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June 29th, 2010, 12:37 PM   #1
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D=RT (Distance = Rate x Time)

Just for clarity, my text is in a red font color and the text book's font color is in black.

[color=#FF4000]I don't understand this math problem. Here it is:[/color]

A long-distance runner started a course running at an average speed of 6 mph. One and one-half hours later, a cyclist traveled the same course at an average speed of 12 mph. How long after the runner started did the cyclist overtake the runner?

[color=#FF4000]Now the book solves it like this:[/color]

>Unknown time for the cyclist: t
>time for the runner: t + 1.5

Runner's rate = 6
Runner's time = t + 1.5
Runner's Distance = 6(t + 1.5)

Cyclist's rate = 12
Cyclist's time = t
Cyclist's Distance = 12t

>The runner and the cyclist traveled the same distance.

>Solution 6(t + 1.5) = 12t
6t + 9 = 12t
9 = 6t
3/2 = t
t + 1.5 = 1.5 + 1.5
= 3
The cyclist overtook the runner 3 h after the runner started.


[color=#FF4000]I just don't understand why the author said the runner's time = t + 1.5
I can't really picture it in my head. Why add t and 1.5? [/color]
Mighty Mouse Jr is offline  
 
June 29th, 2010, 12:46 PM   #2
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Re: D=RT (Distance = Rate x Time)

Hello Mighty Mouse Jr,

t= the time it takes the cyclist to overtake the runner. If you have found t, you will know how long after the cyclist left, he overtook the runner. You don't want to know that, since the question is:
Quote:
How long after the runner started did the cyclist overtake the runner?
The runner started of 1.5 hours earlier than the cyclist did. So the time between the runner started and the cyclist overtook is t+1.5.

Does that make some sense?

Hoempa
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June 29th, 2010, 12:47 PM   #3
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Re: D=RT (Distance = Rate x Time)

(Edit: Hoempa beat me to it, though this might help explain it a bit as well if needed...)

The time t (in hours) is the total time the biker has been biking. The runner started running an hour and a half (1.5 hours) before the biker.

If the biker has been biking for 5 hours, then the runner has been running 6.5 hours.
If the biker has been biking for 0 hours, then the runner has been running 1.5 hours.
If the biker has been biking for t hours, then the runner has been running (t + 1.5) hours.
shynthriir is offline  
June 29th, 2010, 01:49 PM   #4
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Re: D=RT (Distance = Rate x Time)

I think I get it. Whatever the time of the cyclist (t), the runner's time is always t + 1.5. The part that made it partially clear was when the author said "The runner and the cyclist traveled the same distance," he meant that at the point when the runner and cyclist are at equal distances from the same starting point, it is assumed that the cyclist will overtake the runner because his rate is greater than the runner's rate, therefore distance 1 = distance 2, solve for t.

Does that sound right?
Mighty Mouse Jr is offline  
June 29th, 2010, 03:24 PM   #5
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Yes. In the initial 1 hours, the runner (averaging 6mph) covered 9 miles, so the cyclist (averaging 6mph relative to the runner) caught up with and overtook the runner after another 1 hours, i.e., 3 hours after the runner started.
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