# Lavabomb kinematics

1. Jun 22, 2008

### gerry73191

1. The problem statement, all variables and given/known data

A volcano shoots a lava bomb straight upward. Does the displacement of the lava bomb depend on your choice of origin and/or your choice of positive direction?

2. Relevant equations

$$y(t)=vt-\frac{1}{2}gt^2$$

3. The attempt at a solution

first I tried just remembering my old physics lectures. I remember my teacher saying something about having free choice of origin and and positive direction.

But, I tried to prove that by

taking the derivative of $$y(t)=vt-\frac{1}{2}gt^2+a$$; where a is the y coordinate of the origin.

Solving for when velocity is 0, I took that value and subsituted it into y(t).

To me it seems the "a" term changes the value of the displacement.

But I'm not sure

Please help.

Last edited: Jun 22, 2008
2. Jun 22, 2008

### dynamicsolo

The definition of net displacement is the difference between final position and initial position. What happens to your 'a' in calculating displacement according to its definition?

3. Jun 22, 2008

### gerry73191

nothing. all the a term does is to increase/decrease the value of the displacement.

4. Jun 22, 2008

### dynamicsolo

Using your formula for the position of the lavabomb,

$$y(t)=vt-\frac{1}{2}gt^2+a$$

the position at time t = 0 is

y(0) = a .

So a represent the position of the origin in the sense that the origin is a distance a below the starting position of the bomb. At some time T, the bomb will be at

$$y(T)=vT-\frac{1}{2}gT^2+a$$ .

The displacement at time T is given by

y(T) - y(0) . What will that be equal to? Does it depend on a?

5. Jun 23, 2008

### gerry73191

well no in that case you could choose any a and still get the correct displacement.

But is that a legal move? Ive never seen that type of maneuver in a textbook.

Mathematically I know its valid its just that it seems like your just manipulating so that it gives you that answer your looking for.

btw..thanks for your help

6. Jun 23, 2008

### dynamicsolo

It is not a question of what I'm doing being a legitimate "move". This is just an application of the definition of displacement.

The function you are writing, y(t), is a position function. Net displacement is the difference between the final and initial positions of an object. It is only equal in magnitude to the postion in the special case where the starting position is taken to be the origin.

I'm a bit surprised you haven't seen this, since it is discussed in many textbooks. It is the case, though, that problems far more often ask you to calculate positions than displacements.

7. Jun 23, 2008

### gerry73191

wow. I feel dumb. I should have seen that, but thank you. I will be taking AP Physics in the fall so I wanted to enter sharp by prepping over the summer.

I had a very poor physics instruction the previous year. My teacher never taught us the theory and relationship with mathematics of many physics principles. He just told us "tricks" which although valid don't give you the insight into problems that theory does.

Nonetheless, thank you.

8. Jun 23, 2008

### dynamicsolo

There's no reason to "feel dumb" about this: when you described the kind of course you had, it explained something about the approach taken, which I'm afraid is not very effective at giving insight into physical situations (but which is, unfortunately, pretty common nowadays).

You can think about the displacement as a line connecting the starting to the ending position. The direction and length of that line would be the same regardless of what altitude you decided to call "zero" and start measuring positions from, so it is independent of the choice for an origin.

To answer the other part of the question, we started by calling upward "positive", as indicated by your choice of equation

$$y(t)=vt-\frac{1}{2}gt^2+a$$.

If we instead chose downward to be positive, but kept the origin at a distance a below the starting point, then the starting point would be above the origin, which is now in the negative direction from "zero". By the same token, the upward initial velocity is now also negative, and the direction of gravitational acceleration is now positive. So the position equation would become

$$y(t)=-vt+\frac{1}{2}gt^2-a$$ ,

which of course is just the negative of the one you wrote earlier. The displacement between the initial position y(0) and the later position y(T) would now be

$$y(T) - y(0) = [ -vT + \frac{1}{2}gT^2 - a ] - (-a) = -vT + \frac{1}{2}gT^2$$ .

This is the negative of the result you had before for the displacement, which is just what you would expect from reversing the direction of the y-axis. So the displacement is also independent of the direction you choose to call positive (that is to say, the displacement is still upward by the amount indicated, but upward is "negative" here).

This is something to keep in mind in working with many physical quantities: if their values or the physical situation does not depend on the point of view, then the result should be the same regardless of the choice of origin, of axes, etc. (This is the basis of the concept of "physical relativity", which can also extend to choice of the velocity or acceleration of the observer.)

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