Certainly! Let's break down and solve the problem step-by-step.
### Step-by-Step Solution:
Given Values:
1. Initial velocity, [tex]\( V_0 = 60 \)[/tex] feet/second.
2. Road grade, [tex]\( G = 0.05 \)[/tex].
3. Coefficient of friction between the tires and road, [tex]\( f = 0.7 \)[/tex].
4. Acceleration due to gravity, [tex]\( g = 32.2 \)[/tex] feet/second[tex]\(^2\)[/tex].
Formula:
The formula to compute the braking distance is:
[tex]\[ d = \frac{V_0^2}{2 g (f + G)} \][/tex]
### Step 1: Square the Initial Velocity
First, we need to square the initial velocity, [tex]\( V_0 \)[/tex].
[tex]\[ V_0^2 = 60^2 = 3600 \, \text{(feet/second)}^2 \][/tex]
### Step 2: Calculate the Denominator
We need to calculate [tex]\( 2 \cdot g \cdot (f + G) \)[/tex].
First, sum the coefficient of friction and the road grade:
[tex]\[ f + G = 0.7 + 0.05 = 0.75 \][/tex]
Second, calculate [tex]\( 2 \cdot g \cdot 0.75 \)[/tex]:
[tex]\[ 2 \cdot g \cdot 0.75 = 2 \cdot 32.2 \cdot 0.75 \][/tex]
[tex]\[ 2 \cdot 32.2 = 64.4 \][/tex]
[tex]\[ 64.4 \cdot 0.75 = 48.3 \][/tex]
So, the denominator is [tex]\( 48.3 \)[/tex].
### Step 3: Divide the Squared Initial Velocity by the Denominator
Finally, we need to divide [tex]\( V_0^2 \)[/tex] by the result of the previous step:
[tex]\[ d = \frac{3600}{48.3} \][/tex]
### Step 4: Calculate the Braking Distance
[tex]\[ d = 74.53416149068322 \][/tex]
### Conclusion:
The braking distance, given the initial velocity of 60 feet/second, a road grade of 0.05, a coefficient of friction of 0.7, and an acceleration due to gravity of 32.2 feet/second[tex]\(^2\)[/tex], is approximately 74.53 feet.
Thus, the braking distance of the car is [tex]\( \boxed{74.53416149068322} \)[/tex] feet.
