Answer to Question #144017 in Algebra for Savana

Question

An airplane is descending in preparation for a landing at an airport. The height, h, in meters of the airplane above the ground level of the airport is a linear function of time, t, where t is the number of minutes after the plane started to descend. The plane was 3,915 meters above the ground level of the airport 2 minutes into the plane's decent and was 2,025 meters above after 9 minutes. How long will it take the plane to land at the airport after it started to descend

Answer 1

$$Solution$$

After it started to descend, it took the plane exactly $7\ minutes$ to descend from $3915\ meters$ to $2025\ meters$

$Initial\ Height\ (H_0)\ =\ 3915\ m\\ Final\ Height\ (H_1)\ =\ 2025\ m\\ Displacement\ =\ \delta H=1890m\\ Change\ in\ time\ =\ \delta t\ =\ 9-2=7\ minutes\ \times\ 60\ =\ 420\ s\\$

Now, we will calculate the velocity, $\bar v$

$$\bar v=\frac{\delta H}{\delta t}= \frac{1890m}{420s}=4.5ms^{-1}$$

Given $\bar v=4.5ms^{-1}$

$Initial\ Height\ (H_0)\ =\ 2025\ m\\ Final\ Height\ (H_1)\ =\ 0\ m\\ Displacement\ =\ \delta H=2025\ m \\$

$4.5\ ms^{-1}\ \delta t=2025\ m\\ \delta t\ =\ \frac{2025\ m}{4.5 ms^{-1}}=450\ seconds$

Total time it took the plane to land from the time it started to descend.

Total time = $(420 + 450)s=870\ seconds$

$$\therefore T=14.5 \ seconds$$