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Physics

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A motorist drives along a straight road at a constant speed of 16.5 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 3.50 m/s2 to overtake her. Assume that the officer maintains this acceleration.

(a) Determine the time it takes the police officer to reach the motorist.

(b) Find the speed of the officer as he overtakes the motorist.

(c) Find the total displacement of the officer as he overtakes the motorist.

  • Physics -

    Sm = distance travelled by motorist
    Sp = distance travelled by officer
    NOTE: at the time officer catches-up with motorist, Sm = Sp
    t = time taken to catch-up with motorist
    Ap = officer's acceleration
    Up = officer's initial velocity = 0
    Vp = officer's final velocity at the time he caught up with motorist


    Sm = Umt
    Sp = 1/2Apt2

    Sm - Sp = 0 = 16.5t - 1/2x3.5t2

    t(16.5 - 1.75t) = 0

    t = 0 or t = 16.5/1.75 = 9.43s

    Explanation: t = 0 is when the motorist passed the police car; and t = 9.43 is when the police car caught-up with the motorists.

    (b) At the time officer caught up with motorist:
    Up = 0, Ap = 3.5, t = 9.43 Vp = ?

    v = u + at
    Vp = Up + Apt
    Vp = 0 + 3.5 x 9.43 = 33m/s

    (c) Sp = 1/2(Apt^2) ===u = 0
    s = 1/2 x 3.5 x 9.43^2 = 155.6m

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