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.

The average acceleration is (50-30)/(1/6) = 120

Since velocity is continuous, the mean value theorem guarantees that the stipulated condition exists. Maybe everywhere, but surely somewhere.

To show that the acceleration is exactly 120 mi/h^2 at some point between 2:00 PM and 2:10 PM, we can calculate the average acceleration during this time period and check if it equals 120 mi/h^2.

Acceleration is defined as the rate of change of velocity over time. We can use the formula:

Acceleration = (Change in Velocity) / (Change in Time)

In this case, the change in velocity is the difference between the final velocity and the initial velocity, and the change in time is the time interval between the readings.

Given that at 2:00 PM the speedometer reads 30 mi/h, and at 2:10 PM it reads 50 mi/h, we can calculate the change in velocity:

Change in Velocity = 50 mi/h - 30 mi/h = 20 mi/h

The change in time is 10 minutes (or 1/6 hour) because it is the time interval between the two readings.

Change in Time = 2:10 PM - 2:00 PM = 10 minutes = 1/6 hour

Now we can calculate the average acceleration:

Average Acceleration = (Change in Velocity) / (Change in Time)
= (20 mi/h) / (1/6 hour)
= 120 mi/h^2

Therefore, we have shown that the average acceleration between 2:00 PM and 2:10 PM is exactly 120 mi/h^2, which implies that at some point within this time interval, the acceleration must have been exactly 120 mi/h^2.