SOLUTION: i cant figure out how to correctly form an equation for this:
a cyclist leaves his training base for a morning workout riding at a rate of 18mph. one hour later, is support staf
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Question 263249: i cant figure out how to correctly form an equation for this:
a cyclist leaves his training base for a morning workout riding at a rate of 18mph. one hour later, is support staff leaves the base in a car going 45mph in the same direction. how long will it take the support staff o catch up with the cyclist?
thanks for your help
Found 2 solutions by nerdybill, stanbon:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
a cyclist leaves his training base for a morning workout riding at a rate of 18mph. one hour later, is support staff leaves the base in a car going 45mph in the same direction. how long will it take the support staff o catch up with the cyclist?
.
Let t = hours it takes staff to catch up
then
"cyclist distance traveled" = "distance staff traveled"
(t+1)18 = t(45)
18t+18 = 45t
18 = 27t
18/27 = t
2/3 hours = t
or
40 minutes = t
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a cyclist leaves his training base for a morning workout riding at a rate of 18mph.
one hour later, his support staff leaves the base in a car going 45mph in the same direction.
how long will it take the support staff to catch up with the cyclist?
----
Cyclist DATA:
rate = 18 mph ; time = x hrs ; distance = rt = 18x miles
-------------------------
Staff DATA:
rate = 45 mph ; time = x-1 hrs ; distance = rt = 45(x-1) miles
===============================================================
Equation:
cyclist distance = staff distance
18x = 45(x-1)
18x = 45x - 45
27x = 45
x = 5/3 hr (cyclist time)
x-1 = 2/3 hr (staff time)
===============================
Cheers,
Stan H.
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