Can the Bus Cross the Gap?
Date: 9/15/96 at 1:43:11 From: Anonymous Subject: Can the Bus Cross the Gap? Our question is based on the movie Speed. Can a bus cross a 50 ft. gap in the highway? THE QUESTIONS 1) Would it be possible to have the bus take off and land, 50 feet later, having crossed an actual gap and land at the same height? 2) How low would the second side of the freeway have to be if no ramp were involved? 3) How much variation is possible in the calculations (i.e. how unsafe would the stunt be) We have been given the distance formula for any falling object: s(t) = -16t^2 + Vot + So (in feet) [Vo = initial vertical velocity; So = initial vertical position] Question 1 We've been doing derivatives lately, but I don't see how this is related to derivatives. But what we have decided is that we need to use vectors. We have decided that the bus' path is a parabolic curve. However, we're really confused as to what exactly Vo is. In our book, we found another equation: r(t) = (Vo cosX)ti + [h + (Vo sinX)t -.5gt^2]j where the initial height is h, initial speed is Vo, angle of elevation is X, and g is the gravitational constant. To start with, we're unsure how these two equations are related, if at all. Also, for the first equation, we're unsure of how to calculate Vo (we don't think it will be just the speed of the bus, but some sin function or something trigonometric) and don't even know what So represents (we think it's the height of the ramp, but are not sure). The optimal range though is that Vo must be >56.6 for the solution of the first equation to be real....but we don't know what that represents. We know that we're limited by the length angle of the ramp, the bus's speed, and the distance to be crossed. But we are lost now.
Date: 9/16/96 at 21:12:2 From: Doctor Ana Subject: Re: Can the Bus Cross the Gap? This is a good example of a problem where using a "canned" formula can get you in trouble. When an object like a bus, bullet, or cannonball is soaring through the air a la Speed, it will follow an approximately parabolic orbit based on the conditions with which it left the ground. What makes these problems easy to do is that the horizontal and vertical components of motion are UNRELATED. How quickly something is moving forward (unless you're worried about air resistance) has nothing to do with how high it will go. Let's think about how the bus is going to travel vertically first. If the bus just drives off the end of a freeway which is level, it will immediately begin to plummet to the earth because of gravity. In order for this bus to make the 50 ft. jump, it must either land at a lower altitude, or it must be going up when it leaves the pavement. Here are some possible scenarios: (1) The bus is level when it drives off the edge of the freeway. In this case, all of the bus's velocity is horizontal. You must calculate how long it will take the bus to travel the 50 ft, and then calculate how much the bus will drop in this time, and then you can say something like, "If the bus drives off a level freeway doing 80 mph, if will cross the 50ft in x seconds. In that time it will have fallen y feet, so it will only land if in lands on a stretch of freeway that is y (+margin of error) feet below the freeway it left from. Note: in your formula, this corresponds to having v0=0 and s0=0 because there is no initial velocity in the vertical direction. You should be able to derive that formula yourself using integrals, by the way. (2) The bus drives off a ramp that is going up. In this case, the bus's velocity when it leaves the ramp can be divided into 2 components. This is a case where it may be easiest to think of vectors. The bus's horizontal component can be used to tell you how long it will take the bus to travel the 50 horizontal feet it must travel. The bus's vertical component can be used in the formula to tell you, after the bus travels that 50 feet in x seconds, it will be at altitude y. If you don't know the angle of the ramp, I assume you will have to calculate it. You should be able to make a statement saying, "If the angle of the ramp is theta, the horizontal component of velocity will be sufficient to travel the 50 ft gap in x seconds. In this amount of time, the vertical component of velocity will account for the fact that the bus will be at just the right height to land successfully at the other edge of the highway. Any combination of these 2 scenarios might come up. The strategy will be about the same. Find the horizontal component of velocity and use this to calculate the amount of time that the bus will be in the air. Then plug this amount of time into the s(t) function, which will tell you the bus's altitude or height after the gap has been jumped. Then use that information to decide whether the bus has to land on a lower stretch of highway, whether the bus should be at a different speed when it leaves the highway, or whether the angle of the highway should be different. Remember that air resistance, in reality, will be a problem. As well as the fact that the bus isn't a point mass as we're used to thinking. What if the bus's center of gravity leaves the bus going into the pavement either nose-first or on its rear wheels? There's lots to think about. The best thing to do is to understand where the s(t) function comes from, then you will have a better idea of when and how to use it. -Doctor Ana, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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