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# Speed of a Car: Honda Civic Si Commercial

## Abstract

The purpose of this exercise is to analyze the 2006 Honda Civic Si commercial featuring MIA's Galang as the sound track. View the commercial at Google Video or StefMedia.

## Introduction

Since the beginning of the evolution of cars, people have been trying to push cars to their limits. People try to do crazy stunts like burnout 360s and 180s, and let's not mention the drifting done by street racers. In this commercial the Civic Si climbs the walls of a tunnel and comes out on the other side, then jumps off a ramp and makes several questionable jumps off the tops of buildings. The car then goes into a tube suspended from a construction crane, swings around like a tetherball and comes out the other end, finally landing in a garage.

These stunts require a certain amount of speed. Otherwise the car would fall because of the force of gravity when going upside down in the tunnel and when jumping to the top of the building otherwise it would crash into the buildings. The car also needs to decelerate in the tube in order to stay on due to friction.

There are many equations that require consideration. These include the gravitational force or weight

W = mg

centripetal force

Fc = mv2/r

and friction

μ = F/N.

## Analysis

*Note: All values are estimated.

### Part one: the tunnel!

During this part of the commercial the car climbs the walls of the tunnel and comes around the other side. To analyze this part of the commercial we need to determine the radius of the tunnel, which as we see in this screen shot is about 3 car widths plus the space between each car and the walls. We can estimate this to be about 4 car widths. According to Honda, the 2006 Civic Si's width is 68.9 inches, which is 1.75 meters. With this knowledge we can approximate the diameter of the tunnel to be about 7 meters, making the radius 3.5 meters. In order for the car to make it across the top of the tunnel the centripetal force has to be equal to the force of gravity for the car to just make it across the top of the tunnel. Therefore:

mv2/r = mg

The mass cancels and we are left with:

v2/r = g

v = √rg

v = √(3.5m)(9.8m/s2)

v = 5.86 m/s

What this means is the car has to be traveling at 5.86 meters per second (13.11 mph) laterally in order to complete the stunt. This is improbable because in order for the car to attain that speed it would have to accelerate laterally at a rate of 5 m/s2, which is very hard to attain in the one second or so the driver had to turn the wheel.

### Part two: the jumps!

There is no way that the jumps in this commercial can take place. The car jumps from building to building without a ramp and with out a force accelerating it up to make it jump. Therefore it will would not make it across to the first building and during the jump to the next building it would crash into its side. The car also appeared to jump a very high distance about the height of a building which is about 18 to 20 meters. For a car to jump that distance it would require several seconds. The car jumps without an initial upward speed. Therefore:

s = at2/2

t = √2(s/a)

Approximating that its a 20 meter jump.

t = √2(20 m)/(9.8m/s2)

t = 2.02 seconds

In the commercial the jumps take about 8 frames each. This commercial is rated at 15 frames per second. Therefore the jumps took just a little bit more than half a second to take place, which we can certainly say is highly exaggerated.

### Part 3: The Tube of return!

In this part of the commercial the car goes into a tube and undergoes a 270 degree turn and enters into a garage. In order for the car to whip around inside the tube it has to brake inside to tube to create friction to make it stick inside, otherwise it would just drive out the other end. It takes 15 frames for the car to travel from the last building to the tube. The car is moves at a rate of its length every frame. According to Honda, the length of the Civic Si is 147.8 inches which is 3.75 meters. Assuming that the car travels its length every frame, the car upon entrance to the tube is traveling at 56.25 meters per second (125.83 mph):

v = x/t

v = 56.25 m/1 s

v = 56.35 m/s

It takes one frame for the car to stop in the tube. During this frame the tube starts moving as well. This means that's the the car stopped inside the tube within a fifteenth of a second. This is not possible because the car needs at least 94.96 meters to stop:

Fs = mv2/2

μmgs = mv2/2

s = mv2/2μmg

s = (56.25 m/s)2/2(1.7)(9.8)

s = 94.96 m

A car cannot stop 94.96 meters in less than one second. Therefore there this part of the commercial could not have happened as well.

### Part 4: Home Sweet Home

In this part of the commercial the car comes to rest inside a garage right after coming out of the tube. It takes 41 frames after the car comes out of the tube to reach the ground. At a downward acceleration of 9.8 m/s:

v2 = vo2 + 2as

We determined the height to be 20 meters from past scenes. There is no initial velocity downward so we are left with:

v2 = 2as

v = √(9.8 m/s)(20 m)

v = 19.80 m/s

The car was traveling at 19.80 meters per second (44.29 mph) toward the ground, and assuming that the speed was the same as the entrance of the tube, it would be 56.25 meters per second (125.83mph). The true speed of the car itself before impacting the earth is:

vx2 + vy2 = vf2

vf = √(56.25 m/s)2 + (19.80 m/s)2

vf = 59.63 m/s

The final speed of the car as it impacts the ground is 59.63 meters per second (133.39 miles per hour). An impact with the earth at 59.63 meters per second would definitely demolish the vehicle, or at least deploy the airbags. After this collision the vehicle would be rendered undrivable.

## Conclusion

This commercial although extremely fun to watch is very false when it comes to actually replicating the stunts. Honda has produced this commercial in hopes to bring more young and daring drivers to their market. Will it work? Sure why not but for all of you who were wondering this commercial is completely improbably.

Bishoy Emmanuel -- 2006

Physics on Film
1. Commercials
2. Feature Films