To solve these problems, we can use the relationship between speed and stopping distance. The stopping distance of a car is proportional to the square of its speed. This means that if you know the stopping distance at one speed, you can find the stopping distance at a different speed using the following formula:
\[ \frac{d_2}{d_1} = \left( \frac{v_2}{v_1} \right)^2 \]
where:
- \(d_1\) is the stopping distance at speed \(v_1\),
- \(d_2\) is the stopping distance at speed \(v_2\).
(a) Finding the stopping distance at 60 mph
Given:
- \(v_1 = 40\) mph,
- \(d_1 = 112\) feet,
- \(v_2 = 60\) mph.
We will find \(d_2\), the stopping distance at 60 mph.
Using the proportionality:
\[ \frac{d_2}{112} = \left( \frac{60}{40} \right)^2 \]
Calculating \( \left( \frac{60}{40} \right)^2 \):
\[ \frac{60}{40} = 1.5 \] \[ (1.5)^2 = 2.25 \]
Now substituting back:
\[ \frac{d_2}{112} = 2.25 \]
To find \(d_2\):
\[ d_2 = 112 \times 2.25 = 252 \text{ feet} \]
So, the stopping distance of the car traveling at 60 mph is 252 feet.
(b) Finding the speed given a stopping distance of 7 ft
We want to find \(v_1\) given \(d_1 = 7\) feet, while using the stopping distance at 40 mph as our reference point.
We already know the proportionality:
- For 40 mph, \(d_1\) is 112 feet.
Now we want to find \(v_1\) corresponding to \(d_1 = 7\) feet. We use the relationship as before:
\[ \frac{7}{112} = \left( \frac{v_1}{40} \right)^2 \]
Calculating \( \frac{7}{112} \):
\[ \frac{7}{112} = \frac{1}{16} \]
Now we can set up the equation:
\[ \frac{1}{16} = \left( \frac{v_1}{40} \right)^2 \]
Taking the square root of both sides gives us:
\[ \frac{1}{4} = \frac{v_1}{40} \]
Solving for \(v_1\):
\[ v_1 = 40 \times \frac{1}{4} = 10 \text{ mph} \]
Thus, the speed of the car before the brakes were applied was 10 mph.