Answer:
14.2 m/s
Step-by-step explanation:
When the source and receiver are getting closer, we can use the following equation:
![f_o=((v+v_o)/(v-v_s))f_s](https://img.qammunity.org/2023/formulas/physics/college/9x2wpep00wlboidfoc8b6du25bp2yalhef.png)
Where fo is the observed frequency, fs is the emitted frequency, vo is the speed of the observed, vs is the speed of the source, and v is the speed of the sound. Solving for vo, we get:
![\begin{gathered} (f_o)/(f_s)=(v+v_o)/(v-v_s) \\ \\ (f_o)/(f_s)(v-v_s)=v+v_o \\ \\ (f_o)/(f_s)(v-v_s)-v=v_o \\ \\ v_o=(f_o)/(f_s)(v-v_s)-v \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/k1qsorro76zinmh95jv2r20oi7nkxz2ki6.png)
Then, replacing v = 340 m/s, vs = 0 m/s, fs = 960 Hz, and fo = 1000 Hz, we get:
![\begin{gathered} v_r=(1000)/(960)(340-0)-340 \\ \\ v_r=14.2\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/db7hgbt7hvs1s95et3naj6xg33n8i4zy2w.png)
Therefore, the speed of the patrol car is 14.2 m/s