An eagle is flying horizontally at 5.8 m/s with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish's speed doubles? (b) How much additional time would be required for the speed to double again?
To answer these questions, we need to understand the concepts of acceleration, time, and distance.
(a) When the eagle drops the fish, it starts falling freely under the force of gravity. Due to gravity, the fish will accelerate at a constant rate of 9.8 m/s^2 downwards. To find how much time passes before the fish's speed doubles, we can use the formula for distance traveled during constant acceleration:
d = v₀t + (1/2)at²
Where:
d = distance traveled (unknown)
v₀ = initial velocity of the fish (which is 0 m/s)
t = time (unknown)
a = acceleration (gravitational acceleration of 9.8 m/s²)
Since the fish starts at rest, v₀ = 0, and the equation reduces to:
d = (1/2)at²
We know that when the fish's speed doubles, its final velocity will be 2v₀ (where v₀ is the initial velocity). Therefore, we can write:
2v₀ = at
Rearranging this equation, we get:
t = (2v₀) / a
Substituting the values v₀ = 0 m/s and a = 9.8 m/s², we get:
t = (2 * 0) / 9.8
t = 0 seconds
Therefore, no time passes before the fish's speed doubles. This implies that the fish's speed doubles immediately after it is dropped.
(b) To find how much additional time is required for the speed to double again, we need to calculate the total time it takes for the speed to triple.
Using the same equation as before:
t = (2v₀) / a
We can substitute 2v₀ with 3v₀, since the speed triples. Thus, we get:
t = (3v₀) / a
Now, we can calculate the additional time required:
additional time = t - original time
additional time = ((3v₀) / a) - ((2v₀) / a)
additional time = (3v₀ - 2v₀) / a
additional time = (v₀) / a
Substituting v₀ = 0 m/s and a = 9.8 m/s², we get:
additional time = (0) / 9.8
additional time = 0 seconds
So, no additional time is required for the speed to double again.