To model the situation described with the drone's distance over time, we can use a quadratic equation of the form:
\[ y = ax^2 + bx + c \]
where \( y \) represents the distance in feet, \( x \) represents the time in seconds, and \( a \), \( b \), and \( c \) are constants that we need to determine based on the given data points.
Given Data Points
- (0, 15.36)
- (6, 24)
- (10, 20.16)
- (12, 15.36)
- (16, 0)
Step 1: Setting Up Simultaneous Equations
We will substitute the \( x \) and \( y \) values from three of the data points into the quadratic equation to get three simultaneous equations.
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Using point (0, 15.36): \[ c = 15.36 \]
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Using point (6, 24): \[ 24 = 36a + 6b + 15.36 \] \[ 36a + 6b = 24 - 15.36 \] \[ 36a + 6b = 8.64 \] \[ 6a + b = 1.44 \] (Equation 1)
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Using point (10, 20.16): \[ 20.16 = 100a + 10b + 15.36 \] \[ 100a + 10b = 20.16 - 15.36 \] \[ 100a + 10b = 4.8 \] \[ 10a + b = 0.48 \] (Equation 2)
Step 2: Solving the System of Equations
We now have two equations:
- \( 6a + b = 1.44 \)
- \( 10a + b = 0.48 \)
To eliminate \( b \), we can subtract the first equation from the second:
\[ (10a + b) - (6a + b) = 0.48 - 1.44 \] \[ 4a = -0.96 \] \[ a = -0.24 \]
Now, substitute \( a \) back into Equation 1 to find \( b \):
\[ 6(-0.24) + b = 1.44 \] \[ -1.44 + b = 1.44 \] \[ b = 2.88 \]
Step 3: Finding \( c \)
We already know \( c = 15.36 \).
Final Equation
Now we can write the quadratic equation:
\[ y = -0.24x^2 + 2.88x + 15.36 \]
Step 4: Converting to Standard Form
To write it in standard form \( Ax^2 + Bx + C \), we have:
\[ y = -0.24x^2 + 2.88x + 15.36 \]
Step 5: Rounding Values
All values in the equation are rounded to the nearest hundredth, resulting in the final equation:
\[ y = -0.24x^2 + 2.88x + 15.36 \]
Thus, the final answer is:
The equation that models this situation is: \[ y = -0.24x^2 + 2.88x + 15.36 \]