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10 November, 12:02

The formula d = 1.1 t 2 + t + 1 expresses a car's distance (in feet to the north of an intersection, d, in terms of the number of seconds t since the car started to move. As the time t since the car started to move increases from t = 3 to t = 5 seconds, what constant speed must a truck travel to cover the same distance as the car over this 2-second interval?

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  1. 10 November, 14:27
    0
    9.8 ft/s

    Explanation:

    The formula for the distance (d, in ft) traveled by a cer to the north of an intersection is:

    d = 1.1 t² + t + 1

    where,

    t is the time in seconds

    The distance traveled in the first 3 seconds is:

    d (3) = 1.1 * (3) ² + 3 + 1 = 13.9 ft

    The distance traveled in the first 5 seconds is:

    d (5) = 1.1 * (5) ² + 5 + 1 = 33.5 ft

    The distance traveled in the 3-5 s interval is:

    d (5) - d (3) = 33.5 ft - 13.9 ft = 19.6 ft

    A truck should cover the same distance (19.6 ft) in the same time (2s). It must have a constant speed of:

    v = d/t = 19.6 ft/2s = 9.8 ft/s
  2. 10 November, 15:56
    0
    d (t) = 1.1t² + t + 1

    The constant speed required to cover the same distance between t = 3 to t = 5 is the same as the average speed over that same time interval. It is given by:

    v = Δx/Δt

    v = average speed, Δx = change in distance, Δt = elapsed time

    Given values:

    Δx = d (5) - d (3) = 19.6ft

    Δt = 5s - 3s = 2s

    Plug in and solve for v:

    v = 19.6/2

    v = 9.8ft/s
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