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anonymous

  • one year ago

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.2 m/s2 until he catches his friend. (a) How much time does it take until he catches his friend? (b) How far has he traveled in this time? (c) What is his speed when he catches up?

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  1. BAdhi
    • one year ago
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    can you tell where you are stuck with?

  2. BAdhi
    • one year ago
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    |dw:1435553731022:dw| when they meet up the distant each have travelled will be the same. use that..

  3. Michele_Laino
    • one year ago
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    hint: the space traveled by the friend of byciclist, within those 2 seconds is: s=3.6*2=7.2 meters. So they meet when this condition holds: space traveled by the byciclist= space traveled by friend, or, using formula: \[\Large \frac{1}{2}a{t^2} = s + vt\] using your data, we get: \[\Large \frac{1}{2} \times 2.2 \times {t^2} = 7.2 + 3.6 \times t\] which is a quadratic equation for time t. Please solve that equation

  4. Michele_Laino
    • one year ago
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    hint: the final speed of the byciclist, is given by the subsequent formula: \[\Large v = at = 2.2 \times t = ...{\text{meters/seconds}}\]

  5. IrishBoy123
    • one year ago
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    Someone needs to post an "Olde Worlde Galilean Transforme" solution, just to round this off.

  6. anonymous
    • one year ago
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    Thank you all for your help!

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