Question

As mountain climbers know, the higher you go, the cooler the temperature gets. At noon on July 4th last summer, the temperature at the top of Mt. Washington — elevation 6288 feet — was 56◦F. The temperature at base camp in Pinkham Notch — elevation 2041 feet — was 87◦F. It was a clear, still day. At that moment, a group of hikers reached Tuckerman Junction — elevation 5376 feet. To the nearest degree, calculate the temperature the hikers were experiencing at that time and place. When you decided how to model this situation, what assumptions did you make?

Answer 1

a. 63 °F

b. When i decided to model the situation, I assumed that the temperature varied inversely as the elevation and that the change in elevation or temperature was linear.

Explanation:

a. To model this situation, we assume the temperature varies inversely as elevation decreases since at elevation 6288 ft the temperature is 56 °F and at elevation 2041 ft, the temperature is 87 °F

So, we model this as a straight line.

Let m be the gradient of the line.

Let the (6288 ft, 56 F) represent a point on the line and (2041 ft, 87 °F) represent another point on the line.

So m = (6288 ft - 2041 ft)/(56 °F - 87 °F) = 4247 ft/-31 °F = -137 ft/°F

At elevation 5376 ft, let the temperature be T and (5376 ft, T) represent another point on the line.

Since it is a straight line, any of the other two points matched with this point should also give our gradient. Since in the gradient, we took the point (6288 ft, 56 °F) first, we will also take it first in this instant.

So m = -137 ft/ °F = (6288 ft - 5376 ft)/(56 °F - T)

-137 ft/°F = 912 ft/(56 °F - T)

(56 °F - T)/912 ft = -1/(137 ft/ °F)

56 °F - T = -912 ft/(137 ft/°F)

56 °F - T = 6.66 °F

T = 56 °F + 6.66 °F

T = 62.66 °F

T ≅ 62.7 °F

T ≅ 63 °F to the nearest degree

b. When i decided to model the situation, I assumed that the temperature varied inversely as the elevation and that the change in elevation or temperature was linear.