The sum (c−4)+(3c+9) is
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25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path. This can be obtained by considering this as a right angled triangle.
<h3>How fast is the tip of his shadow moving?</h3>Let x be the length between man and the pole, y be the distance between the tip of the shadow and the pole.
Then y - x will be the length between the man and the tip of the shadow.
Since two triangles are similar, we can write
⇒15(y-x) = 6y
15 y - 15 x = 6y
9y = 15x
y = 15/9 x
y = 5/3 x
Differentiate both sides
dy/dt = 5/3 dx/dt
dy/dt is the speed of the tip of the shadow, dx/dt is the speed of the man.
Given that dx/dt = 5 ft/s
Thus dy/dt = (5/3)×5 ft/s
dy/dt = 25/3 ft/s
Hence 25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path.
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