Using Eq. (3.49) where \rm M_1=\frac{v_1}{c}= \frac{v_1}{\sqrt{kRT_1}} :=2.06,

\rm M_2=\left\{\frac{M_1^2+2/(k-1)}{[2k/(k-1)]M_1^2-1} \right\} ^{1/2}                           (3.49)

\mathrm{M}_{2}=\left\lgroup{\frac{\mathrm{M}_{1}^{2}+5}{7\mathrm{M}_{1}^{2}-1}}\right\rgroup^{1/2}:=0.567

To obtain \rm v_2, i.e. \rm v_2=M_2c_2=M_2\sqrt{kRT_2}, we need \rm T_2. Employing Eq. (3.46)

\rm T_1\left\lgroup1+\frac{k-1}{2}M_1^2 \right\rgroup =T_2\left\lgroup1+\frac{k-1}{2}M_2^2 \right\rgroup                      (3.46)

\rm\mathrm{T}_{2}={\frac{\mathrm{T}_{2}\left({{ M}_{1}}^{2}+5\right)(7{ M}_{1}^{2}-1)}{36{{ M}_{1}}^{2}}}:=500\,{ K}

We now can calculate

\rm v_2 = 254 m/s  and  Δv = 254 − 700 = −446 m/s

Comments: A downstream flow with \rm v_2 = 254 m/s is observed from the stationary shock wave where the upstream airflow approaches at \rm v_1 = 700m/s. The induced velocity of 446 m/s is in the same direction as the propagating shock wave (see Sketch).