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 acce
leration 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.
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/
7 tenths times 10 is pretty easy and can be done on a calculator. 7 tenths is 0.7 in standard form.
0.7 × 10 = 7
Because if you think about it... Tenths is like being dropped down by ten... then you multiply by ten which brings it back up to whole number seven. It may not make sense but that's what it is.