You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as . You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as What is the speed of the police car? The speed of sound in a

The complete question is:

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as 1340 Hz. You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as 1300 Hz. What is the speed of the police car? The speed of sound in air is 343m/s.

V_s = 30 m/s

Explanation:

The change in frequency observation occur due to doppler effect is given by the equation;

f_o = [(V ± V_o)/(V ∓ V_s)]f_s

Where;

f_o is observed frequency

f_source is frequency of the source

V is speed of sound

V_o is velocity of the observer

V_s is velocity of the source

Now, When the police is coming to you , you hear a higher frequency and thus, we'll use the positive sign on the numerator and negative sign on denominator.

Thus,

f_o = [(V + V_o)/(V - V_s)]f_s

Plugging in relevant values, we have;

1340 = [(343 + 35)/(343 - V_s)]f_s

1340 = [(378)/(343 - V_s)]f_s - - (eq1)

when the police is passing you , you hear a lesser frequency, and thus, we'll use the negative sign on the numerator and positive sign on denominator. thus;

f_o = [(V - V_o)/(V + V_s)]f_s

Plugging in the relevant values to get;

1300 = [(343 - 35)/(343 + V_s)]f_s

1300 = [(308)/(343 + V_s)]f_s - - eq2

Divide eq2 by eq1 with f_s canceling out to give

1340/1300 = [(378)/(343 - V_s)]/[(308)/(343 + V_s)]

V_s = 30 m/s

i would say

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000