Solve Homework Problem: Min Speed to Catch Bus

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Homework Help Overview

The problem involves a student attempting to catch a bus that is accelerating from rest. The student is initially 40 meters away from the bus, which begins to accelerate at a constant rate of 0.170 m/s². The goal is to determine the minimum speed the student must run to catch the bus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up equations for the positions of both the bus and the student but struggles with eliminating the unknown time variable. Some participants suggest using the quadratic formula to express time in terms of speed, while others propose taking the derivative to find the minimum speed.

Discussion Status

Participants are exploring different mathematical approaches to the problem. There is acknowledgment of the original poster's confusion, and some guidance has been offered regarding potential methods to find the minimum speed.

Contextual Notes

The original poster expresses frustration with the problem and notes that their attempts have not aligned with the expected answer in the textbook. There is an implication of imposed homework constraints, as the problem is part of a homework assignment.

marco0009
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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:
[tex]x_{bus}=1/2*.17*t^2[/tex]
[tex]x_{girl}=-40+v_0t[/tex]
[tex]1/2*.17*t^2=-40+v_0t[/tex]
[tex]v_0=.085t+40/t[/tex]

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

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