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Speed of a Car: Road Trip

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The purpose of this exercise is to determine the speed of the car jump in the movie Road Trip.


The scene analyzed was the one where the four friends take a shortcut but they are stopped due to a broken bridge in the way. They know if they turn back to the Interstate, that would take more time they don't have. E.L (Seann William Scott) and Rubin (Paulo Constanzo) say they could make it across and Rubin gives the idea that the incline of the bridge is about 30 degrees and the weight of the car and a speed up to 50 miles per hour would get them ten feet off the ground. E.L. and Kyle disagree and E.L. shows the bridge is safe on the other side and hocks a spit causing the other side of the bridge to collapse. Rubin then pauses and says they should go 75 miles per hour. They then start their engines and they begin to pick up speed very fast. They make the jump and heavily damage their car crash landing miraculously with no injuries at all. Then they all laugh and the tires break off -- and then, of course, the car explodes. Let's see if they were actually right about their calculations.(Go to and watch the clip.)

To find the velocity the car launches from the bridge, you need to know the distance of the jump and the angle of the inclined bridge. The launch velocity can also be calculated if the maximum height and the distance of the jump with the given angle of inclination. The formula for the horizontal distance and maximum height are …

x = (v02)(sin2θ)/g
h = v02/2g


x is the final displacement,
v0 is the initial velocity,
θ is the angle of inclination,
h is the maximum height of the jump, and
g is the acceleration due to gravity.

Screen Captures


To get the speed of the car, I used a computer with VLC Media Player, a stopwatch, an on-screen protractor and ruler, and a calculator.


To get the speed of the Ford Taurus as it leaves the bridge, I had to get the time where the car is airborne. I reviewed the portion of the film a couple of times to try and figure out where the car leaves the ground and what time it stops after it lands. I found the that it took 5.057 seconds for the car to jump the bridge but I had there was a small section between the 5.057 seconds where the time is exaggerated. So in order to get the actual time, I recorded the time for that split second scene which was 1.181 seconds and subtracted from 5.057 seconds and got 3.876 seconds for the car to be airborne. I then divided the time by 2 to get to the maximum height of the jump.

To get the angle of inclination, I used the on-screen protractor to measure the incline of the bridge in the screen capture and I got 14°.

To get the distance of the jump, I used an on-screen ruler. This internet clip is actually a program that can be configured to the sides of the screen and take measurements of different units. I used the first screen capture again to measure the distance of the jump and also reviewed the clip. I first looked at the clip to take a look at the front of the car to see what model looks like. I looked at to see what model the car was and found out it was a 1991 Ford Taurus and its length was 188.4 inches. So what I did was I looked at the screen capture and took the ruler and measured the car to a certain measurement. Then I took the ruler to the measure from one side of the bridge to the other and I took a proportion and got the distance to be 10.7 m.

The way to get the distance of the jump was a simple proportion.

I took a similar photo of the first screen capture and measured the lengths of the car and the bridge with a standard ruler.

Length of car = 1.125 inches in photo
Length of bridge = 2.5 inches in photo

Length of actual car = 188.4 inches
Length of actual bridge = ?

1.125/188.4 = 2.5/x
x = 34.88 feet or 10.7 m

After reviewing the clip again, the car crashed and move at least 1 meter away so I estimated to 11.7 m.

To get the take off speed, I rearranged to formula for the horizontal distance traveled and substituted in the appropriate values.

v0 = (xg/sin 2θ)½
v0 = ((11.7 m) (9.8 m/s2)/(sin(2 x 14°))½
v0 = 16 m/s or 35 mph

That's the horizontal component of the car's velocity. To ge t the full velocity, use the cosine function.

vf = 16 m/s/cos(14)
vf = 16.5 m/s


I found the velocity of the car to be 16.5 m/s which is 37 mph, which is surprising because it is far short of the 50 mph that Rubin claimed would be needed to jump the gap.

Sources of Error

Arif Hussain -- 2007

Physics on Film
  1. Feature Films
    1. Coefficient of friction for skin: The Incredible Hulk
    2. Compression strength of bone and brick: Sin City
    3. Force of a superhero: Superman Returns
    4. Speed of a car: Road Trip
    5. Speed of a minibike: Jackass Number Two
    6. Speed of a spear: Troy
    7. Speed of a subway: Batman Begins
    8. Speed of superhero: Smallville
  2. Video Clips
    1. Force of a windmill slam dunk: Vince Carter
    2. Force of a windmill slam dunk: Dominique Wilkins
    3. Speed of a retired basketball player
    4. Speed of a cliff diver: Huge Cliff Jump
  3. Video Games
    1. Acceleration due to gravity: Super Mario Brothers
    2. Speed of a football player: Madden NFL 2006