SOLUTION: i need help setting up this equation for rate, time, and distance. at 7:00am joe starts jogging at 6 mi/h. at 7:10am ken starts off after him. how fast must ken run in order

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: i need help setting up this equation for rate, time, and distance. at 7:00am joe starts jogging at 6 mi/h. at 7:10am ken starts off after him. how fast must ken run in order       Log On


   

Question 818575: i need help setting up this equation for rate, time, and distance.

at 7:00am joe starts jogging at 6 mi/h. at 7:10am ken starts off after him. how fast must ken run in order to overtake him at 7:30am ?
thank you
Found 2 solutions by josgarithmetic, TimothyLamb:
Answer by josgarithmetic(39628) About Me  (Show Source):
You can put this solution on YOUR website!
The times are known but Ken's rate is unknown. You know their distances will be equal.

Think in HOURS or fractions of an hour, not in minutes.
Joe goes for h+1/6 hours;
Ken goes for h hours.

R*T=D for Rate, Time, Distance.

6%28h%2B1%2F6%29=r%28h%29 and h%2B1%2F6=%281%2F2%29.
Find the value of h from the second equation and use it in the first equation. This gives you one equation using only the variable, r.

h=1%2F2-1%2F6 but you can see from time on a number line that h is 1%2F3 hour.
6%281%2F2%29=r%281%2F3%29
r=6%281%2F2%29%283%2F1%29=highlight%28r=9%29

Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
at the overtake:
both runners have traveled distance: D
overtake time: 7:30 - 7:00 = 30/60 = (1/2) hour
Joe's time: T = 1/2 hour
Ken's time: (T - 10/60) = (1/2 - 1/6) = (6/12 - 2/12) = (4/12) = (1/3) hour
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s = d / t
d = s * t
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joe:
D = 6(1/2) = 3
D = 3 miles
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ken:
D = s(1/3)
---
D = 3 = s(1/3)
3 = s(1/3)
---
s = 9
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answer:
ken must run at 9 mph to overtake joe at 7:30
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