It’s TV so it’s more dramatic then necessary. I don’t know the drawbar pull of a car but a one ton car should be able to easily move that ship, far less drama if they were willing to wait a bit.
Half a dozen of us once put our backs to a 3000 ton ship and pushed with our legs. We got the ship far enough off the pier to adjust a fender.
Fixed that for you… 
Reminds me of this:
I hate how they stacked the odds in favor of the truck by using a dock that was taller than the tow line attachment point. In the opposite case, it might have gone a bit more like this:
EDIT: Skip to 4:30 for less mindless banter and more boat > car.
The icebreaker in question is Kapitan Dranitsyn, a 22,000 shp diesel-electric shallow-draught polar icebreaker built in 1980.
I shouldn’t have said “easily” either, instead should be “eventually move that ship”.
Would have been pretty easy if he’d just taken a moderate strain and set the brakes while the catenary did the work for him. Rinse and repeat. Basically what he did but he overcooked it. Of course that would be like watching paint dry.
Nice little cog railway he’s got there. 
“Going into the video I honestly wasn’t sure what the outcome would be”
Mike! most here have already said they knew the outcome. Surprised you put that in the blurb
As an aside; please Mike, come into the modern world of the Metric system.
“Ton” is old and could be confused with the volume meaning (still lingering with Gross & Net Tonnage!) or a weight of 2240lbs (or is there another value for a USA ton? )
A special word was invented for everyone to cope with this: “tonne” , meaning 1000kg
Either say “13,000 tonne ship” or “12,390.12 Ton ship” (not sure of how to convert back from metric tonnes to US Tons… hah!)
Neil
Ship:
Approx 100 hp = 1 ton bollard pull.
3000 hp is 30 tons = 7 kts at dead slow
3 tons = 1.5 kt
.3 tons = 0.15 kts (be more, not linear.)
Car:
100 hp = between 0.1 and 0.2 tons of drawbar pull.
At 0.05 kts (1/3 of calculated speed at .3 tons) = 0.08 feet /sec so in 10 seconds the ship will move almost a foot, enough to see it moving.
Also, I’ve seen a ship moved when pushed.
My first thoughts. . . …
Assuming to double speed the force has to increase by a factor of eight. That hurts if you want to go fast but helps if you only need to go very slow.
Assume 30 tons pull by tug could tow that ship at 7 kts or so. Redoing I get under the 0.1 tons capacity for the car at:
0.9 kts = 0.06 tons. So top speed with the car will be around 1 kt.
But the issue here is acceleration. With F= ma and F is small and m is very big than a is going to be small.
Convert 0.13 tons of drawbar force to newtons is about 1300 N, 13,000 tons is 13,000,000 kgs then we will get a in m/s^2
That gives 1300N/13,000,000kg = 0.0001 m/s^2,
0.1 kt which should be easy to see (about a foot ever 5 seconds) is about 0.05 m/s
It will take about five minutes ((300 secs) to get up to 0.1 kt assuming no water resistance.
I’m a bit rusty, look about right @dbeierl?
Feels reasonable to my sanity-checking intuition.
Of course all this changes with a breath of wind or current.
Assume any wind or current is random noise and your friend Clarkson is the signal.
He’s a very noisy signal. 
He gets paid to convert cars to noise and smoke.
In the video he pulls for about 2 minutes.
Current is one for one of course, for ship speed while drifting I use by rule of thumb 10 kts of wind speed = 1 kts through the water. It works very close.
A few zephyrs here and there should be OK
Funny to see mates do math and physics calculations
When the going gets tough the tough get going. Where was Jeremy this day, when his help would have been appreciated?
My tuppence worth.
Distance traveled in one hour ------> d = ½ * a * t^2 = 0,5 * 0.0001 * 3600^2 = 648 m = 0.35 nm. Distance traveled per second is 648/3600 = 0.18 m = 18 cm and in a minute 648/60 = 10.8 m
v = d/t = 648/3600 = 0.18 km/h = 0.097 knots
That is true with zero water resistance. In reality the water resistance at the start is zero because v is zero after that it will build up with v squared.
The drag force Fd = ρ * v^2 * Cd * A/2 where Cd is the drag force coefficient and A the area of the ship’s hull that is perpendicular to the movement of the ship.
Fd is much larger if A represents the broadside area of the hull then if the ship is moving in a forward direction. Also Cd is then much larger as compared to the streamlined bow.
