SOLUTION: A sprinter who average 30ft/sec. completed a race 1 sec before another sprinter who average 28ft/sec.What was the distance of the race.
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Question 144194: A sprinter who average 30ft/sec. completed a race 1 sec before another sprinter who average 28ft/sec.What was the distance of the race. Found 2 solutions by ankor@dixie-net.com, scott8148:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A sprinter who average 30ft/sec. completed a race 1 sec before another sprinter who average 28ft/sec.What was the distance of the race.
:
Find the time of of the 1st sprinter first
Let t = time of the 1st runner to complete the race
then
(t+1) = time of the 2nd runner
:
Write a distance equation
:
Distance = speed * time
:
1st runner dist = 2nd runner dist
30t = 28(t+1)
30t = 28t + 28
30t - 28t = 28
2t = 28
t =
t = 14 sec time of the 1st runner
;
Then, we know dist = speed * time
30 = 14 = 420 ft
:
Check solution using the 2nd runner who took 1 sec longer
28 * 15 = 420 ft
