To determine the radius of the yaw mark, we can use the geometry of a circular segment. A circular segment is formed by a chord (in this case, the yaw mark chord) and an arc. The middle ordinate is the perpendicular distance from the midpoint of the chord to the arc. In this question, you've given the length of the chord as 42.25 feet and the middle ordinate as 6 feet.
The radius of the circle can be determined using the following relation, which can be derived from the geometry of a triangle and a segment of a circle:
\[ r = \frac{h}{2} + \frac{\frac{c^2}{4}}{2h} \]
where
- \( r \) is the radius of the circle,
- \( h \) is the middle ordinate (6 feet),
- \( c \) is the length of the chord (42.25 feet).
Let's calculate it step by step:
1. Calculate \( \frac{c^2}{4} \):
\[ \frac{c^2}{4} = \frac{42.25^2}{4} = \frac{1785.0625}{4} = 446.265625 \text{ feet}^2 \]
2. Now plug in the values in the radius formula:
\[ r = \frac{h}{2} + \frac{\frac{c^2}{4}}{2h} \]
\[ r = \frac{6}{2} + \frac{446.265625}{2 \cdot 6} \]
\[ r = 3 + \frac{446.265625}{12} \]
\[ r = 3 + 37.1888020833 \]
\[ r = 40.1888020833 \text{ feet} \]
Rounding to the nearest tenth gives a radius \( r \) of approximately \( 40.2 \text{ feet} \).
Now, to determine the minimum skid speed (which is essentially the speed at which the vehicle was traveling when it began to skid), we use the formula for the critical speed of a car going around a curve:
\[ v = \sqrt{r \cdot f \cdot g} \]
where
- \( v \) is the speed,
- \( r \) is the radius we just found,
- \( f \) is the drag factor (0.85),
- \( g \) is the acceleration due to gravity (approximately 32.2 feet per second squared).
Plugging in the known values:
\[ v = \sqrt{40.2 \cdot 0.85 \cdot 32.2} \]
\[ v = \sqrt{34.17 \cdot 32.2} \]
\[ v = \sqrt{1100.674} \]
\[ v = 33.155733128 \text{ feet per second} \]
To convert feet per second to miles per hour, we use the conversion factor \( 1 \text{ mph} = 1.46667 \text{ feet per second} \).
\[ v = 33.155733128 \times \frac{1}{1.46667} \]
\[ v = 22.610223 \text{ mph} \]
Rounding to the nearest tenth gives a minimum skid speed \( v \) of approximately \( 22.6 \text{ mph} \).
