Question

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

Answer 1

Answer:

Approximately $722\; \rm m\cdot s^{-1}$.

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$.

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$.

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:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$.

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$.

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$.

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$.

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$.

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$.