A car is being driven at a rate of 40 mph when the brakes are applied.The car decelerates at a constant rate of 10ft/seg^2.How long before the car stops?

To solve this problem, we first need to convert the speed from miles per hour to feet per second since the deceleration rate is given in feet per second squared.

Given:
Speed = 40 mph
Deceleration rate = 10 ft/s^2

We know that 1 mile = 5280 feet and 1 hour = 3600 seconds.

Converting speed from mph to ft/s:
Speed in ft/s = \frac{40 \times 5280}{3600} = \frac{211200}{3600} = 58.67 ft/s

Now, we can use the formula for deceleration to find the time it takes for the car to stop:
v = u + at
where:
v = final velocity (0 ft/s since the car stops)
u = initial velocity (58.67 ft/s)
a = deceleration rate (-10 ft/s^2, negative because it's deceleration)
t = time taken

Substitute the values into the formula:
0 = 58.67 + (-10)t
-58.67 = -10t
t = \frac{-58.67}{-10} = 5.87 seconds

Therefore, the car will stop after approximately 5.87 seconds.

\textbf{Answer:} It will take 5.87 seconds for the car to stop.