Differential Equations Question

A girl lives 500m from school. She starts walking at 2m/s but by the time she has walked x metres, he speed has dropped to (2-(x/400))m/s how long does she take to get to school?

v (speed) = 2 - x/400
Speed can also be written as v = dx/dt

dx/dt = 2 - x/400
Seperating dx and dt we get,
dx / (2 - x/400) = dt

Integrate both sides with limits
x 0---->500
t 0----->t (required time)

On integration we get,

(-400) ln (2 - x/400) = t

On applying limits,

(-400) [ ln 3/4 - ln 2 ] = t - 0

t = 400 ln (8/3) = 392.3 sec

V = dx/dt

dx/dt = [2-x/400]

dx/dt = (1/400)(800-x)

400 dx = (800-x) dt

dt = 400/(800-x) dx

t = int [400/(800-x)] dx

t = - 400 ln(800-x) + C

at t = 0 , x = 0

0 = -400 ln(800-0) + C

C = 400 ln(800)

particular solution:
t = - 400 ln(800-x) + 400 ln(800)

t = 400 ln(800) - 400 ln(800-x)

at x = 500

t = 400 [ ln(800) - ln(800-500) ]

t = 400 [ ln(800) - ln(300) ]

t = 400 ln (800/300)

t = 400 ln(8/3) sec.