## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the speed of the hawk as: $f' = \frac{f}{(1-\frac{v_{source}}{v_{snd}})}$ $1-\frac{v_{source}}{v_{snd}} = \frac{f}{f'}$ $\frac{v_{source}}{v_{snd}} = 1-\frac{f}{f'}$ $v_{source} = (1-\frac{f}{f'})~ v_{snd}$ $v_{source} = (1-\frac{800~Hz}{900~Hz})~343~m/s$ $v_{source} = 38~m/s$ The hawk is approaching with a speed of 38 m/s.