Plot the image of Point Q under the translation (x,y) (x-3, y + 4)
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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/
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/