Question

Susan launches a rocket straight up into the air. The table below gives the height (t) of the rocket (in meters) at a few times t (in seconds) during its flightTimet(seconds)02.26.68.813.2Height HD(meters)0110198440(a) Find the average rate of change for the height from oseconds to 2.2 seconds.Il meters per second(6) Find the average rate of change for the height from6.6 seconds to 13.2 secondsI meters per second

Answer 1

The average rate of change from t1 to t2 where t2>t1 for a function the function H(t) is given as:

`(H(t_2)-H(t_1))/(t_2-t_1)`

Next substitute H(t2)=110, t2=2.2, H(t1)=0, t_1=0 into the formula:

`\begin{gathered} (110-0)/(2.2-0) \\ =(110)/(2.2) \\ =50\text{ m/s} \end{gathered}`

For the "b" part, substitute H(t2)=0, t2=13.2, H(t1)=198, t_1=6.6 into the formula:

`\begin{gathered} (0-198)/(13.2-6.6) \\ =-(198)/(6.6) \\ =-30\text{ m/s} \end{gathered}`

The negative in the rate of change here represent a decrease in height as time increases.