SOLUTION: In a 4 x 100 relay race (each leg 100 meters long), the first place runner has a 0.550 second lead and is running at a constant speed of 8.40 m/s. What minimum speed must the seco

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Question 942534: In a 4 x 100 relay race (each leg 100 meters long), the first place runner has a 0.550 second lead and is running at a constant speed of 8.40 m/s. What minimum speed must the second runner have in order to catch up with the first runner by the end of their leg of the race?
I know that v=d/t and that the formulas will be equal to each other because they finish the leg of the race together. but I'm confused if I add the 0.55 second lead to the constant speed of the first runner or adjust the distance? I drew a picture and I can visually see the problem, just confused about how to deal with the .55 second lead. Thanks in advance for your help.
Found 2 solutions by macston, lwsshak3:
Answer by macston(5194) (Show Source):
You can put this solution on YOUR website!
At 8.4 m/s, the first runner will finish the 100 m leg in (100/8.4)sec. Since he has a 0.550 second start, he has (100/8.4)-0.550 sec left to finish. The second runner must run 100 m /((100/8.4)-.550) to tie the first runner. So the second runner must run at 8.81 m/s

Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
In a 4 x 100 relay race (each leg 100 meters long), the first place runner has a 0.550 second lead and is running at a constant speed of 8.40 m/s. What minimum speed must the second runner have in order to catch up with the first runner by the end of their leg of the race?
***
let x=increase in speed of second runner
8.40+x=catch-up speed of second runner
travel time=distance/speed
..

lcd:8.4(x+8.4)
840+100x-840=.55(8.4x+70.56)
100x=4.62x+38.808
95.38x=38.808
x=0.41
8.4+x=8.81
What minimum speed must the second runner have? 8.81 m/s