Distance and acceleration problem

[quick]

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

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.

 

[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., c_min² - 2ab = 0). Just solve for c_min and plug back into your expression for t_catch from the quadratic formula.

 

[quick]

how do i solve for c_min?

 

[jamesrc]

c_min² - 2ab = 0

 

[quick]

sorry I am retarded, i didn't read your response correctly. thanks for your help!

 

[Nenad]

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}

now your equation is{
\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

there is your answer. Do you understand now?

 

[Chronos]

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

 

[Nenad]

what don't 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

there is your answer. The min value would be:

c^2 - 2ab = 0