Yes, something like that. As speed increases I expect (I don't know, because the Innovia 300 monorail track in my backyard got damaged while I was re-levelling it) that the acceleration is only the initial acceleration (or acceleration at lower speeds), at higher speeds traction power will be limiting. On top of the kinetic energy consideration, losses typically rise with velocity squared. I don't expect a manufacturer to kit out the mechanicals and electricals of a system such that it can do that design acceleration at speeds approaching its maximum.
I have a route simulation for a 30 second dwell time for multiple vehicles running on 75 second headways that gets to ~40 minutes transit time. Add a minute or two for people to climb up and down stairs, chuck in a lateral or vertical curve or two that stop you running at 80 km·h-1 max, vary the station spacing a bit and you'd get to 50 minutes transit time without forcing it too much.
[In terms of derivation, if you want to come at it from kinetic energy point of view (I think this is effectively reversing the definition of kinetic energy):
E = 1/2 . m . v^2
Ignoring losses (all traction power goes into kinetic energy):
P = dE/dt = d/dt (1/2 . m . v^2)
If the driver has remembered to close the doors, then m is constant with respect to time. v is (hopefully) a function of time though.
P = 1/2 . m . d/t (v^2)
P = 1/2 . m . d/dv (v^2) . dv/dt
= 1/2 . m . 2 . v .dv/dt
dv/dt is acceleration.
P = m . v . a
Simpler approach is to consider the unit increment of work, force by distance (a dot product, but hopefully the monorail doesn't vary too far off its path), noting that distance increment is velocity by time increment and Newton sorted out the force to acceleration thing a while back (ignoring losses - traction force is only used for acceleration along the path).
dW = F . dx = F . v . dt = m . a . v . dt
Power is dW/dt, so divide through by dt and you've got the same outcome.]
And this all proves they should be making the NWRL tunnels suit double deckers.