Two cities, Kensington, MD and Reston, VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington. It flies at the speed of 10 km/hour and passes the hiker from its city. When it reaches the Reston hiker, it turns around and flies back to the Kensington hiker. It keeps doing so until the hikers meet. If the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston at the moment the two hikers meet, how many kilometers has the fly flown?

df = 30 km

Explanation:

Given:

- The distance between cities d = 30 km

- The speed of both walkers v1 = v2 = 5 km/h

- speed of the fly vf = 10 km/h

Find:

If the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston at the moment the two hikers meet, how many kilometers has the fly flown?

Solution:

- The distance travelled by individual walkers d1 = d2. The time taken for both hikers to meet.

                                d1 + d2 = 30

                                v1*t + v2*t = 30

                                2*5*t = 30

                                 t = 3 hrs

- The fly flies at a speed of 10 km/h for t = 3 hrs, the total distance travelled would be:

                                df = t*vf

                                df = 3*10

                                df = 30 km

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