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A passenger jet passed through an altitude of 13,000 feet 8 minutes after takeoff and was climbing at a rate of 1,250 feet per minute. Find a linear model for the altitude A (in ft) of the jet t minutes after takeoff.A = __________Use your model to predict the altitude (in ft) of the jet 12 minutes into the flight._________ ft

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A represents the altitude in feet.

t represents the minutes after takeoff

You have to create a linear model for the altitude with respect to the time:


A=mt+b

m → slope

b→ y-intercept

You know that the jet increased its altitude at a rate of 1250ft/min, this value represents the slope of the linear model.

You can express it as follows:


A=1250t+b

After 8 minutes the jet had an altitude of 13000ft, using these values you can determine the y-intercept of the linear model, replace them in the expression:


\begin{gathered} A=1250t+b \\ 13000=1250\cdot8+b \end{gathered}

Solve for b


\begin{gathered} 13000=1250\cdot8+b \\ 13000=10000+b \\ 13000-10000=1000-10000+b \\ 3000=b \end{gathered}

The y-intercept of the model is 3000ft

So the model for the altitude of the yet is


A=1250t+3000

To predict the altitude after 12 minutes you have to replace the model with t=12 and solve for A:


\begin{gathered} A=1250\cdot12+3000 \\ A=15000+3000 \\ A=18000ft \end{gathered}

The altitude of the jet after 12 minutes is 18,000ft

User Dhanveer Thakur
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