If you shout "hello" near a vertical canyon wall, an echo "hello" returns after a short delay. if it returns in 2.0 s and sound travels through air at 340 m/s, then the total distance the signal traveled is 680 m. therefore, the canyon wall must be half that distance from you (340 m). bats use a similar process to hunt their prey. bats travel and locate food at night by emitting ultrasound waves that bounce off prey and return to the bat. the time interval between signal emission and its return to the bat tells the bats which insects are closer than others. the amplitude of the reflected wave indicates the size of the prey−a large animal causes a large amplitude reflected wave. the bat can tell the left-right location of the object by the relative time delays of the return sound between the left and right ears. rows of horizontal folded sound sensors in the bat's ears allow them to determine the up-down location of the prey. finally, the bat can determine if the prey is moving toward it, moving away from it, or is stationary by the doppler shift of the returning signal. the relationship between the source frequency of the bat and the frequency of the reflected sound is given by the equation: fr = fs (v+0/v−vptob) (v+vptob/v−0) = fs (v+vptob/v−vptob)you have a bat-like echolocation system on your car, which emits a 20,000-hz frequency sound that returns to the car in 0.18 s after being reflected by another stationary car. the car is moving at 30 m/s toward the stationary car. find the distance that the moving car travels during the 0.18 s.