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. Using this information how many meters will the object have fallen after 10 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. Using this information how many meters will the object have fallen after 10 seconds
Answers
GPT-4o mini
The scenario you've described involves an object being dropped from a height, influenced by gravitational acceleration. The distance fallen over time can be modeled using a quadratic function since the distance an object falls under the influence of gravity increases over time in a parabolic fashion.
In free fall, the distance \(d\) fallen after \(t\) seconds can be modeled by the equation:
\[
d(t) = \frac{1}{2} g t^2
\]
where \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth). Thus, this becomes:
\[
d(t) = \frac{1}{2} \cdot 9.8 \cdot t^2 = 4.9 t^2
\]
The table you provided matches perfectly with this model, as the distances for each second give the following:
- \(d(1) = 4.9(1^2) = 4.9\)
- \(d(2) = 4.9(2^2) = 19.6\)
- \(d(3) = 4.9(3^2) = 44.1\)
- \(d(4) = 4.9(4^2) = 78.4\)
- \(d(5) = 4.9(5^2) = 122.5\)
- \(d(6) = 4.9(6^2) = 176.4\)
### Finding Distance Fallen After 10 Seconds
Now, to find the distance fallen after 10 seconds, we can substitute \(t = 10\) into the model:
\[
d(10) = 4.9 \cdot (10^2) = 4.9 \cdot 100 = 490 \, \text{meters}
\]
So the object will have fallen **490 meters** after 10 seconds.
In free fall, the distance \(d\) fallen after \(t\) seconds can be modeled by the equation:
\[
d(t) = \frac{1}{2} g t^2
\]
where \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth). Thus, this becomes:
\[
d(t) = \frac{1}{2} \cdot 9.8 \cdot t^2 = 4.9 t^2
\]
The table you provided matches perfectly with this model, as the distances for each second give the following:
- \(d(1) = 4.9(1^2) = 4.9\)
- \(d(2) = 4.9(2^2) = 19.6\)
- \(d(3) = 4.9(3^2) = 44.1\)
- \(d(4) = 4.9(4^2) = 78.4\)
- \(d(5) = 4.9(5^2) = 122.5\)
- \(d(6) = 4.9(6^2) = 176.4\)
### Finding Distance Fallen After 10 Seconds
Now, to find the distance fallen after 10 seconds, we can substitute \(t = 10\) into the model:
\[
d(10) = 4.9 \cdot (10^2) = 4.9 \cdot 100 = 490 \, \text{meters}
\]
So the object will have fallen **490 meters** after 10 seconds.