From one point to another, the cyclist moved at a speed of 36 km / h in 20 minutes

| February 18, 2021 | education

From one point to another, the cyclist moved at a speed of 36 km / h in 20 minutes, he traveled back in 30 minutes. How fast did he move on his way back?

v1 = 36 km / h = 10 m / s – the speed of the cyclist from one point to another;

t1 = 20 minutes = 1200 seconds – the time interval during which the cyclist traveled from one point to another;

t2 = 30 minutes = 1800 seconds – the amount of time the cyclist spent on the return trip.

It is required to determine v2 (m / s) – the speed of the cyclist on the way back.

Let’s find the distance between two points:

S = v1 * t1 = 10 * 1200 = 12000 meters.

Then the speed of the cyclist on the way back will be:

v2 = S / t2 = 12000/1800 = 6.7 m / s.

Answer: on the way back, the cyclist moved at a speed of 6.7 m / s.



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