Two cyclists left simultaneously from cities a and b heading towards each other at constant rates and met in 5 hours. the rate of the cyclist from a was 3 mph less than the rate of the other cyclist. if the cyclist from b had started moving 30 minutes later than the other cyclist, then the two cyclists would have met 31.8 miles away from a. what is the distance between a and b, in miles?

Step-by-step explanation:

Givens

Cyclist A

r = r_a - 3

t = 5 hours.

d = ?

Cyclist B

r = r _a

t = 5 hours - 1/2 hour = 4.5 hours.

d = d - 31.5

Formula

(r - 3)*5 + 5*r = d

r*4.5 = d - 31.5

Explanation

The rate of A is 3 less than the rate of B. Together, they bicycle the entire distance (d). That's the first equation

The second equation is a lot harder. That equation has to do with the one starting off from B. His useful cycling time is 4 1/2 hours because he starts off 1/2 hour later.

He travels d - 31.5 which A travels 31.5

Solution

The total distance is the same. We will use that fact to solve for r first.

(r - 3)*5 + 5r = d

4.5r + 31.5 = d

Remove the brackets in the top equation.

5r - 15 + 5r = d

10r - 15 = 4.5r + 31.5             Add 15 to both sides

10r -15+15 = 4.5r + 31.5+15

10r = 4.5r + 46.5                   Subtract 4.5 r from both sides.

10r-4.5r = 46.5

5.5r = 46.5

r = 8.45 mph

4.5r + 31.5 = d

4.5*8.45 + 31.5 = d

d = 69.53 miles

If this proves to be incorrect, and you have choices, please list them.

the answer is 48m^2!

step-by-step explanation: