Answer:
Approximately [tex]722\; \rm m\cdot s^{-1}[/tex].
Explanation:
The average speed of a vehicle is calculated as:
[tex]\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}[/tex].
In this question, the total distance is [tex]2 \times 560\; \rm km = 1120\; \rm km[/tex].
The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:
[tex]560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m[/tex].
[tex]1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m[/tex].
Time required for the first part of this trip:
[tex]\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s[/tex].
Time required for the second part of this trip:
[tex]\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s[/tex].
The time required for the entire trip would be approximately [tex]930 + 620 = 1550\; \rm s[/tex].
Calculate the average speed of this plane:
[tex]\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}[/tex].
