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7 October, 23:04

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 13 m/s, skates by with the puck. After 1.5 s, the first player makes up his mind to chase his opponent. (a) If he accelerates uniformly at 4.0 m/s2, how long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.) (b) How far as he traveled

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  1. 8 October, 01:48
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    Let us say that:

    1 = 1st player notation

    2 = 2nd player notation (the opponent)

    a. First let us establish the distance travelled by the 2nd player:

    d2 = 13 m/s * (t + 1.5)

    d2 = 13 t + 19.5

    Then the distance of the 1st player:

    d1 = v0 t + 0.5 a t^2 (v0 initial velocity = 0 since he started from rest)

    d1 = 0.5 * 4 m/s^2 * t^2

    d1 = 2 t^2

    The two distances must be equal, d1 = d2:

    2 t^2 = 13 t + 19.5

    t^2 - 6.5 t = 9.75

    Completing the square:

    (t - 3.25) ^2 = 9.75 + ( - 3.25) ^2

    t - 3.25 = ±4.5

    t = - 1.25, 7.75

    Since time cannot be negative, therefore:

    t = 7.75 seconds

    So he catches his opponent after 7.75 seconds.

    b. Using the equation:

    d1 = 2 t^2

    d1 = 2 * (7.75) ^2

    d1 = 120.125 m

    So he travelled about 120.125 meters when he catches up to his opponent.
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