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10 December, 12:27

A car starts from rest with an acceleration of 2ms-2. At the same time a motor cycle starts with an uniform velocity of 20ms-1 in the same path 84m behind the car. How many times will the vehicles meet during their motion?

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  1. 10 December, 15:07
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    Twice

    Explanation:

    Use constant acceleration equation:

    x = x₀ + v₀ t + ½ at²

    where x is the final position,

    x₀ is the initial position,

    v₀ is the initial velocity,

    a is the acceleration,

    and t is time.

    The position of the car is:

    x = (84) + (0) t + ½ (2) t²

    x = 84 + t²

    The position of the motorcycle is:

    x = (0) + (20) t + ½ (0) t²

    x = 20t

    When the positions are equal:

    84 + t² = 20t

    t² - 20t + 84 = 0

    (t - 6) (t - 14) = 0

    t = 6 or 14

    The vehicles meet twice, once after 6 seconds, and once after 14 seconds.
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