15.

Question

At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist heading north is riding 4 km/hour faster than the bicyclist heading south. At 10:30, they are 39 km apart. Find the two bicyclists’ rates.

Answer 1

Let the speed of the bicyclist heading south be $v$ km/hour. Then, the speed of the bicyclist heading north will be $v+4$ km/hour.

Since they pass each other at 9:00 and are 39 km apart at 10:30, the time elapsed between 9:00 and 10:30 is 1.5 hours.

During this time, the distance traveled by each bicyclist can be calculated as follows:

• The distance traveled by the bicyclist heading south: $${\text{Distance}}_{\text{south}}=v\times 1.5$$

• The distance traveled by the bicyclist heading north: $${\text{Distance}}_{\text{north}}=(v+4)\times 1.5$$

The total distance traveled by both bicyclists when they are 39 km apart is: \[ \text{Distance}{\text{south}} + \text{Distance}{\text{north}} = 39 \]

Substituting the distances: $$1.5v+1.5(v+4)=39$$

Simplifying the equation: $$1.5v+1.5v+6=39$$ $$3v+6=39$$

Now, subtract 6 from both sides: $$3v=33$$

Now, divide by 3: $$v=11$$

Now, we can find the speed of the other bicyclist: $$v+4=11+4=15$$

Thus, the speeds of the two bicyclists are:

• The bicyclist heading south: 11 km/hour
• The bicyclist heading north: 15 km/hour