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Thread: Two Cyclists

  1. #1
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    Two Cyclists

    This is a review problem

    Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? I know the answer is 4.5 hours but need to figure an easy way to explain it...
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  2. #2
    Super Member Quacky's Avatar
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    In three hours, the first cyclist will have travelled $\displaystyle 18$m.
    The first is travelling at a speed of $\displaystyle 6m/h$. After $\displaystyle x$ hours, they will have travelled '$\displaystyle 6x$' metres.
    The second is travelling at a speed of $\displaystyle 10m/h$. After $\displaystyle x$ hours, they will have travelled $\displaystyle 10x$ metres.

    Remember that the first has an $\displaystyle 18$ metre headstart, and you can form the following equation:

    $\displaystyle 10x = 18 + 6x$

    And you can quite simply solve for $\displaystyle x$.
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  3. #3
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    HINT:
    distance = speed * time (hope you knew that!)

    1: distance = 6(t)
    2: distance = 10(t - 3)
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    Awesome thanks Those will work...
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  5. #5
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    Hello, Caity!

    I have a back-door approach to these "two cars" problems.


    Two cyclists start biking from a trail's start 3 hours apart.
    The second cyclist travels at 10 mph and starts 3 hours after the first cyclist
    who is traveling at 6 mph.
    How much time will pass before the second cyclist catches up with the first
    from the time the second cyclist started biking?

    I know the answer is 4.5 hours but need to figure an easy way to explain it.

    Cyclist $\displaystyle \,A$ has a $\displaystyle \,3$-hour headstart.
    He is: $\displaystyle 3 \times 6 \,=\,18$ miles ahead.

    Then cyclist $\displaystyle \,B$ starts after him at $\displaystyle \,10$ mph.
    His speed is: $\displaystyle 10-6 \,=\,4$ mph faster than $\displaystyle \,A$'s speed.

    Imagine this . . .
    Cyclist $\displaystyle \,A$ stops . . . and cyclist $\displaystyle \,B$ approaches him at $\displaystyle \,4$ mph.
    How long will it take $\displaystyle \,B$ to make up the $\displaystyle \,18$ miles?

    It will take: .$\displaystyle \dfrac{18}{4} \,=\,4.5$ hours.


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    This techniques with cars moving in opposite directions.

    . . Two cars start from the same place and at the same time.
    . . Car $\displaystyle \,A$ drives east at $\displaystyle \,45$ mph.
    . . Car $\displaystyle \,B$ drives west at $\displaystyle \,35$ mph.
    . . How long will it take for them to be $\displaystyle \,180$ miles apart?

    Their combine speed is: $\displaystyle 45 + 35\,=\,80$ mph.

    It is as if car $\displaystyle \,A$ is stationary and car $\displaystyle \,B$ drives away at $\displaystyle \,80$ mph.
    When will they be $\displaystyle \,180$ miles apart?

    It will take: .$\displaystyle \dfrac{180}{80} \,=\,2.25$ hours.

    . . See?

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