A street light is mounted at the top of a 15-ft-tall pole. a man 6 ft tall walks away from the pole with a speed of 4 ft/s along
a straight path. how fast is the tip of his shadow moving when he is 30 ft from the pole?
1 answer:
Answer: 2.67 ft/s
Explanation:
1) The diagram with the triangle that represents the situation is in the image attached.
2) Similarity properties
(x + y) / 15 = y/6
⇒ 6(x+y) = 15y
⇒ 6x + 6y = 15y
⇒ 6x = 15y - 6y
⇒ 6x = 9y
⇒ y = 6x / 9
⇒ y = 2x / 3
3) Velocity is the derivative respecto to time. Then, find the derivative respect to time, t, on both sides
dy/dt = (2/3)dx/dt
4) The statement tells v = dx/dt = 4 ft/s
⇒ dy/dt = (2/3) 4 ft/s = (8/3)m/s ≈ 2.67 ft/s
Notice that this speed is constant, it does not depends upon the distance of 30 ft.
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Complete Question
A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave frequency is
Answer:
The value is
Explanation:
From the question we are told that
The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)
Generally the sinusoidal equation representing the motion of a wave is mathematically represented as
Where w is the angular frequency
Now comparing this equation with that given we see that
Answer:
V = 3.54 m/s
Explanation:
Using the conservation of energy:
so:
where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.
So, the mass of kelly is:
m = 425N/9.8 = 43.36 Kg
and h is:
h = 1m-0.36m =0.64m
then, replacing values, we get:
Solving for v:
V = 3.54 m/s