Ask Question
17 April, 21:32

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of her shadow changing when she is 15 ft from the base of the light?

+5
Answers (2)
  1. 17 April, 21:33
    0
    By applying similar triangles rule,

    22/5.5 = (x + s) / s

    s = 5.5/22 * (x + s)

    ds/dt = 5.5/22 * (dx/dt + ds/dt)

    16.5/22 * ds/dt = 5.5/22 * dx/dt

    16.5/22 * ds/dt = 5.5/22 * - 6

    ds/st = - 2 ft/s.
  2. 17 April, 21:39
    0
    The length of her shadow is changing at the rate - 2 m/s

    Explanation:

    Let the height oh the street light, h = 22 ft

    Let the height of the woman, w = 5.5 ft

    Horizontal distance to the street light = l

    length of shadow = x

    h/w = (l + x) / x

    22/5.5 = (l + x) / x

    4x = l + x

    3x = l

    x = 1/3 l

    taking the derivative with respect to t of both sides

    dx/dt = 1/3 dl/dt

    dl/dt = - 6 ft/sec (since the woman is walking towards the street light, the value of l is decreasing with time)

    dx/dt = 1/3 * (-6)

    dx/dt = - 2 m/s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of her shadow ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers