Wanted: Equation for Distance Traveled Given Time

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Homework Help Overview

The discussion revolves around a physics problem involving a crate acted upon by a net horizontal force, requiring the calculation of distance traveled over a specified time given its acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine the distance traveled by the crate using its initial conditions and constant acceleration. Some participants provide equations related to motion and discuss integration methods to derive the position function.

Discussion Status

Participants are actively engaging with the problem, exploring different approaches to derive the distance traveled. Some guidance has been offered regarding the use of calculus to find the position function, but no consensus or final solution has been reached.

Contextual Notes

The original poster expresses difficulty with the problem, indicating a potential overload from previous calculus and physics work. The problem setup includes specific values for mass, force, and time, but further details on the context or constraints are not provided.

iamhumble
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Hi --

I know this is so easy but right now I am a box of rocks. I think I am doing too much calculus and physics problems and just can not think of the simple things.

What I am trying to do is solve this problem.

Problem:

A crate with mass 34.0 initially at rest on a warehouse floor is acted on by a net horizontal force of 140

1) What acceleration is produced?

Answer is 4.12 m/s^2

2) How far does the crate travel in 14.0 s?


I know the initial position is 0 since the crate is at rest, and the acceleration is constant here so I need to figure the final position after 14.0 seconds at this acceleration.

Any help is appreciated. Thanks.
 
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Google saved the day for me.

x = v_i(t) + .5a(t^2)
 
well d^2(x)/dt^2 = a right?
integrate once to get dx/dt = at + c . We know v(0) = 0 since it was initially at rest dx/dt = at

integrate again to get position: x(t) = 1/2*a*t^2 + c. we can use the original starting point as x(0) = 0 so c = 0

You are given a time and an acceleration so... plug it in. Don't worry it happens to the best of us (me not being the best but shrug)
 
thanks for the calculus approach. i think now i can remember what to do using calculus :smile:
 

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