SOLUTION: In a first race, A gives b a Headstart of 100 feet and overtakes him in four minutes. In the second race he gains 750 feet on B when B runs 9000 feet. Find the rate at which each r

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: In a first race, A gives b a Headstart of 100 feet and overtakes him in four minutes. In the second race he gains 750 feet on B when B runs 9000 feet. Find the rate at which each r      Log On


   



Question 893076: In a first race, A gives b a Headstart of 100 feet and overtakes him in four minutes. In the second race he gains 750 feet on B when B runs 9000 feet. Find the rate at which each runs.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the runners A and B
Let their speeds be +a+ and +b+ in feet / minute
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Start a stopwatch when A leaves
In the 1st race:
Let +d+ = the distance A runs when he overtakes B
A's equation:
(1) +d+=+a%2A4+
B's equation:
(2) +d+-+100+=+b%2A4+
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In the 2nd race:
I think they both leave at the same time, so
Let +t+ = time in minutes until A is +750+ ft
ahead of B
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A's equation:
(3) +9000+%2B+750+=+a%2At+
B's equation:
(4) +9000=+b%2At+
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This is actually 4 equations with 4 unknowns,
so it is solvable
Substitute (1) into (2)
(2) +a%2A4+-+100+=+b%2A4+
(2) +4a+-+4b+=+100+
(2) +a+-+b+=+25+
(2) +a+=+b+%2B+25+
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Substitute (4) into (3)
(4) +t+=+9000%2Fb+
(3) +9000+%2B+750+=+a%2A%28+9000%2Fb+%29+
(3) +9750+=++a%2A%28+9000%2Fb+%29+
(3) +9750b+=+9000a+
Divide both sides by +750+
(3) +13b+=+12a+
Substitute (2) int (3)
(3) +13b+=+12%2A%28+b+%2B+25+%29+
(3) +13b+=+12b+%2B+300+
(3) +b+=+300+
and, since
(2) +a+=+b+%2B+25+
(2) +a+=+325+
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A's speed is 325 ft/min
B's speed is 300 ft/min
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check:
(1) +d+=+a%2A4+
(1) +d+=+325%2A4+
(1) +d+=+1300+ ft
and
(2) +d+-+100+=+b%2A4+
(2) +d+-+100+=+300%2A4+
(2) +d+=+1200+%2B+100+
(2) +d+=+1300+ ft
OK
You can check (3) and (4)