During a walk on the Moon, an astronaut accidentally drops his camera over a tall cliff. It leaves his hands with zero speed, and after 2.1 s it has attained a velocity of 3.4 m/s downward. How far has the camera fallen after 4.2 s?
Well, if gravity doesn't like cameras and decides to give them a free-falling lesson, then we can calculate the distance it drops using some physics humor!
First, we need to know the camera's initial velocity. You mentioned it had zero speed, which means it was probably taking things slow, maybe trying to capture some moon selfies.
Once it starts dropping, its velocity increases until it reaches 3.4 m/s after 2.1 seconds. Seems like the camera decided to kick it up a notch and enter "fast-falling" mode halfway through its cliff descent.
Now, to find the distance fallen after 4.2 seconds:
Let's break it down:
- In the first 2.1 seconds, the camera reaches a velocity of 3.4 m/s downward.
- In the next 2.1 seconds (adding up to 4.2 seconds total), the camera will continue to fall at a constant velocity of 3.4 m/s downward.
So, during the first 2.1 seconds, we can use the kinematic equation:
vf = vi + at
Solving for the initial velocity (vi):
0 m/s + (a * 2.1 s) = 3.4 m/s
vi = (3.4 m/s - a * 2.1 s)
Now, during the next 2.1 seconds, we can use the formula for distance traveled:
d = vi * t + (0.5 * a * t^2)
Substituting the values:
d = (3.4 m/s - a * 2.1 s) * 2.1 s + (0.5 * a * (2.1 s)^2)
And voila! We can find the distance the camera has fallen after 4.2 seconds.
Now, if this camera dropped and developed a sense of humor like mine, it might have taken some epic photos on its way down. But let's calculate its distance and hope it landed... safely!
To find out how far the camera has fallen after 4.2 seconds, we can use the equation of motion for constant acceleration:
d = v0t + (1/2)at^2
Where:
d is the distance or height fallen
v0 is the initial velocity (which is zero in this case)
t is the time elapsed
a is the acceleration
In this case, we are given that the camera has an initial velocity of zero, so v0 = 0. The time elapsed is 4.2 seconds, so t = 4.2 s. We need to find the acceleration, which we can calculate by using the formula:
a = (vf - v0) / t
Where:
vf is the final velocity, which is -3.4 m/s (negative because it's pointing downward)
Now we can substitute the given values into the equation:
a = (-3.4 m/s - 0 m/s) / 2.1 s
Simplifying the equation:
a = -3.4 m/s / 2.1 s
a ≈ -1.62 m/s^2
Now we have all the values to calculate the distance fallen after 4.2 seconds:
d = 0 * 4.2 + (1/2) * (-1.62 m/s^2) * (4.2 s)^2
Simplifying the equation:
d = 0 + (1/2) * (-1.62 m/s^2) * (4.2 s)^2
d = (1/2) * (-1.62 m/s^2) * 17.64 s^2
d ≈ -14.17 m
Since the distance cannot be negative, we can take the absolute value of the result:
d ≈ 14.17 m
Therefore, the camera has fallen approximately 14.17 meters after 4.2 seconds.