Speed - the bus jump in the movie

AI Thread Summary
In the movie Speed, a city bus must jump a gap in a bridge while maintaining a speed above 50 miles per hour to avoid detonation. Calculations reveal that the bus would fall short of the jump, landing at a negative height, indicating a crash. The discussion highlights the need for an initial upward velocity to achieve a successful jump, which complicates the calculations due to the unknown angle of the slope. Participants express confusion over determining the minimum velocity required for a safe landing, emphasizing the importance of both speed and angle. The conversation underscores the challenges of applying physics concepts to the film's dramatic scenario.
Phalon88
Messages
2
Reaction score
0
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:
Physics news on Phys.org


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?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top