Question
A small object is dropped from the top of a cliff. A measuring device on the object records the distance fallen every second. The following table shows the results for the first 6 seconds.
Time (seconds) 0 1 2 3 4 5 6
Distance Fallen (meters) 0 4.9 19.6 44.1 78.4 122.5 176.4
A. Model this scenario with a function. Explain how you chose which type of function to use.
Time (seconds) 0 1 2 3 4 5 6
Distance Fallen (meters) 0 4.9 19.6 44.1 78.4 122.5 176.4
A. Model this scenario with a function. Explain how you chose which type of function to use.
Answers
GPT-4o mini
To model the distance fallen by the object dropped from the cliff, we can use a quadratic function. The reason for choosing a quadratic model arises from the fact that the distance fallen by an object in free fall under the influence of gravity is proportionate to the square of the time elapsed.
In the case of free fall, the equation for motion under constant acceleration (due to gravity, g ≈ 9.8 m/s²) can be expressed as:
\[
s(t) = \frac{1}{2} g t^2
\]
Where:
- \( s(t) \) is the distance fallen after time \( t \),
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).
Thus, our model would look like this:
\[
s(t) = \frac{1}{2} \cdot 9.8 \cdot t^2 \implies s(t) = 4.9 t^2
\]
### Explanation of the Model:
1. **Quadratic Form**: The function \( s(t) = 4.9 t^2 \) is a quadratic function because it describes distance \( s(t) \) as a polynomial of degree 2 in terms of \( t \).
2. **Proportionality to Time Squared**: The table shows that as time increases, the distance fallen does not increase linearly but rather is proportional to the square of time—as confirmed by the pattern seen in the distance values. For instance:
- At \( t = 1 \) second, \( s(1) = 4.9(1^2) = 4.9 \) meters
- At \( t = 2 \) seconds, \( s(2) = 4.9(2^2) = 19.6 \) meters
- At \( t = 3 \) seconds, \( s(3) = 4.9(3^2) = 44.1 \) meters
- And so on…
3. **Validation with Data**: Each value in the distance fallen at each second corresponds precisely to the formula \( s(t) = 4.9 t^2 \). This consistency across multiple time intervals supports the validity of our quadratic model.
### Conclusion:
The best fit for modeling the distance fallen over time in free fall is given by the quadratic function \( s(t) = 4.9 t^2 \). This function demonstrates how the distance increases with the square of the time, reflecting the physics of gravitational acceleration.
In the case of free fall, the equation for motion under constant acceleration (due to gravity, g ≈ 9.8 m/s²) can be expressed as:
\[
s(t) = \frac{1}{2} g t^2
\]
Where:
- \( s(t) \) is the distance fallen after time \( t \),
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).
Thus, our model would look like this:
\[
s(t) = \frac{1}{2} \cdot 9.8 \cdot t^2 \implies s(t) = 4.9 t^2
\]
### Explanation of the Model:
1. **Quadratic Form**: The function \( s(t) = 4.9 t^2 \) is a quadratic function because it describes distance \( s(t) \) as a polynomial of degree 2 in terms of \( t \).
2. **Proportionality to Time Squared**: The table shows that as time increases, the distance fallen does not increase linearly but rather is proportional to the square of time—as confirmed by the pattern seen in the distance values. For instance:
- At \( t = 1 \) second, \( s(1) = 4.9(1^2) = 4.9 \) meters
- At \( t = 2 \) seconds, \( s(2) = 4.9(2^2) = 19.6 \) meters
- At \( t = 3 \) seconds, \( s(3) = 4.9(3^2) = 44.1 \) meters
- And so on…
3. **Validation with Data**: Each value in the distance fallen at each second corresponds precisely to the formula \( s(t) = 4.9 t^2 \). This consistency across multiple time intervals supports the validity of our quadratic model.
### Conclusion:
The best fit for modeling the distance fallen over time in free fall is given by the quadratic function \( s(t) = 4.9 t^2 \). This function demonstrates how the distance increases with the square of the time, reflecting the physics of gravitational acceleration.