# Distance and acceleration problem

1. Aug 31, 2004

### quick

this picture shows the problem:
http://s93755476.onlinehome.us/phys.jpg [Broken]

i realize the discriminant must be positive for t_catch to have a real value. i just can't figure out how to write an expression with a and b. thanks in advance.

Last edited by a moderator: May 1, 2017
2. Aug 31, 2004

### jamesrc

Realizing that the discriminant must be greater than or equal to 0 is really the whole problem. The minimum speed, c, will give a discriminant of 0 (i.e., cmin2 - 2ab = 0). Just solve for cmin and plug back into your expression for tcatch from the quadratic formula.

3. Aug 31, 2004

### quick

how do i solve for c_min?

4. Aug 31, 2004

### jamesrc

cmin2 - 2ab = 0

5. Aug 31, 2004

### quick

6. Aug 31, 2004

if you have a quadratic formula:
$$ax^{2} + bx + c = 0$$
the solution for its roots is:

$$x = \frac {-b {+-} \sqrt{b^{2} - 4ac}}{2a}$$

$$\frac{1}{2}at^{2} - ct + b = 0$$

so the solution to have a positive x walue would be easy to find, just sub in your constants in the equation above:

$$x = \frac {-(-c) {+-} \sqrt{(-c)^{2} - 4((\frac{1}{2}a)(b)}}{2(\frac{1}{2}a)}$$

$$x = \frac {c {+-} \sqrt{c^{2} - 2ab}}{a}$$

so in order to have a positive x value, there must be a positive root,
$$c^2 - 2ab > 0$$

7. Sep 1, 2004

### Chronos

Hmm, I don't see the connection to the original question.

8. Sep 1, 2004

what dont you see about it, the question is asking to express the minimum values of the mans speed in terms of a and b. As long as

$$c^2 - 2ab > 0$$

$$c^2 - 2ab = 0$$