A man is walking beside a 30-m street lamp. After walking 25 m away from the street lamp, he sunk into a quicksand. At the time when 2 m of him are above the ground, he is sinking at 1 m/s. At what rate is the length of his shadow changing at that moment?
2 answers
Can any one help me to answer?
if his shadow has length s, and his height is x meters, then we have
s/x = (s+25)/30
or,
s = 25x/(30-x)
ds/dt = 750/(30-x)^2 dx/dt
so, when x=2,
ds/dt = 750/28^2 (-1) = -0.9566 m/s
s/x = (s+25)/30
or,
s = 25x/(30-x)
ds/dt = 750/(30-x)^2 dx/dt
so, when x=2,
ds/dt = 750/28^2 (-1) = -0.9566 m/s
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