Answer to Question #134832 in Mechanics | Relativity for Jaida Shipp

Question

a reconnaissance plane flies 424 km away from its base 578 m/s, then flies back to its base at 867 m/s. What is its average speed, answer in units of m/s

Answer 1

By the definition, the average speed is the total distance traveled divided by the total time:

$$v_{avg} = \dfrac{d_{tot}}{t_{tot}}.$$

Let the distance between the base and the destination point where the reconnaissance plane flies be $d$. Then, we can find the total distance:

$$d_{tot} = s_1 + s_2 = d+d = 2d.$$

Then, we can find the time that reconnaissance plane takes to fly from the base to the destination point:

$$t_1 = \dfrac{s_1}{v_1} = \dfrac{d}{578 \ ms^{-1}}.$$

Similarly, we can find the time that reconnaissance plane takes to fly back to its base:

$$t_2 = \dfrac{s_2}{v_2} = \dfrac{d}{867 \ ms^{-1}}.$$

Then, we can find the total time:

$$t_{tot} = t_1 + t_2 = \dfrac{d}{578 \ ms^{-1}} + \dfrac{d}{867 \ ms^{-1}}.$$

Finally, we can find the average speed of the plane:

$$v_{avg} = \dfrac{d_{tot}}{t_{tot}} = \dfrac{2d}{\dfrac{d}{578 \ ms^{-1}} + \dfrac{d}{867 \ ms^{-1}}},$$ $$v_{avg} = \dfrac{2 \cdot 424000 \ m}{\dfrac{424000 \ m}{578 \ ms^{-1}} + \dfrac{424000 \ m}{867 \ ms^{-1}}} = 694 \ ms^{-1}.$$

Answer:

$v_{avg} = 694 \ ms^{-1}.$