Minimum speed for student to catch bus 40m away

[marco0009]

I'm on the last part of a homework problem and I'm not doing something right...

A student is running to catch a bus. When the student is 40m away, the bus begins accelerating from rest with a constant acceleration of .170m/s^2. What is the minimum speed she must run to catch the bus?

This is what I've done so far:
$$x_{bus}=1/2*.17*t^2$$
$$x_{girl}=-40+v_0t$$
$$1/2*.17*t^2=-40+v_0t$$
$$v_0=.085t+40/t$$

And this is where I get stuck. I'm having a hell of a time getting rid of that unknown time. I've tried integrating that and setting it equal to the position of the bus but nothing is matching up with the correct answer in the back of the book. Any ideas?

 

[James R]

Use the quadratic formula to get an expression for t in terms of v from your third line, above. If there is a positive solution for t, then v is "fast enough". For what values of v is there a positive solution?

 

[moose]

Can't you take the derivative of it, so when the slope=0, that's your minimum?

 

[marco0009]

Thank you moose, I don't know why I didn't do that earlier. I s'pose looking at the problem for too long had me wanting to do it one way so I ignored the correct way. Thanks a ton!