Question

If a free-falling object is dropped from a height of 100 feet, its height "h" at time "t" is given by:

h=-16t^2+100

where "h" measured in feet and "t" is measured om seconds. Find the average rate of change of the height over the following intervals:

a) [1,2] b) [1,1.5] c) [1,1.1]

Answer 1

rate of change of height would be\[dh \div dt= -16d(t ^{2})+d(100)\]
\[dh \div dt=-16\times2t\]
\[dh \div dt=-32t\]
now put values of t to find rate of change of height as per particular interval of time. . .

Answer 2

\[H(average)=[h(t_{2})-h(t_{1})]/(t_{2}-t_{1})\]

No need to find the derivative. The derivative dh/dt will give you the instantaneous speed not the average height change (average speed)

Answer 3

you can get average of all instantaneous speeds after that.

Answer 4

Yes, integrating the derivative, which is the original function as per my formula

Answer 5

\[v(t)=dh(t)/dt \]
and
\[AvgSpeed=1/(t_2-t_1)\int\limits_{t_1}^{t_2}v(t)dt=\]
\[1/(t_2-t_1)\int\limits_{t_1}^{t_2}(dh(t)/dt)dt=1/(t_2-t_1)\int\limits_{t_1}^{t_2}dh(t)=[h(t_2)-h(t_1)]/(t_2-t_1)\]

quod erat demostrandum