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.

Answer 1

Answer: If we define 2:00pm as our 0 in time; then:

at t= 0. the velocity is 30 mi/h.

then at t = 10m (or 1/6 hours) the velocity is 50mi/h

Then, if we think in the "mean acceleration" as the slope between the two velocities, we can find the slope as:

a= (y2 - y1)/(x2 - x1) = (50 mi/h - 30 mi/h)/(1/6h - 0h) = 20*6mi/(h*h) = 120mi/$h^{2}$

Now, this is the slope of the mean acceleration between t= 0h and t = 1/6h, then we can use the mean value theorem; who says that if F is a differentiable function on the interval (a,b), then exist at least one point c between a and b where F'(c) = (F(b) - F(a))/(b - a)

So if v is differentiable, then there is a time T between 0h and 1/6h where v(T) = 120mi/$h^{2}$