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Position of an object after a certain amount of time?

  • Thread starter Loppyfoot
  • Start date
  • #1
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Homework Statement


Your location is at the origin. When an object passes you, it is traveling 5 m/s in the −z direction. If the object's mass is 200 kg, and the net force on the object remains constant at ‹ 0, 0, −30 › N, what is the position of the object 20 seconds after it passes you?


Homework Equations


The Change in momentum = Force * deltat


The Attempt at a Solution


I solved for the impulse, and got the vector, <0,0,-600>N. This is equal to the final momentum minus the initial momentum. The initial momentum is <0,0,-8>m/s * 200kg. And the final momentum is pf = vf * 200kg. I solved for vf, using vf = vi + Fnet*deltat, but I am stuck.

I keep getting the wrong answer, and I am unsure of where I've went wrong and what I should do next. Thanks.
 

Answers and Replies

  • #2
gneill
Mentor
20,781
2,759

Homework Statement


... what is the position of the object 20 seconds after it passes you?

...

The Attempt at a Solution


I solved for the impulse, and got the vector, <0,0,-600>N. This is equal to the final momentum minus the initial momentum. The initial momentum is <0,0,-8>m/s * 200kg. And the final momentum is pf = vf * 200kg. I solved for vf, using vf = vi + Fnet*deltat, but I am stuck.

I keep getting the wrong answer, and I am unsure of where I've went wrong and what I should do next. Thanks.
You're looking for the final position, not the final velocity. Consider using a different kinematic equation, one that relates distance traveled to force applied to a mass (acceleration).
 
  • #3
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So I'll use:
deltax = vi*t + 0.5*a*t^2.
I plug in the proper values, and I don't get the correct answer. I am unsure of where I am going wrong. The initial position is, <0,0,0>, so I don't need to worry about that.
 
  • #4
gneill
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20,781
2,759
Show your work.
 
  • #5
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Ok,
xf = xi + vi*t + 0.5*a*t^2
xf = <0,0,0>m + <0,0,5>*20 + 0.5*(<0,0,-30>N/200kg)*(20s^2)
 
  • #6
gneill
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20,781
2,759
Check your velocity vector against what the original problem states.
 
  • #7
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So, the velocity would be <0,0,-5> since it is in the -z direction?
 
  • #8
gneill
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20,781
2,759
So, the velocity would be <0,0,-5> since it is in the -z direction?
What do you think?
 
  • #9
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Yes, sir. I got the correct answer. Thanks for your guidance!
 
  • #10
1,137
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you could have also used 2nd eqn of motion
s = ut + (1/2)at^2
 

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