Daniela and Jack are hiking a steady incline. They use their GPS device to determine their elevation every 15 minutes. At 15 minutes and 30 minutes they were at elevations of 10,300 and 10,900 feet, respectively. Write an equation expressing their elevation in relation to time.

Answer:

[tex]h=40t+9700[/tex]

where the elevation, h, is in feet and time, t, is in minutes.

Step-by-step explanation:

At the time, [tex]t_1=15[/tex] min, the elevation, [tex]h_1= 10,300[/tex] feet and

at the time [tex]t_2= 30[/tex] min, the elevation, [tex]h_2=10,900[/tex] feet.

As they are hiking a steady incline, so the change in the elevation with respect to time will be constant.

So, there will be a linear relationship between the elevation and the time.

Let [tex]h[/tex] be the elevation at any time instant [tex]t[/tex], so the linear relation among these quantities is

[tex]h=mt+C_0\;\cdots(i)[/tex]

where [tex]m[/tex] is the rate of change of elevation with respect to time and [tex]C_0[/tex] is constant.

The change in the elevation, [tex]\Delta h=h_2-h_1= 10,900-10,300=600[/tex] feet.

and the change in time, [tex]\Delta t=t_2-1_1=30-15=15[/tex] min.

So, change in the elevation in unit time,

[tex]m=\frac{\Delat h}{\Delta t}=\frac{600}{15}=40[/tex] feet/min.

Now, from equation (i)

[tex]h=40t+C_0[/tex]

As the elevation, [tex]h=h_1[/tex] at time [tex]t=t_1[/tex], so

[tex]10,300=40\times15+C_0[/tex]

[tex]\Rightarrow C_0=9700[/tex]

Hence, the required equation is

[tex]h=40t+9700[/tex]

where the elevation, [tex]h[/tex], is in feet and time, [tex]t[/tex], is in minutes.

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