To solve this problem, we will use the kinematic equations for uniformly accelerated motion. Since the rock is dropped from rest, we can denote its initial velocity (\(v_0\)) as 0 m/s and the acceleration due to gravity (\(g\)) as approximately 9.81 m/s² (downward).
Kinematic Equations
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Distance Fallen (d): The distance fallen after time \(t\) seconds is given by the equation: \[ d = v_0 t + \frac{1}{2} g t^2 \] Since \(v_0 = 0\), this simplifies to: \[ d = \frac{1}{2} g t^2 \]
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Instantaneous Speed (v): The instantaneous speed after time \(t\) seconds is given by: \[ v = v_0 + g t \] Again, since \(v_0 = 0\), this simplifies to: \[ v = g t \]
Calculations for Each Second
We'll calculate the distance fallen and the instantaneous speed for the first 5 seconds.
Using \(g = 9.81 , \text{m/s}^2\):
At \(t = 1s\)
- Distance fallen: \[ d = \frac{1}{2} \cdot 9.81 \cdot (1)^2 = \frac{9.81}{2} \approx 4.91 , \text{m} , \text{(rounded to 5 m)} \]
- Instantaneous speed: \[ v = 9.81 \cdot 1 \approx 9.81 , \text{m/s} , \text{(rounded to 10 m/s)} \]
At \(t = 2s\)
- Distance fallen: \[ d = \frac{1}{2} \cdot 9.81 \cdot (2)^2 = \frac{9.81 \cdot 4}{2} = 19.62 , \text{m} , \text{(rounded to 20 m)} \]
- Instantaneous speed: \[ v = 9.81 \cdot 2 \approx 19.62 , \text{m/s} , \text{(rounded to 20 m/s)} \]
At \(t = 3s\)
- Distance fallen: \[ d = \frac{1}{2} \cdot 9.81 \cdot (3)^2 = \frac{9.81 \cdot 9}{2} \approx 44.145 , \text{m} , \text{(rounded to 44 m)} \]
- Instantaneous speed: \[ v = 9.81 \cdot 3 \approx 29.43 , \text{m/s} , \text{(rounded to 29 m/s)} \]
At \(t = 4s\)
- Distance fallen: \[ d = \frac{1}{2} \cdot 9.81 \cdot (4)^2 = \frac{9.81 \cdot 16}{2} = 78.48 , \text{m} , \text{(rounded to 78 m)} \]
- Instantaneous speed: \[ v = 9.81 \cdot 4 \approx 39.24 , \text{m/s} , \text{(rounded to 39 m/s)} \]
At \(t = 5s\)
- Distance fallen: \[ d = \frac{1}{2} \cdot 9.81 \cdot (5)^2 = \frac{9.81 \cdot 25}{2} \approx 122.625 , \text{m} , \text{(rounded to 123 m)} \]
- Instantaneous speed: \[ v = 9.81 \cdot 5 \approx 49.05 , \text{m/s} , \text{(rounded to 49 m/s)} \]
Summary of Results
| Time (s) | Distance Fallen (m) | Instantaneous Speed (m/s) | |----------|----------------------|---------------------------| | 1 | 5 | 10 | | 2 | 20 | 20 | | 3 | 44 | 29 | | 4 | 78 | 39 | | 5 | 123 | 49 |
These calculations provide a clear view of how the distance and speed change with time during the rock’s free fall.