A ball was tossed into the air. its height, in metres, is given by h(t)= - 4.9t^2 + 6t + 11, where t is time, in seconds.
Write an expression that represents the average rate of change over the interval 2<t<2 + h.
h(2) = -4.9(4) + 6(2) + 11 = 3.4
h(2+h)
= -4.9(2+h)^2 + 6(2+h) + 11
= -4.9(4 + 4h + h^2 + 12 + 6h + 11
= -19.6 - 19.6h - 4.9h^2 + h^2 + 6h + 23
= 3.4 - 3.9h^2 - 13.6h
average rate of change = (3.4 - 3.9h^2 - 13.6h - 3.4)/(2+h - 2)
= (-3.9h^2 - 13.6h)/h
= -3.9h - 13.6 , h ≠ 0
To find the average rate of change over the given interval, we need to evaluate the expression for h(t) at the upper and lower bounds of the interval and then calculate the difference between the two values.
First, let's find h(2 + h) by substituting 2 + h into the expression for h(t):
h(2 + h) = -4.9(2 + h)^2 + 6(2 + h) + 11
Simplifying the expression:
h(2 + h) = -4.9(4 + 4h + h^2) + 12 + 6h + 11
= -19.6 - 19.6h - 4.9h^2 + 12 + 6h + 11
= -4.9h^2 - 13h + 3.4
Next, let's find h(2) by substituting 2 into the expression for h(t):
h(2) = -4.9(2)^2 + 6(2) + 11
= -19.6 + 12 + 11
= 3.4
Now, we can calculate the average rate of change by finding the difference between h(2 + h) and h(2) and dividing it by h:
Average rate of change = (h(2 + h) - h(2)) / h
= (-4.9h^2 - 13h + 3.4 - 3.4) / h
= (-4.9h^2 - 13h) / h
= -4.9h - 13
Therefore, the expression that represents the average rate of change over the interval 2<t<2 + h is -4.9h - 13.