Finding the constant of this retarding force

[Like Tony Stark]

Homework Statement

A block of mass $100 kg$ is moving with velocity $27,7 \frac{m}{s}$, which reduces to $15 \frac{m}{s}$ after a distance of $200 m$. This change in velocity is caused by a force $Fr=-cv^2$ where $c$ is a constant and $v$ the velocity.
Find the value of the constant, the time that it takes to move the distance given and an expression for velocity in function of position.

Relevant Equations

$-Fr=m.a$

$-Fr=m.a$
$-cv^2=m.a$
$-cv^2=m.\frac{dv}{dt}$
$dt=-\frac{m}{cv^2} dv$

After integrating, I get
$t=\frac{m}{c.v}-\frac{m}{c.v_0}$
Then, solving for $v$ we get
$v=\frac{m.v_0}{v_0.t.c+m}$
$\frac{dx}{dt} = \frac{m.v_0}{v_0.t.c+m}$

After integrating that, I get an expression for $x(t)$.
But how can I get the constant and the time? Because they are unknowns and if I try to use $t(v)$ and $x(t)$ I get an equation which I can't solve.

 

[Antarres]

Since you haven't been given the time here, it might be smarter to find the velocity as a function of distance, instead of finding it as a function of time. There is a well-known trick that is frequently used in these kinds of problems:
$$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}$$

 

[Delta2]

If you follow the hint by @Antarres you ll be able to find $x(v)$ which will be much simpler and will allow you to determine the constant c from the data given (200m,27.7m/s,15m/s). Then you can find the time t by the equation you have already found in the OP.