At 2:00 PM a car's speedometer reads 30 mi/h. At 2:10 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi/h^2.

Answer 1

This is an application of the mean value theorem, which says that, at some instant in the interval (2:00, 2:10),

d/dt (speed) = [speed at time(2:10) - speed at time(2:00)]/[duration of the interval]

= (50 mi/hr - 30 mi/hr ) / (1/6 of an hr) = 6 * 20 mi/hr² = 120 mi/hr²

Since acceleration d/dt (speed), we have acceleration = 120 mi/hr² at some instant in (2:00, 2:10). (The mean value theorem doesn't tell us exactly when.)

Answer 2

Assuming that the velocity function f(t) is continuous on the closed time interval and is differentiable on the open time interval, then, by the Mean Value Theorem, there exists a value of c on the open time interval for which f'(c) = [50 - 30] / (1/6)hr = 120 mi/hr². Since the derivative of velocity is acceleration, we are done.

Answer 3

acceleration = change of speed / time

change of speed = 50 -20 = 20 mph

time = 10 minutes = 1/6 hour

20 divided by 1/6 = 20 * 6/1 = 20*6 = 120 mi/hr^2