Force from engine = 1.4*10^5 - 3.9 t
The cars do not move apart, so can be considered one rigid body twice as massive. m = 1.6 * 10^4 kg
Slope = 24 deg so component of weight down track = m g sin 24 = 1.6*10^4 * 9.8 * .407
= 6.38 *10^4 N down track
F = m a
F = 1.4*10^5 - 3.9 t -6.38*10^4 = (1.6*10^4) a
1.6 * 10^4 a = 7.62 * 10^4 -3.9 t
a = d^2x/dt^2 = 4.76 - 2.44*10^-4 t
now if it were constant acceleration, I would assume that v = Vo + a t
so I will try something like that to solve this differential equation
v = Vo + B t + C t^2
then a = dv/dt = B + 2C t
well that is kind of encouraging
I will use B = 4.76 and 2C = -2.44 *10^-4
v = 2.9 + 4.76 t - 1.22 *10^-4 t^2
the train stops when v = 0
0 = 2.9 + 4.76 t - 1.22 *10^-4 t^2
solve that quadratic equation for t
Check my arithmetic (and your given decrease of 3.9 N/second) because I get about 3.9*10^4 seconds which is around 10 hours.
I think you have a typo, throttle should be backed off like 3.9*10^4 N/s or something. It takes forever to back off from 1.4*10^5 N at only 3.9 N/s
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