# Solving equations without number values

1. Feb 19, 2006

### quaker

I'm having trouble wrapping my head around problems without any given number values. All of the book problems I've done so far have had number values for position or velocity etc and those fit nicely into equations that could be solved for other variables. I was recently given a set of sample quiz questions and they are all like this:

1. An object moving along the x axis has a constant negative acceleration of a1 and an initial velocity of +v1. A second object located +d meters ahead of the first one also has a constant acceleration of +a2 with an initial velocity if +v2. Find the condition(s) on v1 and v2 such that a collision will not occur.

How would I go about solving something like this?

2. Feb 19, 2006

### phucnv87

$$x_1=v_1t-\frac{1}{2}a_1t^2$$
$$x_2=d+v_2t+\frac{1}{2}a_2t^2$$
The collision will not occur if $$x_2>x_1$$

3. Feb 20, 2006

### quaker

Well, I get those equations plugging in what's given, but how would I then solve for the conditions on v1 and v2? This is so confusing without numbers to work with.

4. Feb 20, 2006

### DaveC426913

You *did* take algebra as a prerequisite to physics, right?

5. Feb 20, 2006

### quaker

Yes, but I'm not sure how to turn the resulting equations for $v_1$ and $v_2$ into their conditions for no collision.

6. Feb 20, 2006

### nrqed

what you can do is to use the two equations given by the other poster and set them equal to one another, $$x_1 = x_2$$. This woul dgive the condition such that there *would* be a collision. Now, solve for the time at which the collision would occur (you would get a quadratic formula for t).

This is the time at which a collision would occur *if* there was one. Now, if you do NOT want a collision to occur, you want this equation to have no solution. You can make sure that there is no solution by imposing that the term under the square root ($$b^2 - 4 a c$$) is negative. That will give you a condition on the speeds.

Pat

7. Feb 20, 2006

### quaker

I get it now, that makes sense. Thanks for your help!