Answer :
Let's solve the problem step-by-step.
First, we are given:
- Initial velocity [tex]\( v = 40 \)[/tex] mph
- Stopping distance [tex]\( d = 138 \)[/tex] feet
The general equation for stopping distance [tex]\( d(v) \)[/tex] with respect to [tex]\( v \)[/tex] and [tex]\( f \)[/tex] is:
[tex]\[ d(v) = \frac{2.15 \, v^2}{f} \][/tex]
We are asked to determine which of the given models fits the given scenario:
1. [tex]\( d(v) = \frac{2.15 \, v^2}{0.039} \)[/tex]
2. [tex]\( d(v) = \frac{2.15 \, v^2}{64.79} \)[/tex]
3. [tex]\( d(v) = \frac{2.15 \, v^2}{25.116} \)[/tex]
To do this, we'll compare stopping distance calculation for each case by calculating the friction factor [tex]\( f \)[/tex] using the given conditions.
1. For the model [tex]\( d(v) = \frac{2.15 \, v^2}{0.039} \)[/tex]:
[tex]\[ d = 138 \, \text{feet} \][/tex]
[tex]\[ v = 40 \, \text{mph} \][/tex]
Rearrange the equation [tex]\( d(v) = \frac{2.15 \, v^2}{f} \)[/tex] to solve for [tex]\( f \)[/tex]:
[tex]\[ f = \frac{2.15 \, v^2}{d} \][/tex]
Substitute in the values:
[tex]\[ f = \frac{2.15 \cdot (40)^2}{138} \][/tex]
Calculate [tex]\( f \)[/tex]:
[tex]\[ 2.15 \cdot 1600 = 3440 \][/tex]
[tex]\[ f = \frac{3440}{138} \][/tex]
When computed, the value of [tex]\( f \)[/tex]:
[tex]\[ f \approx 24.93 \][/tex]
Comparing with the given models, we see that none of the given denominators exactly match [tex]\( 24.93 \)[/tex], showing that the correct quadratic model does not align with the given options.
Therefore, the correct choice here is:
[tex]\[ \boxed{\text{None}} \][/tex]
First, we are given:
- Initial velocity [tex]\( v = 40 \)[/tex] mph
- Stopping distance [tex]\( d = 138 \)[/tex] feet
The general equation for stopping distance [tex]\( d(v) \)[/tex] with respect to [tex]\( v \)[/tex] and [tex]\( f \)[/tex] is:
[tex]\[ d(v) = \frac{2.15 \, v^2}{f} \][/tex]
We are asked to determine which of the given models fits the given scenario:
1. [tex]\( d(v) = \frac{2.15 \, v^2}{0.039} \)[/tex]
2. [tex]\( d(v) = \frac{2.15 \, v^2}{64.79} \)[/tex]
3. [tex]\( d(v) = \frac{2.15 \, v^2}{25.116} \)[/tex]
To do this, we'll compare stopping distance calculation for each case by calculating the friction factor [tex]\( f \)[/tex] using the given conditions.
1. For the model [tex]\( d(v) = \frac{2.15 \, v^2}{0.039} \)[/tex]:
[tex]\[ d = 138 \, \text{feet} \][/tex]
[tex]\[ v = 40 \, \text{mph} \][/tex]
Rearrange the equation [tex]\( d(v) = \frac{2.15 \, v^2}{f} \)[/tex] to solve for [tex]\( f \)[/tex]:
[tex]\[ f = \frac{2.15 \, v^2}{d} \][/tex]
Substitute in the values:
[tex]\[ f = \frac{2.15 \cdot (40)^2}{138} \][/tex]
Calculate [tex]\( f \)[/tex]:
[tex]\[ 2.15 \cdot 1600 = 3440 \][/tex]
[tex]\[ f = \frac{3440}{138} \][/tex]
When computed, the value of [tex]\( f \)[/tex]:
[tex]\[ f \approx 24.93 \][/tex]
Comparing with the given models, we see that none of the given denominators exactly match [tex]\( 24.93 \)[/tex], showing that the correct quadratic model does not align with the given options.
Therefore, the correct choice here is:
[tex]\[ \boxed{\text{None}} \][/tex]
