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Homework Help: 2 Fast 2 Furious Scene

  1. May 27, 2004 #1
    I have a special effects physics project that I'm doing and I need some help. In this movie, there's a scene where a part of the road is elevated to become a ramp. Two cars who are racing approach this ramp. Car 1 is in the lead and Car 2 is behind him. Car 1 goes off the ramp and Car 2 accelerates and goes of the ramp second. While car 2 is in the air it apparently had a higher velocity when it went off the ramp so it goes higher and faster than Car 1 and goes over Car 1 in midair. Car 2 lands first ahead of Car 1, Car 1 lands second. What i need to do is find any kind of physics going on here, solve for it and prove it. I need some help to find something in physics to solve for. It could be force, work and energy, momentum and collision, or projectile motion. One thing my teacher has suggested is to look up the engine specifics and see if it would be possible for that car 2 to accelerate to that velocity in that amount of time. How would I do this? Thank you very much for any help you can offer.
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  3. May 27, 2004 #2


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    At reasonable speeds something like 35mph, cars go appreciably faster on the ground than in the air. In order to pass a car while jumping, the back car would have to be going *much* faster than the front car when it goes off the jump. Unless there is some premium for a long jump such as avoiding an obstacle, there is an optimal speed, and going off the jump faster than it looses time. Assuming that both cars are racing full out, and they have coparable speeds, jumping higher isn't going to be any appreciable advantage.

    If the cars have wings (that provide downforce), then it's possible (if insane) for the bottom car to pass the top car, especially if there is a steep downhill following the jump. It's also possible for the bottom car to get an advantage by jumping over the second in a pair of double jumps. In general, barring obstacles the car that takes the *shorter* jump is at an advantage.

    The first thing to look into is projectile motion, and ignore the air resistance, and see how much faster the back car has to be going in order to perform the 'air pass'. Adding air resistance makes the air pass more difficult.
  4. May 27, 2004 #3


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    I am having trouble with this stunt. Was it shown in a single long range shot so both cars were in view through the entire stunt or were there cuts between cars and various phases of the stunt?

    Why? Given the same ramp each car has the same initial angle so the faster a car is going the HIGHER and further it will go. Because the faster car goes higher it will spent MORE time in the air. It is physically impossible for the second, faster, car to hit the ground before the leading slower car. The slower lead car is in the air first and spends less time in the air, it MUST land first. I am making no statement about the comparative speeds or the distance traveled, this is only looking at the TIME in the air which is a result of simple free fall physics.

    The wonders of movies defy the laws of Physics, this is very common.
  5. May 28, 2004 #4


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    Just elaborating on Integral's comments:

    1.If the initial angle is the same, and the ground height is the same for both cars (i.e, they are landing on a strictly horizontal road), this stunt is impossible by the laws of free-fall physics (i.e, the first car must land first)

    2. Conceivably, you might be able to pull off the trick by either:
    a) Reducing the angle/height of the ramp in the interval between the 1. and 2. car,
    b) Letting the ground be sufficiently "curved" such that the impact point of the second car is higher than the impact point of the first car.

    3. However, I am confident that both of these "tricks", or "real special effects" would be noticeable, that is, the viewer wouldn't be fooled into believing they saw two cars leaving at the same height/angle and landing on a horizontal road (with the second landing first).

    You'll need other forms of trickery to pull off this stunt..
  6. May 28, 2004 #5


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    Ah, everyone's getting too hung up on who lands first. Of course the slower car will land first, but that's a good thing.

    The key is your position equation

    [tex]s_f = s_i + v_i*t + \frac{1}{2}a*t^2[/tex]

    You have two components two each car's motion. A vertical component and a horizontal component. You have to calculate each separately. The vertical speed is the car's speed (preferably converted to meters/sec or feet/sec) times the sine of the angle of the ramp. The horizontal speed is the car's speed times the cosine of the angle of the ramp.

    Once the car leaves the ramp, there can only be acceleration in the vertical component. The initial velocity ([tex]v_i[/tex] is up (positive) and the acceleration (a) is down (negative). The initial position ([tex]s_i[/tex] is the height of the end of the ramp. Your final vertical position should be 0. You solve your position equation just like a quadratic equation (in fact, it is a quadratic equation).

    The horizontal component is the car's horizontal speed times however long the car spends in the air (the end of the ramp is zero and you probably don't want to get into computing the negative acceleration of wind resistance).

    The faster car has to land after the first car. Going faster, it has a higher vertical speed so it will spend more time in the air. Plus, it left the ground after the first car. That's actually a good thing. That way, the guys in the slower car have time to look in their rear view mirror in befuddlement wondering where the chase car went only to have it suddenly appear on the road in front of them.
  7. May 29, 2004 #6
    Thanks for the suggestions! If anyone has anymore please let me know.
  8. May 30, 2004 #7
    Position Equasion:
    [tex]s_f = s_i + v_it + \frac{1}{2}at^2[/tex]

    There is a solution in the parabolic trajectories formula:

    [tex]R = \frac{v_0^2}{g} \sin 2 \theta_i[/tex]

    Assuming that Range is [tex]R = (s_f - s_i)[/tex] then:

    [tex]R = v_it + \frac{1}{2}at^2 = \frac{v_i^2}{g} \sin 2 \theta_i[/tex]


    [tex]R = \frac{1}{2}(v_i + v)t = \frac{v_i^2}{g} \sin 2 \theta_i[/tex]

    [tex]\theta_1 = \theta_2[/tex] ???
    [tex]v_1 < v_2[/tex]
    [tex]R_1 < R_2[/tex]

    However it is interesting to note what happens when [tex]R_1 = R_2[/tex]

    However note that the 'stunt' is probably a 'computer generated effect' than actual physics.
  9. May 31, 2004 #8
    Do this via energy calculation. Compute the gain in kinetic and potential energy the car achieves at the ramp. You know more or less the initial and final velocity of the car by looking at its trajectory, its mass is maybe about 1000 kg, then do a rough estimate on the height of the ramp. Now divide this energy by the time the acceleration takes to get the power of the engine. If the value you get is above 350 kW (500 hp) you've won.
  10. May 31, 2004 #9
    The question is, if the two cars are initially racing neck to neck, can car 2 accelerate fast enough after car 1 jumps but before hitting the ramp? If they are initially going at the same speed, and car 1 is only a few feet in front, it might not be possible. The inital setup of the cars might be that car 1 is very far ahead, but driving much slowly than car 2, so when they actually hit the ramp (and start fliming) the distance between them is small but car is much faster than car 1 already.
  11. Jun 6, 2004 #10
    Interesting teacher!

    I have no time to provide physics forumlas here, but there is something in real life road racing called 'drafting'. This occurs when the first car moves at a fast enough velocity to create a vacuum behind it. This reduces the air friction on the second car, and thereby makes the second accelerate significantly faster than the first car.

    While on the bridge, if the second car has enough horsepower (which he did, using his NOS) he can accelerate his car very quickly within this vacuum, and pass on top of the first car.
    Last edited: Jun 6, 2004
  12. Jun 6, 2004 #11
    The second car accelerated in car1's draft EXACTLY when the first car left the bridge, thereby increasing its initial velocity to be faster than the first car.

    All this happens within the length of the first car.

    The downforce on the second car makes it land faster than the first car.

    THis stunt is possible, though extremely unlikely to be executed properly.
    Last edited: Jun 6, 2004
  13. Jun 7, 2004 #12
    Just wanted to say thanks for all the help. Invinc1ble, Car 2 began accelerating long before the 1st car went off the bridge. Does this still make it possible? Just curious, I'm done with the project.
  14. Jun 7, 2004 #13


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    If you need 'research' material for your next stunt, check out this site:

    http://www.wagoneers.com/pages/RocketCar/rockit.html [Broken]

    Last edited by a moderator: May 1, 2017
  15. Jun 8, 2004 #14

    This stunt is not possible, Integral is absolutely right.
  16. Jun 8, 2004 #15
    Ok, I did my project and we got a C on it. My teacher gave me an option of either keeping this grade or redoing it and fixing our solution on horsepower. We calculated the horsepower, however he said we solved it for 0-160. I believe what we need to do is calculate it from 0-100 and then from 100-160 (mph of course) and compare the two. Can someone give me some equations that will work for this, thanks.
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