Final answer:
It takes approximately 1.34 seconds for a car moving at 30.8 feet per second to stop, assuming a deceleration rate of 7.00 m/s², typical for dry concrete.
Step-by-step explanation:
To determine how long it takes for a car moving at a speed of 30.8 feet per second (which is approximately 9.38 m/s) to stop, we need information about the deceleration rate. However, since that information is not provided in the question, we'll use a typical deceleration rate of 7.00 m/s² (the rate for dry concrete as per the provided reference). Using the formula to find time (t) from the final velocity (v_f), initial velocity (v_i), and acceleration (a):
t = (v_f - v_i) / a, we assume the final velocity (v_f) is 0 m/s since the car comes to a stop, and the initial velocity (v_i) is 9.38 m/s (the equivalent of 30.8 feet per second).
The deceleration rate (a) is -7.00 m/s² (negative because it is deceleration). Plugging the values into the formula gives:
t = (0 m/s - 9.38 m/s) / (-7.00 m/s²) = 1.34 seconds
So, it would take approximately 1.34 seconds for the car to stop if it could decelerate at a rate of 7.00 m/s².