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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?

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 

(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
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Associated Topics:
High School Physics/Chemistry

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