Answer
(a) $0.33KHz$
(b) (i) bicyclist A speeds up.
Work Step by Step
(a) We know that
$f_1^{\prime}=(\frac{1+u_{\circ}/v}{1-u_s/v})f$
We plug in the known values to obtain:
$f_1^{\prime}=(\frac{1+8.50/343}{1-8.50/343})(315)=0.33KHz$
(b) We know that for equal changes in speeds, the Doppler shift due to a moving source is greater than that due to a moving observer. Thus, the greater increase in frequency heard by bicyclist B is (i) bicyclist A speeds up by 1.5 m/s.
