QUESTION IMAGE
Question
the graph below shows the altitude of an airplane during a 5 - hour flight.
- find the rate of change in the first 30 minutes after take off.
- find the rate of change from one hour and thirty minutes to three hours.
- find the rate of change from four hours and thirty minutes to when the plane landed.
- at 6:00 a.m., the temperature is 58°. at 2:00 p.m., the temperature is 76°. find the rate of change in degrees per hour during this time.
Step1: Identify points for Q7
First 30 mins = 0.5 hours. Points: $(0, 0)$ and $(0.5, 24)$ (altitude in thousands of ft).
Rate of change formula: $\frac{\text{Change in altitude}}{\text{Change in time}}$
Step2: Calculate Q7 rate
$\frac{24 - 0}{0.5 - 0} = \frac{24}{0.5} = 48$ (thousand ft per hour)
Step3: Identify points for Q8
1.5 hours to 3 hours. Points: $(1.5, 36)$ and $(3, 36)$.
Step4: Calculate Q8 rate
$\frac{36 - 36}{3 - 1.5} = \frac{0}{1.5} = 0$ (thousand ft per hour)
Step5: Identify points for Q9
4.5 hours to 5 hours. Points: $(4.5, 16)$ and $(5, 0)$.
Step6: Calculate Q9 rate
$\frac{0 - 16}{5 - 4.5} = \frac{-16}{0.5} = -32$ (thousand ft per hour)
Step7: Identify values for Q10
Time difference: 2:00 p.m. - 6:00 a.m. = 8 hours. Temp change: $76^\circ - 58^\circ = 18^\circ$.
Step8: Calculate Q10 rate
$\frac{76 - 58}{8} = \frac{18}{8} = 2.25$ (degrees per hour)
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- 48,000 feet per hour
- 0 feet per hour
- -32,000 feet per hour
- $2.25^\circ$ per hour