 Solution

In an ideal continuous-price English auction, staying active until one's private valuation is optimal: dropping earlier can forgo positive surplus, while staying later risks paying more than the car is worth. The highest-valued buyer therefore wins, paying the price at which the other buyer drops out, namely $\min (V_1,V_2)$.

For $0\le t\le 1$, independence gives $\mathrm{Pr}(\min (V_1,V_2)>t)=(1-t{)}^2$. Integrating this tail probability gives“Pays his bid” in this ascending auction does not mean that the winning buyer pays his full valuation; the competing buyer's dropout determines the winning price.