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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Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:
where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as
where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:
Answer:
a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.
Explanation:
Statement is incomplete. The complete description is now described below:
<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>
<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>
<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>
Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (), measured in meters per second, is:
Where:
- Initial velocity in +x direction, measured in meters per second.
- Acceleration in +x direction, measured in meter per square second.
- Time, measured in seconds.
- Velocity in +y direction, measured in meters per second.
If we know that ,
,
and
, the final velocity of the satellite is:
a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:
The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:
The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.