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6 September, 11:49

How long (distance and time) would it take a vehicle to accelerate from a stop to 203 mph with an average acceleration of 2.93ft/s²

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  1. 6 September, 15:19
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    t = 1.68 min

    S = 2.82 miles

    Explanation:

    Given,

    The initial velocity of the vehicle, u = 0

    The final velocity of the vehicle, v = 203 mph

    The average acceleration of the vehicle, a = 2.93 ft/s²

    = 7191.82 miles/h²

    Using the first equation of motion

    v = u + at

    t = (v - u) / a

    = (203 - 0) / 7191.82

    = 0.028 h

    = 100.8 s

    Thus, the time taken by the vehicle to reach the final velocity is, t = 100.8 s

    Using the third equations of motion

    S = ut + 1/2 at²

    Substituting the values

    S = 0.5 x 7191.82 x (0.028) ²

    = 2.82 miles

    Hence, the distance traveled by the vehicle, S = 2.82 miles
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