Andrea, a 59.0 kg sprinter, starts a race with an acceleration of 3.900 m/s2. If she accelerates at that rate for 12.00 m and then maintains

Question

Andrea, a 59.0 kg sprinter, starts a race with an acceleration of 3.900 m/s2. If she accelerates at that rate for 12.00 m and then maintains that velocity for the remainder of a 100.00 m dash, what will her time (in s) be for the race

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Thiên Di 4 years 2021-07-23T10:20:50+00:00 1 Answers 8 views 0

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  1. thucuc
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    2021-07-23T10:21:53+00:00
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    Answer:

    her total time taken for the race is 11.58 s

    Explanation:

    mass of the sprinter, m = 59 kg

    acceleration of the sprinter, a = 3.9 m/s²

    initial distance covered by the sprinter, d₁ = 12 m

    Time taken to cover the first 12 m;

    d₁ = ut + ¹/₂at²

    where;

    u is her initial velocity = 0

    d₁ = ¹/₂at²

    2d₁ = at²

    t_1 = \sqrt{\frac{2d_1}{a} } \\\\t_1 = \sqrt{\frac{2\times 12}{3.9} }\\\\t_1 = 2.48 \ s

    The velocity of the sprinter at the end of the 12 m;

    v = u + at

    v = 0 + at

    v = at

    v = 3.9 x 2.48

    v = 9.67 m/s

    The last distance covered by the sprinter at the constant velocity of 9.67 m/s = (100 – 12)m = 88 m

    The time taken to cover 88 m with an initial velocity of 9.67 m/s is calculated as;

    d₂ = vt + ¹/₂at²

    Since the velocity is constant, the acceleration, a = 0

    d₂ = vt₂

    t₂ = d₂/v

    t₂ = 88 / 9.67

    t₂ = 9.1 s

    The total time taken for the race = t₁ + t₂

                                                           = 2.48 s  +  9.1 s

                                                            = 11.58 s

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