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21 September, 15:22

A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 44 km/h at the 74-m mark. He then maintains this speed for the next 85 meters before uniformly slowing to a final speed of 36 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur

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  1. 21 September, 15:29
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    The maximum acceleration occurs from the zero mark to the 74m mark with a magnitude equal to 1.01m/s²

    Explanation:

    The sprinter accelerates in two instances

    1. From the 0m to the 74m mark

    2. From the 159m (85 + 74) mark to the finish line.

    Acceleration is a vector quantity having both magnitude and direction.

    For the first case,

    Initial velocity is zero at the sprinter starts from rest,

    Final velocity is v = 44km/h = 44*1000/3600m/s = 12.2m/s

    From the equations for constant acceleration motion,

    v² = u² + 2as

    Where u = initial velocity, a = acceleration, and s = distance covered,

    12.2² = 0² + 2a*74

    148.84 = 148a

    a = 148.84/148 = 1.01m/s²

    Second instance of acceleration,

    The sprinter slows down from an initial speed of 12.2 m/s to 10m/s (36km/hr) from the 159m mark to the finish line or over a distance of 41m (200 - 159).

    So using the same equation above with u = 12.2m/s and v = 10m/s we have

    10² = 12.2² + 2a*41

    100 = 148.84 + 82a

    82a = 100 - 148.84

    82a = - 48.84

    a = - 48.84/82 = 0.59m/s²

    So the maximum acceleration occurs from the zero mark to the 74m mark with a magnitude equal to 1.01m/s²
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