Recall that the average velocity between times t1and t2 is calculated using
vave = y(t2) − y(t1)/t2 − t1.
We are given that
y = 14t − 1.86t^2.
Evaluating at
t = 1,
we have
y(1) = 14(1) − 1.86(1)2 = 12.14.
Thus, the average velocity between times 1 and 1 + h is as follows.
vave = y(1 + h) − y(1)/(1 + h) − 1
= [14(1 + h) − 1.86(1 + h)^2] − 12.14/h
=10.28h − _____h^2/h
= ____ − _______h, h ≠ 0
(10.28h - 3,46h^2)/h
10.28 - 3.46h
Man, how could you get that far, and be stumped by 1.86^2?
To find the average velocity between times 1 and 1 + h, we need to substitute the values of y(1 + h) and y(1) into the average velocity formula vave = (y(1 + h) - y(1))/(1 + h - 1).
Given that y = 14t - 1.86t^2, we can evaluate y(1 + h) and y(1) using this equation.
First, substitute t = 1 + h into the equation for y:
y(1 + h) = 14(1 + h) - 1.86(1 + h)^2
Next, substitute t = 1 into the equation for y:
y(1) = 14(1) - 1.86(1)^2
Calculating these values will give us the result for y(1 + h) and y(1).
Now, substitute these values into the average velocity formula:
vave = (y(1 + h) - y(1))/(1 + h - 1)
Simplifying the equation will give us the expression for the average velocity between times 1 and 1 + h.