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Posted by on Saturday, September 7, 2013 at 2:55pm.

As soon as a traffic light turns green, a car speeds up from rest to 45.0 mi/h with constant acceleration 8.50 mi/h/s. In the adjoining bike lane, a cyclist speeds up from rest to 15.0 mi/h with constant acceleration 12.50 mi/h/s. Each vehicle maintains constant velocity after reaching its cruising speed.


(a) For what time interval is the bicycle ahead of the car?


(b) By what maximum distance does the bicycle lead the car?

  • Physics - , Sunday, September 8, 2013 at 5:40pm

    CAR:
    V1 = 45mi/h * 1600m/mi * 1h/3600s=20 m/s
    a=8.5mi/h/s*1600m/mi*1h/3600s=3.78m/s^2
    a*t = 20 m/s
    3.78t = 20
    t = 5.29 s. to reach max velocity.
    d1 = 0.5a*t^2
    d1 = 0.5*3.78*5.29^2 = 52.9 m = Dist.@
    which max. velocity is reached.

    BIKE:
    V2 = (15/45)*20 m/s = 6.67 m/s = max.
    velocity.

    a = (12.5/8.5)*3.78 = 5.56 m/s^2.

    a*t = 6.67 m/s
    5.56t = 6.67
    t = 1.2 s to reach max velocity.

    d2 = 0.5a*t^2
    d2 = 0.5*5.56*1.2^2 = 4.0 m = Dist. @
    which max. velocity is reached.

    d1 = 0.5*3.78*1.2^2 = 2.72 m. = Dist.of
    the car after 1.2 s.

    D = d2-d1 = 4-2.72 = 1.28 m = Dist. the
    by which the bike leads after 1.2 s.

    When the car catches up:
    a. d1 = d2 + 1.28 m.
    0,5*a*t^2 = V2*t + 1.28
    0.5*3.78*t^2 = 6.67*t+1.28
    1.89t^2-6.67t-1.28 = 0
    t = 3.71 s.(Use Quadratic Formula).

    b. D = d2-d1 = 4-2.72 = 1.28 m.
























































    a. t = (V-Vo)/a = (15-0)/12.5 = 1.2 s.

    b. a=12.5mi/h/s*1600m/mi*1h/3600s=5.56m/s^2
    = Acceleration of bike.
    Db = 0.5*5.56*1.2^2 = 4 m. = Dist. of bike after 1.2 s.

    a = (8.5/12.5) * 5.56 = 3.78 m/s^2 =
    Acceleration of the car.
    Dc = 0.5*3,78*1.2^2 = 2.72 m. = Dist. of the car after 1.2 s.

    D = Db - Dc = 4 - 2.72 = 1.28 m

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