The average person passes out at an acceleration of 7g (that is, seven times the gravitational acceleration on Earth). Suppose a car is designed to accelerate at this rate. How much time would be required for the car to accelerate from rest to 59.1 miles per hour?

To find the time required for the car to accelerate from rest to 59.1 miles per hour, we need to use the equation of motion that relates acceleration, initial velocity, final velocity, and time.

The equation we can use in this case is:

v = u + at,

where:
v = final velocity (59.1 miles per hour),
u = initial velocity (0 miles per hour, since the car starts from rest),
a = acceleration (7g, or 7 times the gravitational acceleration on Earth),
t = time.

First, we need to convert the final velocity from miles per hour to a more convenient unit, such as meters per second. Since 1 mile equals 1609.34 meters and 1 hour equals 3600 seconds, we can convert 59.1 miles per hour to meters per second as follows:

(59.1 miles/hour) * (1609.34 meters/mile) * (1 hour/3600 seconds) ≈ 26.42 meters/second

Now, we can substitute the known values into the equation:

26.42 = 0 + 7gt.

Since the car starts from rest (0 initial velocity), the equation simplifies to:

26.42 = 7gt.

To find the time, we need to isolate t. Divide both sides of the equation by 7g:

t = 26.42 / (7g).

To calculate the value of g, which is the gravitational acceleration on Earth, we can simply use the average value of approximately 9.8 meters per second squared.

Now, substitute the value of g and calculate the time required:

t = 26.42 / (7 * 9.8) ≈ 0.376 seconds.

Therefore, it would take approximately 0.376 seconds for the car to accelerate from rest to 59.1 miles per hour if it maintains an acceleration of 7g.