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anonymous

  • one year ago

racing car traveling with constant acceleration increases its speed from 10 m/s to 50 m/s over a distance of 60 m. How long does this take?

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  1. IrishBoy123
    • one year ago
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    use: \(v^2 = u^2 + 2 a x\) to get 'a' then: \(v = u + at\) to get t

  2. anonymous
    • one year ago
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    I had also done exactly like that but this can't give the appropriate answer

  3. IrishBoy123
    • one year ago
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    from (1) \(a= \frac{50^2 - 10^2}{2 \times 60} = 20m/s^2\) and (2) \(t = \frac{50 - 10}{20} = 2 s\) to be sure, check using final equation of motion: \(x = ut + \frac{1}{2}a t^2 = 10(2) + \frac{1}{2} (20) 2^2 = 60m \ \huge \checkmark\)

  4. anonymous
    • one year ago
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    Hmmm thanks brother...me too got the same answer but the book from which i isolated this question have 4s answer in it thats why i was just confirming it

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