Answered

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Keith launches a rocket straight up into the air. The table below gives the height H () of the rocket (in meters) at a few times 1 (in seconds) during its flightTime !Height H (1)(seconds) (meters)02.85.6014042011.2(a) Find the average rate of change for the height from oseconds to 5.6 seconds.0 meters per second(b) Find the average rate of change for the height from8.4 seconds to 11.2 secondsmeters per second

Sagot :

Given data:

The given table.

a)

The expression for average rate change of height is,

[tex]\begin{gathered} h_{avg}=\frac{140-0}{5.6-0} \\ =25\text{ m/s} \end{gathered}[/tex]

Thus, the average rate of change for the height from 0 seconds to 5.6 seconds is 25 m/s.

b)

The expression for average rate change of height is,

[tex]\begin{gathered} h^{\prime}_{avg}=\frac{0-42}{11.2-8.4} \\ =\text{ -15 m/s} \end{gathered}[/tex]

Thus, the average rate of change for the height from 8.4 seconds to 11.2 seconds is - 15 m/s.

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