Terminator 2 Car Chase

Launch worksheet

Launch solution 

The Movie

Terminator 2: Judgement Day is the 1991 sequel to the original 1984 film The Terminator, with the iconic Arnold Schwarzenegger starring as the unstoppable, time travelling terminator cyborg. Car chases are common in all the terminator films – and there’s a multiple great ones in Terminator 2 including between a police van and a police helicopter.

Math Situation 1: Motorbike Jump Onto The Helicopter

Near the end of the movie, the Terminator and Sarah and John Connor flee the Cyberdyne building in a police van. The relentless T1000 liquid terminator jumps out a building on a motorbike and crashes into a police helicopter, chucks the pilot out and then chases after them.

To make the jump it looks like the bike accelerates to about 50 km/hr out of the building window. A reasonable police bike takes about 5 seconds to accelerate from a standing start to 100 km/hr.

How long a run up does the T1000 need to reach the 50 km/hr speed?

First we need to calculate how long the police bike needs to accelerate for:

Acceleration duration = required speed / acceleration capability

Acceleration duration = 50 km/hr / (100 km/hr/5s)

Acceleration duration = 50 km/hr / (20 km/hr/s)

Acceleration duration = 2.5 s

So the bike needs a 2.5 second acceleration period. How far will it travel in this time?

Distance = 0.5 * a * t^2

Distance = 0.5 * 20 km/hr/s * 2.5^2

Distance = 0.5 * 20000 m/ 3600s /s * 2.5^2

Distance = 17.36 m

Given the size of the office floor in the movie that the terminator rides the bike out of, this seems like a plausible distance.

Math Situation 2: Helicopter Under the Bridge

After capturing the helicopter, the T1000 flies after the Connors in their police van. At one stage the T1000 makes a crazy maneuver flying under a low bridge to keep chasing them (and yes they really did do this, it’s not CGI).

A modern police helicopter such as the AS350 B3 has a height of about 3.14 metres. In Terminator 2, the bridge looks to have a clearance height of about 6 metres.

Let’s say the helicopter is flying forward at 80 km/hr and can jitter up or downwards at a rate of 2.5 m/s.

If the pilot flies the helicopter midway between the ground and the bridge, how quickly must they react to up or downwards jitter to stop it hitting either the ground or the bridge?

Margin at top and bottom = 0.5 * (bridge height – helicopter height)

Margin at top and bottom = 0.5 * (6 – 3.14)

Margin at top and bottom = 1.43 m

Now we can calculate how quickly the helicopter can jerk upwards or downwards and hit something:

Collision time = margin / maximum jitter rate

Collision time = 1.43 / 2.5

Collision time = 0.57 s

If the bridge is 50 metres wide, how long will the pilot need to be super vigilant for?

Time flying under bridge = bridge width / horizontal speed

Time flying under bridge = 50 m / 80 km/hr

Time flying under bridge = 50 m / (80 km/hr * 1000 m/km / 3600 s/hr)

Time flying under bridge = 2.25 s

Of interest is that this has happens all the time in real life, with varying levels of clearance.

Real Life Example – Drone Racing

First person view drone racing has exploded as a popular sport, where people fly drones such as quadcopters at high speed using cameras on the drone that return what the drone sees to goggles worn by the racer.

Much of the flying is through forests or around parking garages and abandoned industrial estates. 

Drone pilots need to continuously make many mental calculations regarding how tight a space they can fly their drone through, how quickly their drone can accelerate or decelerate, how quickly it can turn around tight corners, and many others.

With enough training the human brain is still very good at learning how to perform these calculations very quickly as can be seen in these impressive videos.