A hiker is going up a mountain trail 30 degree incline at a

A hiker is going up a mountain trail (30 degree incline) at a speed of 2000ft/hr (along incline) starting at 7am:

Please answer both parts of the question and provide explanations!

RECALL FROM LECTUPE AND BOOK SECTION 4-2 FOR ANY QUANTITY f TOTAL OR MATERIAL DERIVATIve d b t è LOCALCONVECTIVE DERIV. DERIV A HIKER IS GOING UP A MOUNTAIN TRAIL (30 INCLINE AT A SPEED OF 2000 FEET / HOUR (ALONG INCLINE) \"TARTIN6 ATAM . ATIO - AS SUN COMES UP, THE WHOLE HILLsiDe WARMS AT 3c/HoUR, E IN HOURS a) what totol rate of temperature change would the hiker observe op b) Houu fost shouldl a hiker go in order to remain ; thermae ? 30° TEMPERATURE DOES NOT CHANGE ALONG x or

Solution

a)

Speed of hiker on inclined = 2000ft/hr

angle of incline = 30^o

Vertical component of hiker\'s speed = 2000 x sin 30^o ft/hr = 2000 x 1/2 = 1000 ft/hr

Hiker is starting at 7 am;

assuming that the temperature at 7 am is T₇

temperature due to elevation = T (z,t) = T_A(t) - 0.001z (given in the question)

z = vertical distance covered by the hiker in time t = vertical component of speed x time t = 1000 x t ft/hr

Thus, temeperature at time t = T_A(t) - 0.001 x 1000 x t = T₇ - t

so, the decrease in temperature is t units after t hours from 7 am

or , rate of decrease in temperature = 1 unit per hour

Now, increase in temperature due to sun is 3⁰ C per hour

therefore, temperature at any time t from the start = T (z,t) = T(z,7) + 3(t-7) = T₇ + 3t - 21

increase in temperture = 3t - 21 units after t hours

Net change in temeprature = (3t - 21) - t = 2t-21 units after t hours

differentiating 2t-21 with respect to t; we get,

rate of change of temperature = 2 units per hour (increase)

this rate can also be simply found out as we already calculated that the decrease in temperature is 1⁰ C per hour and the questionclearly states that there is a 3⁰ C increase in temperaature per hour due to hillwarming

thus the net rate of increase in temperature = 3-1 = 2⁰ C per hour

b)

To remain isothermal , lets assume that the hiker should maintain a speed of s km/hr

vertical ccompenent of speed = s x sin30 = s/2

as in the previous part (a);

temperatue at any time t (decreased temperature due to elevation) = T_A (t)- 0.001z

T₇ - 0.001 x s/2 x t

thus, decrease in temperature after time t = 0.001 x s/2 x t

and rate of change in temperature = 0.001 x s/2 ⁰C per hour  

now, increase in temperature due to warming = 3^o C per hour;

to remain isothermal;

rate of decrease in temperature = rate of increase in temperature

thus, 0.001 x s/2 = 3

s = 3/0.001 x 2 = 6000 ft/per hour

therefore, the hiker should go at a speed of 6000 ft/hr to remain isothermal.