SOLUTION: A person of height 1.68 m is walking away from a lamp-post at 1.5 m/s. The light on the lamp-post is 5.3 m above the ground.
At what rate (in m/s to the nearest cm/s) is the le
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-> SOLUTION: A person of height 1.68 m is walking away from a lamp-post at 1.5 m/s. The light on the lamp-post is 5.3 m above the ground.
At what rate (in m/s to the nearest cm/s) is the le
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Question 989337: A person of height 1.68 m is walking away from a lamp-post at 1.5 m/s. The light on the lamp-post is 5.3 m above the ground.
At what rate (in m/s to the nearest cm/s) is the length of the person's shadow changing?
At what speed (in m/s to the nearest cm/s) is the tip of the person's shadow moving across the ground?
I'm really stuck
THANK YOU Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! man is p from pole
shadow tip is s from pole
we want ds/dt
s/5.3 = (s-p)/1.68
(similar triangles)
1.68 s = 5.3 s - 5.3 p
3.62 s = 5.3 p
s = 1.464088398 p approx 1.46 p
ds/dt = 1.46 dp/dt
but dp/dt = 1.5
so
ds/dt = 2.19 m/s
this tells us that the length of the person's shadow is changing at the rate of 2.19 m/s
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the speed (in m/s to the nearest cm/s) that the tip of the person's shadow moving across the ground is 1.5 + 2.19 = 3.69 m/s
