Grand Prix Race- Differential Equations

Question

Driver A has boon leading archrival B for a while by a steady 3 miles. Only 2 miles from the finish, driver A ran out of gas and decelerated thereafter at ta rate proportional to the square of his remaining speed. One mile later,driver A's speed was exactly halved.If driver B's speed remained constant,who won the race?

Answer 1

Let $A(t)$ be the distances from the finish of $A$ at time $t$ from when A runs out of gas, and the initialy velocity of $A$ and $B$ be $v$. Then $B$ takes a time of $\frac{5}{v}$ to reach the finish.

$A(0)=0 \ , \ A'(0)=-v$

$\frac{\mathrm{d}^2A}{\mathrm{d}t^2}=k(\frac{\mathrm{d}A}{\mathrm{d}t})^2\\$

Seperation of variables to find $\frac{\mathrm{d}A}{\mathrm{d}t}$ followed by integration gives $A(t)=c_1t+\frac{kt^4}{12}+c_2$.

From the boundary conditions, $c_2=2 \ \ c_1=-v$.

Where $A(t)=1, \ \frac{\mathrm{d}A}{\mathrm{d}t}=\frac{v}{2}$. Hence $v=\frac{2kt^3}{3}, \ t^4=\frac{12}{7k}$ and $k^4=0.534v$.

I cannot answer the question without knowing $v$, but the time taken by $a$ is just the solution to $A(t)=0$.