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23 August, 22:51

A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pushed 2.5 m straight backwards in 1.65 s. He breaks the tackle and runs straight forward another 24 m in 5.2 s.

Part (a) Calculate the wide receiver's average velocity in the horizontal direction during the first interval, in meters per second.

Part (b) Calculate the wide receiver's average velocity in the horizontal direction during the second interval, in meters per second.

Part (c) Calculate the wide receiver's average velocity in the horizontal direction during the third interval, in meters per second.

Part (d) Calculate the wide receiver's average velocity in the horizontal direction for the entire motion, in meters per second

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Answers (1)
  1. 24 August, 02:47
    0
    a) v1 = 5.52m/s

    b) v2 = - 1.52m/s

    c) v3 = 4.62m/s

    d) vt = 3.85m/s

    Explanation:

    The velocity of the football wide receiver is his displacement per unit time.

    Velocity v = (displacement d) / time t

    v = d/t ... 1

    For each of the cases, equation 1 would be used to calculate the velocity.

    a) v1 = d1/t1

    d1 = 16m

    t1 = 2.9s

    v1 = 16m/2.9s

    v1 = 5.52m/s

    b) v2 = d2/t2

    d2 = - 2.5m

    t2 = 1.65s

    v2 = - 2.5/1.65

    v2 = - 1.52m/s

    c) v3 = d3/t3

    d3 = 24m

    t3 = 5.2s

    v3 = 24/5.2

    v3 = 4.62m/s

    d) vt = dt/tt

    dt = 16m - 2.5m + 24m = 37.5m

    tt = 2.9 + 1.65 + 5.2 = 9.75s

    vt = 37.5/9.75

    vt = 3.85m/s
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Home » Physics » A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pushed 2.5 m straight backwards in 1.65 s. He breaks the tackle and runs straight forward another 24 m in 5.2 s.
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