# Distance and acceleration problem

• quick

#### 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:

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.

how do i solve for c_min?

cmin2 - 2ab = 0

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

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