Respuesta :
Answer:
a)[tex]\Delta T= 100^{\circ}C[/tex]
b)[tex]\bigtriangledown T=1^{\circ}C.mile^{-1} [/tex]
c)[tex]\bigtriangledown T_4=1^{\circ}C.mile^{-1} [/tex]
d)[tex]\bigtriangledown T_4=1^{\circ}C.mile^{-1} [/tex]
Explanation:
Given is the data of variation of temperature with respect to the distance traveled:
Temperature T as a function of distance d:
[tex]T=(d+30) ^{\circ}C[/tex]...................................(1)
(a)
Total change in temperature from the start till the end of the journey:
[tex]\Delta T= T_f-T_i[/tex]..............................(2)
where:
[tex]T_f[/tex]= final temperature
[tex]T_i[/tex]= initial temperature
∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.
So, correspondingly we have the eq. (2) & (1) as:
[tex]\Delta T= (100+30)-(0+30)[/tex]
[tex]\Delta T= 100^{\circ}C[/tex]
(b)
Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:
[tex]\bigtriangledown T=\frac{\Delta T}{\Delta d}[/tex]......................(3)
where:
[tex]\Delta d[/tex]= change in distance
[tex]\bigtriangledown T=[/tex]change in temperature with respect to distance
putting the respective values in eq. (3)
[tex]\bigtriangledown T=\frac{100}{100}[/tex]
[tex]\bigtriangledown T=1^{\circ}C.mile^{-1} [/tex]
(c)
comparing the given function of the temperature with the general equation of a straight line:
[tex]y=m.x+c[/tex]
We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.
[tex]\bigtriangledown T_4=1^{\circ}C.mile^{-1} [/tex]
(d)
comparing the given function of the temperature with the general equation of a straight line:
[tex]y=m.x+c[/tex]
We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.
[tex]\bigtriangledown T_4=1^{\circ}C.mile^{-1} [/tex]