d=at^2+V(sub i)t+d(sub i)

V(sub i) and d(sub i) are the initial Velocity and position of the train. For the purposes of this discussion, we will set both of these values to 0. so:

d=at^2

a=(V(sub t) - V(sub i))/t or a=V(sub t)/t: V(sub t) is 88 Miles per hour, d is 3 miles and t is what we are trying to calculate. so:

3=(88/t)t^2 or 3=88t

From this we can calculate that the entire trip should take approximately 0.0341 hours or a just 2 minutes.

Plug this value in for t in the acceleration calculation and it comes to just over 2,581 m/h^2

Now for the power requirements: Force= Mass Times Acceleration or F=ma

m=600 Tons or 450,000 Kg (based on built in the 1940s)
a=2,581 m/h^2 or 1,147 m/s^2
F=450,000 kg * 1,147 m/s^2 = 516,266,666 N
W=F*d
d=3 mi or 4800 m
W=516,266,666 N * 4800 m = 2,478,080,000,000 J
P=W/t
P=2,478,080,000,000 J/ just over 2 minutes or 122 Seconds=20,191,762,962 W or approximately 20,191,763 kW

Big Boy could run at 6000 HP or a measly 4,474 kW.

In Conclusion. The first tip is that it only takes 2 minutes to do this acceleration. In the movie, even accounting for temporal compression, the trip takes much longer. The second problem is that the power requirements are not present to push such a missive object that quickly. I was using data from a locomotive that was not a classic western steam locomotive, but the math doesn't lie. We can also assume that Doc's logs can help to accelerate the locomotive, but I am not sure that could account for an increase of 15,00 kWs.