Two cars A and B are moving with a velocity of $30\mathrm{kmph}$ in the same direction. They are separated by $10 \mathrm{km}$. The speed of another car C moving in the opposite direction, if it meets these two cars at an interval of eight minutes is

Two cars A and B are moving at the same speed therefore, the relative velocity of both cars with respect to the other is zero.

Now, a car C is moving in the opposite direction let's say with velocity $v_C$then, the relative velocity of car C with respect to A and B is $v_{\mathrm{CB}}=v_{\mathrm{CA}}=v_{\mathrm{C}}-v_{\mathrm{A}}$

Given that distance of separation between cars A and B is $10 \mathrm{km}$and speed of cars is $30 \mathrm{kmph}$

Also, given that car C crossed B and A in $8 \min \text{or} \frac{8}{60} \mathrm{hrs}$then, it means that car C travels a distance of $10 \mathrm{km}$in $\frac{2}{15} \mathrm{hrs}$with relative velocity $v_{\mathrm{CA}}$

$$\begin{gathered} \text{Distance = speed ×time} \\ \Rightarrow 10 \mathrm{km}=v_{\mathrm{CA}}\times \frac{2}{15} \mathrm{hrs} \\ \Rightarrow v_{\mathrm{CA}}=75 \mathrm{kmph} \\ \Rightarrow v_{\mathrm{C}}-v_{\mathrm{A}}=75 \mathrm{kmph} \\ \Rightarrow v_{\mathrm{C}}=75+v_{\mathrm{A}} \\ \Rightarrow v_{\mathrm{C}}=75+\left(-30\right)=45 \mathrm{kmph} \end{gathered}$$