Speed - the bus jump in the movie

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SUMMARY

The discussion focuses on the physics of the bus jump scene in the movie "Speed". Participants analyze the projectile motion involved in the jump, calculating the necessary velocity and trajectory to successfully clear the gap in the bridge. Key equations utilized include delta y = VoyT + 1/2aT^2 and delta x = VoxT + 1/2at^2. The conclusion drawn is that the bus would not make the jump without crashing, as the calculations indicate a negative delta y, suggesting a downward trajectory.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic knowledge of physics concepts such as velocity and acceleration
  • Ability to perform calculations involving time, distance, and speed
NEXT STEPS
  • Study the effects of air resistance on projectile motion
  • Learn about the physics of inclined planes and their impact on projectile trajectories
  • Explore advanced kinematic equations for varying acceleration
  • Investigate real-world applications of projectile motion in engineering
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Students studying physics, educators teaching projectile motion, and film enthusiasts analyzing the realism of action sequences in movies.

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Speed --- the bus jump in the movie!

Homework Statement



In the movie Speed1, a loaded city bus is rigged with an explosive that will detonate and kill everyone aboard if its speed goes below 50 miles per hour. As the bus is moving, the people on board are informed that a bridge ahead in the highway they are on is still under construction and is missing a center segment. They can’t slow down so decide to jump it, so they begin to accelerate.



You must watch a film clip below from the movie to get a good overview of the problem at hand. Some information you need is given in the clip also (observe and listen).

It is then up to you to determine, given the information you hear and observe, if their plan will work.



Q1. Just before the jump, using the info you have and projectile motion (i.e. no air friction) with simple calculations, where would bus be once it reaches other side of unfinished section of bridge (also include a trajectory diagram)?



Q2. What’s about the jump scene is not consistent with the previous scenes?



Q3. Given the info you have, what would be the minimum velocity (magnitude and direction) required to make that jump (include another trajectory diagram and clearly show your calculations)?



Q4. What about the bus might lead you to alter your answers to Q3 a bit for a safe landing?


------


Homework Equations


Vy = Voy + aT ... Vx = Vox + aT
delta y = VoyT + 1/2aT^2 ... Delta x = VoxT + 1/2at^2
delta y = 1/2(voy + Vy)T ...Deta x - 1/2 (vox + Vx)T
Vy^2 = Voy^2 + 2a(delta) ...Vx^2 = vox^2 + 2a(deltax)


The Attempt at a Solution


delta x = 50 ft = 15.24 m
To find T I used V = (delta x)/(delta t)
29.78 m/s = (15.24 m)/t
t=0.511 s
I then used the formula:
delta y = VoyT + 1/2aT^2
I knew that Voy = 0, a = -9.8 m/s^2
so I got -1.279 m = delta y.
So from the negative I know that the bus fell DOWN. so there is no way it made the jump without crashing. If it was > 0 it would have made it just fine.

I'm stumped for question 3...
I have all my components 'figured' out:
Y axis:
Voy = 0 m/s
Ay = -9.8 m/s^2
delta y = -1.279m
T = 0.511s

x axis:
Vox = Vx = 29.78 m/s
Ax = 0
delta x = 15.24 m
T = 0.511 m

I really need help with this. thanks
 
Last edited by a moderator:
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There should be an upslope, so the initial velocity in the y direction is not 0. No value for the angle of the slope is given, but you can work out the minimum value from the speed of the bus and the size of the gap,
 


How?
Do you mean I calculate the min velocity needed using the speed and distance of gap?
Or do you mean the angle?
I can't figure out how to solve the angle to get initial V in the y direction, because then I wouldn't know delta y and therefore would only know one side of the triangle --> delta x?
 

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