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19 November, 11:34

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 seconds, coasts for 2.4 s, and then slows down at a rate of 1.5m/s2 for the next stop sign.

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  1. 19 November, 12:21
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    d = 112.8m

    the stop signs are 112.8m apart.

    A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 seconds, coasts for 2.4 s, and then slows down at a rate of 1.5m/s2 for the next stop sign. How far apart are the stop signs?

    Explanation:

    The motion are in three stages;

    1. It accelerates at 2.0 m/s2 for 6.0 seconds

    2. It coasts for 2.4 s

    3. then slows down at a rate of 1.5m/s2 for the next stop sign.

    1. It accelerates at 2.0 m/s2 for 6.0 seconds;

    distance d1 = ut + 0.5at^2

    u = 0

    t = 6.0 s

    a = 2.0 m/s^2

    d1 = 0 + 0.5*2.0 (6^2)

    d1 = 36 m

    2. It coasts for 2.4 s;

    Final Speed v2 = at = 2 * 6

    v2 = 12 m/s

    Distance d2 = vt = v2 (t2)

    T2 = 2.4 s

    Substituting the values;

    d2 = 12*2.4 = 28.8 m

    3. then slows down at a rate of 1.5m/s2 for the next stop sign.

    Time taken to decelerate t3 = ∆v/a = 12/1.5 = 8 s

    Distance d3 = ut + 0.5at^2

    u = 12 m/s

    a = - 1.5 m/s^2

    t = 8 s

    d3 = 12 (8) + 0.5 (-1.5 * 8^2)

    d3 = 48 m

    Total distance d = d1 + d2 + d3

    d = 36m + 28.8m + 48m

    d = 112.8m

    the stop signs are 112.8m apart.
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