Homework Help: Car Ramping

1. Nov 29, 2004

Nonok

Ok, a simple vector problem. I found all the data, but have no idea to set it up. Here is what I am trying to do...

I want to find it Car1 can ramp over the top of Car2, without hitting it. Both cars are ramping off the same ramp, and lets assume air resistance will not be significant.

Car1
v = 72 m/s
Weight = 1666 kgs
Length = 4.6 m

Car2
v = 34 m/s
Weight = 1590 kgs
Length = 4.5 m
Height = 1.27 m

I am going to assume both cars are ramping off the ramp at the same time, and the angle of their ramp is 13 degrees. Also, the 'v' value is when they are leaving the ramp.

2. Nov 29, 2004

Justin Lazear

What is 'ramping'? And what value exactly are you asking for?

--J

3. Nov 29, 2004

Nonok

I have all the information, I just need to draw out the vectors, and see if there is more then a 1.27 clearance over the length of the cars. I just cannot figure out how to draw vectors anymore. The only variable you need to find is how high both the cars are during the ramping.

4. Nov 29, 2004

Justin Lazear

Maybe I'm still confused about what you're asking (hint: I am. And thanks for quoting, but I can read what you've already written just fine. If I'd understood it the first time, I wouldn't have asked about it. It's not like I randomly skipped that part just for kicks.)

Supposing that you're talking about driving two cars off the same ramp and (somehow) at the same time, but with two different velocities, then there's no guarantee that one car will even be directly over the other car, as the horizontal velocities are different.

Are you familiar with 2-d kinematics?

--J

5. Nov 29, 2004

Nonok

I apologize, I thought that would clear it up. Let me refraise the whole question.

Tomarrow I do something our teacher likes to call 'Physics in the Real World.' For my presentation I decided to do a scene from the movie '2 Fast 2 Furious,' where one car ramps over another, both in mid air, off the same ramp. (I said they were going off at the same time for simplicity, but now i see that wont work) I want to figure out if it is actually possible with the given speeds and weights if the car will actually be able to ramp over the other one, or will they just collide in mid air.

Here is all the data I have:

Car1
v = 72 m/s
Weight = 1666 kgs
Length = 4.6 m

Car2

v = 34 m/s
Weight = 1590 kgs
Length = 4.5 m
Height = 1.27 m

The ramp is 13 degrees and the velosity is when they hit the end of the ramp. If we need a time for how long between the two cars ramping, then I suppose 1 second would work fine.

Again, I apologize for being so rude, and thank you for your prompt help.

6. Nov 30, 2004

Justin Lazear

All right. Now we have something we can solve.

You need to model the motion of the two cars as a function of time. This means you need equations for the x position of each car as a function of time and equations for the y position of each car as a function of time.

Once you have that, you have two conditions that must be satisfied:

1) For all times during which one car is above the other, the difference in y positions must be greater than the height of the car on bottom.

2) When the faster car lands (it will land after the slower car, since the slower car takes off earlier and spends less time in the air), the difference in x positions must be greater than the length of the car.

Are you familiar with 2-d kinematics and modelling motion like this? I'm going to guess that you're not, since you've listed masses, which are not important for this problem.

https://www.physicsforums.com/archive/topic/t-27913_2_Fast_2_Furious_Scene.html

--J

7. Nov 30, 2004

Nonok

I do have troubles with 2-d kinematics and modelling motion. In a vector that is a perfect square. You would model the vector to be, velocity as the x value, and g as the y value. Well we are on earth so g will be 9.8. And velocity I have. I have no idea how to bring the angle into the equation.

For the square vector, I would be going horizontal at the velocity value, and vertical as the 'g' value. So it would be...

For car1, t = 1 (after leaving ramp). The coordinates would be, (72 m, -9.8 m). For t = 2, the coordinates would be (144 m, -19.6 m).

But I have a few problems, those coordinates do not factor in the angle of the ramp. Nor the fact that as the car is descending, it is speeding up. I do not know how to do either of these.

Thanks for the help so far.

8. Nov 30, 2004

Justin Lazear

I'm really not sure what this first sentence means, but I'm getting the feeling that you're not quite sure what a vector is.

The second sentence doesn't make any kind of physical sense. Velocity in the x direction (velocity is a vector in and of itself, by the way) and acceleration (another vector) in the y direction... Velocity and acceleration do not mix directly.

I think your best bet is to pull out that physics book and review 2-d kinematics a bit before you try to solve your problem. You're having some pretty serious conceptual difficulties here, and this problem is not for somebody who is not comfortable with the subject. If your project's due tomorrow, then, well, I guess you have a long night ahead of you.

I'll be happy to help you with some slightly more minor difficulties like implementation, but I'm afraid I don't have the time to walk you through the whole concept. And I'm not being paid enough.

--J

9. Nov 30, 2004

BobG

The cars will never collide in mid-air. The slower car, leaving first, will follow a lower trajectory and will never be in the path of the faster car (the link Justin Lazear explains how to use the vectors to determine how long each car will be in the air and how high each car will get above the ramp). One thing to keep in mind is that the slower car won't stop once it hits the ground. You need to use the fast car's time in the air (plus the slow car's lead) to determine how much horizontal distance the slow covered. While the cars will never collide in the air, you do have to worry about the second car landing on top of the first car.

With the fast car going over twice as fast as the first car, you have plenty of leeway.

Of course, actually driving the fast car after it lands is a little bigger problem. With a 13 degree ramp and traveling at 72m/s (160 mph), the car's going to reach a max height a little over 13 meters above the ramp .... then it's going to fall a height of (at least) 13 meters back to the ground. That's like driving off the top of a four-story building. Better have some really good shocks.