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/h2. Let v(t) be the velocity of the car t hours after 2:00 PM. Then (v(1/6) − v(0))/(1/6 − 0) = ? .

Question

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/h2. Let v(t) be the velocity of the car t hours after 2:00 PM. Then (v(1/6) − v(0))/(1/6 − 0) =  ? .

By the Mean Value Theorem, there is a number c such that 0 < c <  ?   with v'(c) =  ? . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 120 mi/h2

oobleck · Accepted Answer

0 < c < 1/6 v'(c) is given by the formula at the end of paragraph one.