Wanted: Equation for Distance Traveled Given Time

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SUMMARY

The discussion focuses on calculating the distance traveled by a crate under constant acceleration. Given a mass of 34.0 kg and a net horizontal force of 140 N, the resulting acceleration is determined to be 4.12 m/s². Using the kinematic equation x(t) = 1/2 * a * t², the distance traveled in 14.0 seconds can be calculated. The initial position is zero, and the integration of acceleration leads to the final position formula.

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  • Understanding of Newton's Second Law of Motion
  • Familiarity with kinematic equations
  • Basic calculus concepts, including integration
  • Knowledge of units of measurement in physics (e.g., meters, seconds)
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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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