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31 December, 09:26

Suppose a plane accelerates from rest for 32.3, achieving a takeoff speed of 47.1 m/s after traveling a distance of 607 m down the runway. A smaller plane with the same acceleration has a takeoff speed of 28.2 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed?

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  1. 31 December, 09:49
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    217.28 m/s

    Explanation:

    u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

    Let a be the acceleration.

    Use third equation of motion.

    v^2 = u^2 + 2 a s

    47.1 x 47.1 = 0 + 2 a x 607

    a = 1.83 m/s^2

    For small plane

    a = 1.83 m/s^2, v = 28.2 m/s, u = 0, Let teh distance be s.

    Use third equation of motion

    28.2^2 = 0 + 2 x 1.83 x s

    s = 217.28 m/s
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