Talk:Speed (movie)
Have to point this out, the formula for final displacement before and after a period of acceleration is D = u*t + 1/2*a*t2, where u is the initial velocity. Now, when the bus approaches the gap it is traveling up a slope. Any object traveling up a slope is going to have a tangential velocity which is comprised of both a horizontal (forward) component and a vertical (upwards) component, the only time when this is not the case is when the object is traveling parallel to the frame of reference, in which case it will not have any vertical velocity. Hence the same inertia that keeps the bus moving forward across the gap is going to be what propels it off the end of one side of the gap, the bus will retain the vertical component of its velocity as it leaves the gap causing it to continue traveling upwards, allowing it to negate the downwards acceleration of gravity.
Because we're cool we can work out the exact angle of the slope required for the bus to clear the gap. In order to do this, consider that if the two sides of the gap are at the same height then the vertical displacement of the bus must be greater than 0, otherwise it would fall below the road line and crash and explode and so on. Now, we'll have to work in metres because I'm too tired to convert to the imperial system, and let's assume that the bus travels at 100km/h, or 27 7/9 m/s (we don't won't won't to be endorsing a movie that promotes the breaking of road rules, and 100km/h is the maximum speed for buses in Australia). Using a right angled triangle with the hypotenuse as the tangential velocity of the bus, the base as the horizontal velocity and the third side as the vertical velocity we can calculate that the vertical velocity of the bus is equal to V*sinA, where V is the tangential velocity (or speed) or the bus and A is the angle of inclination of the slope. Now, let's consider the formula:
D = v*t + 1/2 * a * t2
Subbing in v=V*sinA, t=0.5 (from Ed Poor's measurements), a=-9.8m/s (acceleration due to gravity in metric terms) then we get
D = V*sinA*0.5 - 4.9*0.52
We estimated that V = 27 2/9 m/s so we'll sub this in, and we also know that D must be greater than or equal to 0 (we'll let it equal 0 to find the absolute minimum angle of inclination).
D = (27 7/9) * sinA * 0.5 - 1.225 = 0
Therefore
125/9 * sinA - 1.225 = 0 125/9 * sinA = 1.225 sinA = 411/5000 A = 5.06 degrees
Now, provided I haven't made an error in my calculations (and please correct me if I have) then this angle seems more than reasonable to have on a sloped freeway, and seems accurately portrayed. Now, even if we double the time taken for our bus to cross the gap our answer is still practical:
D = v*t + 1/2 * a * t2 D = 250/9*sinA - 4.9 = 0 (assuming t=1) 250/9 * sinA = 4.9 sinA = 0.1764 A = 10.16 degrees
Admittedly this may seem a bit extreme, however consider that some roads have sharper gradients, and given its representation in the movie I am inclined (haha, get the pun) to say that this is very possible. Remember, that even if the bus took 1 second to cross the gap it would still cover 27 7/9 * cos(10.16) = 27.34 metres, which by memory appears to be an accurate portrayal of the gap.
Now, we have to consider that the bus was more than likely traveling faster than 100km/h, therefore its vertical and horizontal velocities are going to be much larger, therefore it will not only be able to cover a larger gap, but also require less of an angle to stay in the crucial zone of positive vertical displacement. If I get time I'll do a calculation for 150km/h and see the different results that are possible from that.
So in short, while there are many examples of the liberal media distorting scientific fact in order to sensationalise reality and pass it off as entertainment, this is more than likely not one of them. TheGuy 07:15, 10 January 2008 (EST)
- I don't remember that scene, but you don't need to consider time for your calculations. Given the horizontal distance to be overcome, and the possible positive or negative height difference between 'takeoff' and 'landing', it's quite easy to compute the necessary 'takeoff' slope given speed or vice versa (aerodynamic drag is negligible). But I agree with you, it is certainly doable as long as there is some slope - or negative height difference (landing lower than takeoff). --Leopeo 07:43, 10 January 2008 (EST)
- (EDIT)Ok, I didn't read closely enough. You considered time (1 s) and calculated the horizontal gap, I'd take the gap as initial data, but it's the same. --Leopeo 07:46, 10 January 2008 (EST)
- If a student submitted work like yours, TheGuy, I'd grade his paper with an F. You don't even say what you figure the bus's upward velocity was. Did you see the movie? Was it going up a ramp when it went into the air? I thought it went straight forward, but maybe I am remembering it wrong.