formula per se. You just have to logically work through the problem.

Distance = speed x time
That's about hte only formula you'll need.

Let t = time after the second cyclist starts
When t = 0, the first cyclist has already been travelling for 3 hours, thus has travelled 6*3 = 18 miles

When t = T, they meet
So first cyclist travelled an additional 6T miles for a total of
6T + 18
The second travelled 10T miles

So 10T = 6T + 18
T = 4.5 hours
This is from the time hte second cyclist starts.
That would be 7.5 hours from the first.

Answer 2

Cyclist A: 6mph
Cyclist B: 10 mph

After 3 hours cyclist A is 18 miles from start.
Relative speed of cyclists is 4 mph
Assume cyclist A at rest and cyclist B travelling at 4 mph
t = 18 / 4 h = 4 h 30 min

Answer 3

its a system of linear equations dude

let x = the number of hours after the second biker started going
let y = the distance traveled by each biker

the first bikers equation is such:

3 hours * 6 miles per hour = 18 miles
the guy cycled 18 miles before the second biker started

his equation is

y = 18 + 6x

the second bikers equation is:

y = 10 X

what we need to find is where the two bikers distance traveled is equal so we use substitution

18 + 6x = 10x
18 = 4x
x = 4.5

4.5 hours will pass before the second biker catches the first biker , after the second biker starts going

it may also help to think of it graphically. put the two linear equations in to your graphing calculator and take a look.