To find the car's acceleration when its brakes are applied, you can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for acceleration is:
\[a = \frac{F}{m}\]
Where:
- \(a\) is the acceleration (in meters per second squared, m/s²).
- \(F\) is the net force (in newtons, N).
- \(m\) is the mass of the car (in kilograms, kg).
In your case, the car has a mass (\(m\)) of 1000 kg and a net force (\(F\)) of 5000 N. Plugging these values into the formula, you get:
\[a = \frac{5000\, N}{1000\, kg} = 5\, m/s^2\]
So, the car's acceleration when its brakes are applied is \(5\, m/s²\).
To find the minimum distance required for the car to stop, you can use the following equation of motion:
\[v^2 = u^2 + 2as\]
Where:
- \(v\) is the final velocity (0 m/s since the car comes to a stop).
- \(u\) is the initial velocity (55 mph, which we need to convert to m/s).
- \(a\) is the acceleration (5 m/s², as calculated above).
- \(s\) is the distance traveled.
First, let's convert the initial velocity from mph to m/s. There are approximately 0.44704 m/s in 1 mph, so:
\(u = 55\, mph \times 0.44704\, m/s/mph \approx 24.59\, m/s\)
Now, we can plug in the values into the equation:
\[0^2 = (24.59\, m/s)^2 + 2 \times 5\, m/s^2 \times s\]
Simplify the equation:
\[0 = 604.48\, m^2/s^2 + 10\, m/s^2 \times s\]
Now, we can solve for \(s\):
\[10\, m/s^2 \times s = -604.48\, m^2/s^2\]
\[s = \frac{-604.48\, m^2/s^2}{10\, m/s^2} \approx -60.45\, m^2\]
Since distance cannot be negative, the minimum distance required for the car to stop is approximately 60.45 meters.