Suppose that a police car is moving to the right at 27.5 m/s, while a speeder is

Question

Suppose that a police car is moving to the right at 27.5 m/s, while a speeder is coming up from behind at a speed of 37.9 m/s, both speeds being with respect to the ground. Assume that the electromagnetic wave emitted by the radar gun has a frequency of 7.87×109 Hz.

a.)Calculate the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car.

Now suppose a speeder is pulling directly away and increasing his distance from a police car that is moving at 26.2 m/s with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of 6.87×109 Hz. The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be 312 Hz less than the emitted frequency.

b.)Calculate the speeder's speed with respect to the ground.

Explanation / Answer

doppler's effect:

frequency observed by the observer=(c+vr)*f0/(c+vs)

where c=speed of wave

vr=speed of the receiver, positive if receiver is moving towards the source

vs=speed of the source, negative if source is moving towards the receiver

f0=frequency of the emitted wave

part a:

frequency received by the speeder's car:

f0=7.87*10^9 Hz

c=3*10^8 m/s

vr=37.9 m/s

vs=27.5 m/s

frequency received by the speeder=f1=(3*10^8+37.9)*7.87*10^9/(3*10^8+27.5)

=7870000272.82664 Hz

this frequency is reflected from the speeder's car.

in that case, vr=-27.5 m/s

vs=-37.9 m/s

then frequency received by the police car=(3*10^8-27.5)*7870000272.82664/(3*10^8-37.9)

=7870000545.65335 Hz

difference in frequency=7870000545.65335-7870000272.82664=272.8267 Hz

part b:

let speeder's speed be v.

for frequency received by the speer's car:

vr=-v m/s

vs=-26.2 m/s

then frequency received by the speeder's car=f1=(3*10^8-v)*6.87*10^9/(3*10^8-26.2)

for the wave relfected from the speeder's car:

vr=26.2 m/s

vs=v m/s

then frequency received by the police car

=(3*10^8+26.2)*f1/(3*10^8+v)

=6870001199.9601*(3*10^8-v)/(3*10^8+v)

given this frequency=6.87*10^9-312

==>6870001199.9601*(3*10^8-v)/(3*10^8+v) =6869999688

==>(3*10^8-v)/(3*10^8+v)=0.99999977991851

==>3*10^8-v=299999933.975553+0.99999977991851*v

==>v=(3*10^8-299999933.975553)/(1+0.99999977991851)=33.012 m/s

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