How do I incorporate the initial velocity into the equation?

[hydr]

Ok here's my problem:

The acceleration of a car is proportional to the difference between 250 km/h and the velocity of the car. If this machine can accelerate from rest to 100 km/h in 10s, how long will it take for the car to accelerate from rest to 200 km/h?

Here is what I've done so far:

dv/dt = k(250-v)
integrating that...
\intdv/(250-v) = \intkdt which equates to
-ln|250-v| = kt + c solving for v gives me
v = e^c * e^(-kt) - 250

Now, i proceeded to solve for k, making e^c = 1 since i assumed c = vnaught and vnaught = 0

When i solved for t i ended up getting like 7.2 seconds which isn't right if it takes 10 seconds to accelerate to 100 km/h. My main question is: how do i equate vnaught into the equation? Because obviously i was mistaken to think vnaught = c.

I have also tried to make e^c = B, but once again am stuck as to how to equate B to vnaught. Any suggestions would be helpful, my book doesn't really explain much.

 

[Defennder]

Now, i proceeded to solve for k, making e^c = 1 since i assumed c = vnaught and vnaught = 0

v₀ is the initial velocity = 0. That's where t=0. That doesn't make e^c = 1 in any way.