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Matt71

  • 3 years ago

An athlete's usual running speed in still air is 8 m/s. On one windy day, the time he took to run 100m, running with the wind, was 6 2/3 seconds less than the time he took to run 100m, running against the wind. a. Given that the speed of the wind that day was "w m/s", write down an expression in terms of w for the time he took to run 100m when running with the wind. b. Form an equation in w and show that it reduces to w^2+30w-64=0 c. Solve the equation to find the speed of the wind

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  1. thewildones
    • 3 years ago
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    i say b..

  2. Matt71
    • 3 years ago
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    Ummm... It's not a multiple choice question, those are actually the questions, you know, a,b and c

  3. ganeshie8
    • 3 years ago
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    (time taken to travel against wind) - (time take to travel with wind) = 6 2/3 seconds

  4. ganeshie8
    • 3 years ago
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    time = distance/speed

  5. ganeshie8
    • 3 years ago
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    can you calculate both the times, and plug in ?

  6. FoolAroundMath
    • 3 years ago
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    Speed of the man running with the wind = (8+w) Speed when running against the wind = (8-w) Time taken for 100m running with the wind = 100/(8+w) Time taken for 100m running against the wind = 100/(8-w) Given: \[\frac{100}{8+w} = \frac{100}{8-w} - 6\frac{2}{3}\] Can you proceed further ?

  7. matricked
    • 3 years ago
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    the equation can be deduced from ((100/(8-w)) - ((100/(8+w)) =20/3 which reduces to 2w/((8-w)(8+w)) =1/15 and hence w^2+30w-64=0 solving we have (w+32)(w-2)=0 w=2 (w=-32 ignored) spped of wind =2m/s

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