SOLUTION: Running at an average rate of 6 meters per second, a sprinter ran to the end of a track. The sprinter then jogged back to the starting point at an average rate of 2 meters per seco

Question 132952This question is from textbook Applied College Algebra
: Running at an average rate of 6 meters per second, a sprinter ran to the end of a track. The sprinter then jogged back to the starting point at an average rate of 2 meters per second. The total time for the sprint and the jog back was 2 minutes 40 seconds. Find the length of the track. This question is from textbook Applied College Algebra

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let L=length of track
Time spent running to end of track= L/6
Time spent jogging back=L/2
And we are told that these two times add up to 2min 40sec or 160sec (let's deal in seconds so we don't get confused). So:
L/6 + L/2=160 multiply each term by 6
L+3L=960
4L=960 divide both sides by 4
L=240 meters------------------------length of track
(Note in the equation above, both L/6 and L/2 are expressed in (meters/meters/sec)or sec).
CK
240/6+240/2=160 sec
40+120=160
160=160

Hope this helps----ptaylor

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