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
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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
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
