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vampirchik [111]
3 years ago
5

Two cyclists leave towns 120 km apart at the same time and travel toward each other. One cyclist travels 10 km/h slower than the

other. If they meet in 2 hours, what is the rate of each cyclist?
Physics
2 answers:
stiv31 [10]3 years ago
5 0

Answer:

35 km/hr and 25 km/hr

Explanation:

Total distance is D.

Let speed of one cyclist be X km/hr.

Speed of second cyclist = ( X-10) km/hr

Now,

                        Distance = Speed * Time

Total distance D = Distance covered by slower cyclist + Distance covered by                                     faster cyclist.

                       D =  2* X + 2(X-10)

                       120 =  2X + 2X -20

                       120 = 4X - 20

                       140 = 4X

                         X = 35 km/hr

Therefore, Speed of faster cyclist = 35 km/hr

                  Speed of slower cyclist = 35-10

                                                          = 25km/hr

yawa3891 [41]3 years ago
3 0

Answer:

Exp35 km/h , 25 km/hlanation:

Let the speed of one cyclist is v and the speed of another cyclist is v - 10.

Distance traveled by the first cyclist in 2 hours = v x 2

Distance traveled by the second cyclist in 2 hours = (v - 10) x 2

They meet after 2 hours.

So the distance traveled by the first and the second cyclist is equal to 120 km.

2 v + 2 (v -10) = 120

2v + 2v - 20 = 120

4v = 140

v = 35 km/h

The speed of first cyclist is 35 km/h and the speed of another cyclist is 35 - 10 = 25 km/h.

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