SOLUTION: A cyclist rode the first 22​-mile portion of his workout at a constant speed. For the 16​-mile cooldown portion of his​ workout, he reduced his speed by 33 miles
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Question 1038880: A cyclist rode the first 22-mile portion of his workout at a constant speed. For the 16-mile cooldown portion of his workout, he reduced his speed by 33 miles per hour. Each portion of the workout took the same time. Find the cyclist's speed during the first portion and find his speed during the cooldown portion.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The formula we will use here is
D = r*t
where
D = distance
r = rate (aka speed)
t = time
--------------------------------
"A cyclist rode the first 22 mile portion of his workout at a constant speed"
means that we know D = 22, so
D = r*t
22 = r*t
Let's solve for t to get
t = 22/r
-----------------------------------
"For the 16 mile cooldown portion of his? workout, he reduced his speed by 33 miles per hour"
the distance is now D = 16 and the speed goes from r to r-33
D = r*t
16 = (r-33)*t
t = 16/(r-33)
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We have these two equations
t = 22/r
t = 16/(r-33)
set the two right sides equal to each other and solve for r
22/r = 16/(r-33)
22(r-33) = 16r
22r-726 = 16r
22r-726-22r = 16r-22r
-726 = -6r
-6r = -726
-6r/(-6) = -726/(-6)
r = 121
So his initial speed is 121 miles per hour which seems to be way too fast to be realistic.
The speed during the cool-down period is 121 - 33 = 88 miles per hour which is also very unrealistic for a bicyclist.
There's a possibility there could be a typo in the problem.
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