How long did it take for the box to travel 4.00 meters?

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Homework Help Overview

The problem involves a box sliding along a surface with an initial positive velocity and experiencing a negative acceleration. The task is to determine the time taken for the box to travel a distance of 4.00 meters, given its final velocity after this distance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the kinematic equation and the application of the quadratic formula. There are questions about the initial velocity and the signs of acceleration. Some participants express uncertainty about the correctness of the original poster's calculations and setup.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach and calculations. There is no clear consensus, but some guidance has been offered regarding the setup of the equations and the application of the quadratic formula.

Contextual Notes

There is a lack of explicit information regarding the initial velocity, which is a point of contention in the discussion. Participants are also questioning the signs used for acceleration in the equations.

hatcheezy
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Homework Statement


A box slides along a surface with a positive initial velocity. It the experiences an acceleration of -0.25m/s^2. After traveling 4.00 meters, its velocity is +0.50m/s. How long did it take for the boxto travel the 4.00 meters?


Homework Equations


[tex]\Delta[/tex]x=volt + 1/2 a t^2 ??


The Attempt at a Solution


4.00m=(.50m/s)t+1/2(.25m/s^2)(t^2)

(.125m/s^2)(t^2)+(.50m/s)(t)-(4.00m)

-(.50) + - [tex]\sqrt{(.50)^2-4(.125)(-4)/2(.125)}[/tex]

[tex]\sqrt{2}[/tex]-.50 = .9seconds?
 
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Not from my calculations.

From you problematic, [tex]v_0[/tex] is the initial velocity of the box, which is not equal to [tex]0.5m/s[/tex]

Cheers
 
You got the quadratic formula wrong. Everything is over 2a, not just 4ac.

I had to double-check your initial set-up, because you had acceleration as positive, but after a bit of algebra it's correct. I'm not sure if that was by accident or intentional. You didn't show that step, if it was intentional. V-initial isn't given, so you have to solve for V-initial in terms of V-final, a, and t, which happens to be what you got.
 
Jack21222 said:
You got the quadratic formula wrong. Everything is over 2a, not just 4ac.

I had to double-check your initial set-up, because you had acceleration as positive, but after a bit of algebra it's correct. I'm not sure if that was by accident or intentional. You didn't show that step, if it was intentional. V-initial isn't given, so you have to solve for V-initial in terms of V-final, a, and t, which happens to be what you got.

i set it over 2a...im just wondering if i set the previous equation up properly
 
hatcheezy said:
i set it over 2a...im just wondering if i set the previous equation up properly

You did, but I suspect it was due to shear luck, because you didn't show how you went from Vo to Vf, and you didn't show how you went from 'a' being negative to 'a' being positive. According to my early-morning algebra, the set-up is right, however.

And no, you didn't set it all over 2a. If you had, you would have gotten the correct answer.
 

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