Two cyclists, 36 mi apart, start riding towards each other at the same time. One cycles twice as fast as the other. If they meet 1 hr later, at what average speed is each cyclist traveling?
slower cyclist
mi/hr
faster cyclist
mi/hr
Let's assume the speed of the slower cyclist is x mi/hr.
Since the faster cyclist is cycling twice as fast, their speed would be 2x mi/hr.
In 1 hour, the slower cyclist would have traveled x mi, and the faster cyclist would have traveled 2x mi.
Since they start 36 mi apart and meet after 1 hour, the distance traveled by the slower cyclist and the faster cyclist should add up to 36 mi.
So, x + 2x = 36
3x = 36
x = 12
Therefore, the slower cyclist is traveling at 12 mi/hr and the faster cyclist is traveling at 24 mi/hr.
Let's assume the speed of the slower cyclist is x mi/hr. According to the problem, the faster cyclist is cycling twice as fast, so their speed would be 2x mi/hr.
Since they are riding towards each other, their speeds will add up to cover the distance of 36 mi.
Using the formula Distance = Speed × Time, we can set up the equation:
Distance = Speed × Time
36 mi = (x mi/hr + 2x mi/hr) × 1 hr
Simplifying the equation, we get:
36 mi = 3x mi/hr
To find the speed of each cyclist, we need to solve for x:
36 mi = 3x mi/hr
36 mi / 3 = x mi/hr
12 mi/hr = x mi/hr
Therefore, the speed of the slower cyclist is 12 mi/hr, and the speed of the faster cyclist is twice that, which is 24 mi/hr.