Question 562516: Jack and Edson decide they will go jogging from West hills athletic club. THey arrive at west hills and start jogging at different time so, as the clock strikes noon, Jack is 250 yards from the starting point and Edson is 400 yards from the start. Jack jogs at the constant pace of 3 yards/sec. Edson jogs at the constant pace of 2.5 yards/sec. How long will it take Jack to catch up to Edson?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! as the clock strikes noon, Jack is 250 yards from the starting point and Edson is 400 yards from the start.
Jack jogs at the constant pace of 3 yards/sec.
Edson jogs at the constant pace of 2.5 yards/sec.
How long will it take Jack to catch up to Edson?
:
From the information given, Ed has 400-250 = 150 yd head start (at noon)
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Let t = time required for Jack to catch Ed
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Write a distance equation, dist = speed * time
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J's dist = Ed's dist + 150 yds
3t = 2.5t + 150
3t - 2.5t = 150
.5t = 150
Multiply both sides by 2
t = 300 seconds or 5 min, 12:05 would be the time
:
:
Check this by finding the dist each ran
3(300) = 900 yds
2.5(300)=750 yds
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differs: 150 yds
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