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1 June, 00:23

A motorcycle starts from rest at and travels along a straight road with a constant acceleration of 2 6 / ft s until it reaches a speed of 50 / ft s. Afterwards it maintains this speed. Also, when t s  0, a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 / ft s. Determine the time and the distance traveled by the motorcycle when they pass each other.

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  1. 1 June, 00:39
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    Time = 77.63 s

    Distance = 3673.3 ft

    Explanation:

    Using equation of motion

    v = u + at'

    50 = 6 * t'

    t' = 50 / 6

    t' = 8.33 s

    v² = u² + 2a (s - s•)

    50² = 0² + 2 * 6 * (s - 0)

    2500 = 12 * s

    s' = 2500 / 12

    s' = 208.3 ft

    At t' = 8.33 the distance traveled by the car is

    s'' = v• * t'

    s'' = 30 * 8.33

    s'' = 250 ft

    Now, the distance between the motorcycle and car is

    6000 - 208.33 - 250 = 5541.67

    For motorcycle, when passing occurs,

    s = v• * t, x = 50 * t''

    For the car, when passing occurs,

    s = v• * t, 5541.67 - x = 30 * t''

    Solving both simultaneously, we have

    5541.67 - [50 * t''] = 30 * t''

    5541.67 = 30t'' + 50t''

    5541.67 = 80t''

    t'' = 5541.67 / 80

    t'' = 69.3 s, substituting t'' = 69.3 back, we have

    x = 50 * 69.3

    x = 3465 ft

    Therefore for motorcycle,

    t = 69.3 + 8.33

    t = 77.63 s

    S = 3465 + 208.3

    s = 3673.3 ft
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