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

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SUMMARY

The discussion centers on calculating the time it takes for a box to travel 4.00 meters while experiencing a negative acceleration of -0.25 m/s² and reaching a final velocity of +0.50 m/s. The equation used was Δx = v₀t + 1/2 at², where the initial velocity (v₀) was not provided. Participants identified errors in the application of the quadratic formula and the sign of acceleration, leading to confusion in the calculations. The correct approach involves ensuring all terms are accurately represented in the quadratic formula.

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  • Understanding of kinematic equations, specifically Δx = v₀t + 1/2 at²
  • Familiarity with the quadratic formula for solving equations
  • Basic knowledge of acceleration and its effects on motion
  • Ability to manipulate algebraic expressions and equations
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  • Review the derivation and application of kinematic equations in physics
  • Practice solving quadratic equations using the quadratic formula
  • Study the concepts of initial and final velocity in uniformly accelerated motion
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Students studying physics, educators teaching kinematics, and anyone seeking to improve their problem-solving skills in motion-related scenarios.

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


\Deltax=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) + - \sqrt{(.50)^2-4(.125)(-4)/2(.125)}

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

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

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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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