Verified answer

First, convert 85.1 km/h to m/s units. Turns out to be 23.64 m/s.

Now, what's the acceleration (which is constant) that the locomotive can exert? F = ma, so a = F/m

= (6.7 x 10^5 N)/(1.4 x 10^7 kg) = 0.047857 m/s²

Now we know the train starts from rest, we know the initial velocity is 0 m/s. We also know the final velocity and the acceleration. We need to know the time.

u = 0 m/s

v = 23.64 m/s

a = 0.047857 m/s²

t = ?

v = u + at. Since u = 0, this reduces to v = at and rearranges to t = v/a = (23.64 m/s)/(0.047857 m/s²) = 494 seconds.

10 minutes. F = m * a F = 7.7 * 10^5 m = a million.9 * 10^7 so we get, a = 0.04 m/s2 Now, a = (v - u)/t (the region a = acceleration, v is extremely very final velocity, u is preliminary velocity, t is time) consequently t = (v - u)/a substituting u = 0, v = 88*one thousand/3600 and a = 0.04 we get t = 600 secs = 10 minutes