SOLUTION: -Applications of Quadratic Equations during a bicycle race, suppose that on one particular day, the racers must complete 210 mi. One Cyclist, traveling 10 mph faster than the seco

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Question 419407: -Applications of Quadratic Equations
during a bicycle race, suppose that on one particular day, the racers must complete 210 mi. One Cyclist, traveling 10 mph faster than the second cyclist, covers this distance in 2.4 h less time than the second cyclist. Find the rate of the first cyclist.
I am so confused on how to do word problems dealing with quadratic equations...
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Always remember that with d = r*t problems, each thing that is
moving must have it's own equation, and each equation must
refer to 1 and only 1 person or thing that is moving.
Both cyclists go the same distance: +210+ mi
Let t = time for 2nd cyclist
Let s = speed for the 2nd cyclist
-----------------------
1st cyclist:
(1) 210+=+%28s+%2B+10%29%2A%28t+-+2.4%29
2nd cyclist:
(2) 210+=+s%2At
----------------
Note that there are 2 equations and 2 unknowns,
so this is solvable.
(1) 210+=+%28s+%2B+10%29%2A%28t+-+2.4%29
(1) 210+=+s%2At+%2B+10t+-+2.4s+-+24
and, from (2)
(2) t+=+210%2Fs
-----------------
Substituting from (2) into (1):
(1) 210+=+210+%2B+10%2A%28210%2Fs%29+-+2.4s+-+24
(1) +2.4s+=+2100%2Fs+-+24+
Multiply both sides by s
(1) +2.4s%5E2+=+2100+-+24s
(1) +2.4s%5E2+%2B+24s+-+2100+=+0
Divide through by 2.4
(1) +s%5E2+%2B+10s+-+875+=+0+
Use the quadratic equation
s+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+10
c+=+-875
x+=+%28-10+%2B-+sqrt%28+10%5E2-4%2A1%2A%28-875%29+%29%29%2F%282%2A1%29+
x+=+%28-10+%2B-+sqrt%28+100+%2B+3500+%29%29%2F+2+
x+=+%28-10+%2B-+sqrt%28+3600+%29%29%2F+2+
x+=+%28-10+%2B+60%29%2F+2+ (note: the negative root gives a negative answer-can't use it)
s+=+50%2F2
s+=+25
The 1st cyclist's rate is
s+%2B+10+=+25+%2B+10
s+%2B+10+=+35
The 1st cyclist's speed is 35 mi/hr
check answer:
(1) 210+=+%28s+%2B+10%29%2A%28t+-+2.4%29
(1) 210+=+%2825+%2B+10%29%2A%28t+-+2.4%29+
(1) +210%2F35+=+t+-+2.4+
(1) t+=+6+%2B+2.4+
(1) t+=+8.4
-------------
(2) 210+=+s%2At
(2) 210+=+25%2A8.4+
(2) 210+=+210+
OK