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The most freakish change in temperature ever recorded was from [tex]-4^{\circ} F[/tex] to [tex]45^{\circ} F[/tex] between [tex]7:30 \, \text{am}[/tex] and [tex]7:32 \, \text{am}[/tex] on January 22, 1943, at Spearfish, South Dakota. What was the average rate of change of the temperature for this time period?



Answer :

GinnyAnswer
To determine the average rate of change of the temperature over the given time period, let's follow these steps:

1. Identify the Initial and Final Temperatures:
- Initial temperature ([tex]\( T_{initial} \)[/tex]): [tex]\(-4^\circ \text{F}\)[/tex]
- Final temperature ([tex]\( T_{final} \)[/tex]): [tex]\(45^\circ \text{F}\)[/tex]

2. Identify the Initial and Final Times:
- Initial time ([tex]\( t_{initial} \)[/tex]): [tex]\(7:30 \text{ am}\)[/tex]
- Final time ([tex]\( t_{final} \)[/tex]): [tex]\(7:32 \text{ am}\)[/tex]

3. Convert the Times into Decimal Hours:
- Time is usually easier to work with in decimal form rather than hours and minutes separately.
- Initial time ([tex]\( t_{initial} \)[/tex]):
[tex]\[ 7 \text{ hours } + \frac{30 \text{ minutes}}{60} = 7 + \frac{1}{2} = 7.5 \text{ hours} \][/tex]
- Final time ([tex]\( t_{final} \)[/tex]):
[tex]\[ 7 \text{ hours } + \frac{32 \text{ minutes}}{60} = 7 + \frac{32}{60} = 7 + \frac{8}{15} \approx 7.5333 \text{ hours} \][/tex]

4. Calculate the Temperature Change:
- The change in temperature ([tex]\( \Delta T \)[/tex]) is the difference between the final and initial temperatures:
[tex]\[ \Delta T = T_{final} - T_{initial} = 45^\circ \text{F} - (-4^\circ \text{F}) = 45^\circ \text{F} + 4^\circ \text{F} = 49^\circ \text{F} \][/tex]

5. Calculate the Time Change:
- The change in time ([tex]\( \Delta t \)[/tex]) is the difference between the final and initial times in hours:
[tex]\[ \Delta t = t_{final} - t_{initial} \approx 7.5333 \text{ hours} - 7.5 \text{ hours} = 0.0333 \text{ hours} \][/tex]

6. Calculate the Average Rate of Change:
- The average rate of change of the temperature is given by the change in temperature divided by the change in time:
[tex]\[ \text{Average rate of change} = \frac{\Delta T}{\Delta t} = \frac{49^\circ \text{F}}{0.0333 \text{ hours}} \approx 1470 \text{ degrees per hour} \][/tex]

So, the average rate of change of the temperature for this period was approximately [tex]\(1470\)[/tex] degrees Fahrenheit per hour.

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