($A$) $v_P+v_R=2 v_Q$
($B$) The rate of change in beat frequency is maximum when the car passes through $Q$
($C$) The plot below represents schematically the variation of beat frequency with time
(image)
($D$) The plot below represents schematically the variation of beat frequency with time
(image)
$f _1=118\left(\frac{V+V_0 \cos \theta}{V}\right)$
$f _2=121\left(\frac{V+V_0 \cos \theta}{V}\right)$
$Image$
No. of beats $n=\Delta f=f_2-f_1$
$n=3\left(\frac{V+V_0 \cos \theta}{V}\right)$
$n=3\left(1+\frac{V_0}{V} \cos \theta\right)$
As $\theta \uparrow, \cos \theta \downarrow, n \perp$
Rate of change of beat frequency
$\frac{d n}{d \theta}=3\left[\frac{V_0}{V}(-\sin \theta)\right]$
$\frac{d n}{d \theta}$ is maximum when $\sin \theta=1 ; \theta=90^{\circ}$ $i.e.$ car is at point $Q$ .
$v_p=3\left(1+\frac{v_0}{V} \cos \theta\right)$
$v_R=3\left(1-\frac{v_0}{V} \cos \theta\right)$
At $Q$
No. of beats $v_Q=121-118=3$
$v_Q=\frac{v_P+v_R}{2}$
