A runner decides to run out in the country. He begins to run at an average rate of 9mph. He runs a certain distance and then turns around and returns along the same route at an average rate of 6mph. If the round trip took 2 and a half hours, how far did the runner travel before turning around? Every other distance/rate/time problem I've looked at hasn't helped me figure this one out. thanks
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Another word problem
- Thread starter mkay
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A runner decides to run out in the country. He begins to run at an average rate of 9mph. He runs a certain distance and then turns around and returns along the same route at an average rate of 6mph. If the round trip took 2 and a half hours, how far did the runner travel before turning around? Every other distance/rate/time problem I've looked at hasn't helped me figure this one out. thanks
1. Use the definition of speed:
\(\displaystyle \displaystyle{speed = \frac{distance}{time}~\implies~distance = speed \cdot time}\)
2. Let \(\displaystyle t_1\) denotes the time for the 1st part of the run and \(\displaystyle t_2\) the time for the 2nd part. Then you know:
\(\displaystyle \left|\begin{array}{rcl}t_1+t_2&=&\frac52 \\ 9t_1&=& 6t_2\end{array}\right.\)
3. Solve for \(\displaystyle t_1\) and \(\displaystyle t_2\) and consequently determine the corresponding distances.
Last edited:
Hello, mkay!
\(\displaystyle \text{Formula: } \:\text{Time} \:=\:\dfrac{\text{Distance}}{\text{Speed}}\)
Let \(\displaystyle x\) = distance (one way).
He ran \(\displaystyle x\) miles at 9 mph.
. . This took \(\displaystyle \dfrac{x}{9}\) hours.
He ran \(\displaystyle x\) miles at 6 mph.
. . This took \(\displaystyle \dfrac{x}{6}\) hours.
His total time was \(\displaystyle 2\frac{1}{2}\) hours.
There is our equation! . \(\displaystyle \hdots \;\;\dfrac{x}{9} + \dfrac{x}{6} \:=\:\dfrac{5}{2}\)
Go for it!
A runner decides to run out in the country.
He begins to run at an average rate of 9mph.
He runs a certain distance and then turns around
. . and returns along the same route at an average rate of 6mph.
If the round trip took 2 and a half hours,
. . how far did the runner travel before turning around?
\(\displaystyle \text{Formula: } \:\text{Time} \:=\:\dfrac{\text{Distance}}{\text{Speed}}\)
Let \(\displaystyle x\) = distance (one way).
He ran \(\displaystyle x\) miles at 9 mph.
. . This took \(\displaystyle \dfrac{x}{9}\) hours.
He ran \(\displaystyle x\) miles at 6 mph.
. . This took \(\displaystyle \dfrac{x}{6}\) hours.
His total time was \(\displaystyle 2\frac{1}{2}\) hours.
There is our equation! . \(\displaystyle \hdots \;\;\dfrac{x}{9} + \dfrac{x}{6} \:=\:\dfrac{5}{2}\)
Go for it!