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Homework Help: Displacement and distance

  1. Aug 1, 2015 #1
    If the displacement of a particle is given by x=(t-2)(t-2) if x is in mtrs and t in seconds the distance covered by the particle in 4 seconds is?

    I am not sure how to start.

    I know is S=ut+1/2at*t

    But not sure how to proceed.

    I am new member, would be grateful for your help.
     
  2. jcsd
  3. Aug 1, 2015 #2
    Orignal quesiton
     

    Attached Files:

  4. Aug 1, 2015 #3
    What do you think the function tells you exactly?
     
  5. Aug 1, 2015 #4

    CWatters

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    Welcome to the forum. Next time please use the homework template (forum rules).

    Have you considered sketching the graph of x=(t-2)2 ?
     
  6. Aug 1, 2015 #5
    I think the particle is moving in two direction. At 2 seconds the displacement is 0, means it changes its direction.
    I am unfortunately, not a physics expert, I read several different books, could not find a explanation for the question.
    thank you for your time.
     
  7. Aug 1, 2015 #6
    Sorry, I did not do graph, I will try. Thank you
     
  8. Aug 1, 2015 #7
    Or definite integration?
     
  9. Aug 1, 2015 #8
    I got this graph, but unable to calculate the distance travelled.

    upload_2015-8-1_20-13-37.png
     

    Attached Files:

  10. Aug 1, 2015 #9
    Alright.Now,since distance is asked,you could break up the motion into two parts.Depending on the direction in which it is moving.Then you could solve them seperately.
     
  11. Aug 1, 2015 #10
    Why not? What does y-axis say?
     
  12. Aug 1, 2015 #11
    thank you. But I am not able to calculate the distance in first 2 seconds and next 2 seconds with a formula.
    If I replace t with 2 in the formula it becomes zero.
     
  13. Aug 1, 2015 #12
    Y axis shows: at 0 second 4, at 1 sec 1 and similarly at 2,3, 4 it says 0, 1, 4 again.

    Does that mean total distance is: ?10m
     
  14. Aug 1, 2015 #13
    Yes,what does it tell you about the directions of the motions?
    Though it would be more efficient if you understood and used the graph method.
     
  15. Aug 1, 2015 #14
    Thank you. Actually this is for my niece who is in India, has no access to internet. I am an established doctor, I was trying to help them, by posting in the forum here. I studied this 23 years back, tried to read cambridge material, finally put down to this forum to clever people to help me to help others.

    So do we agree the answer is 10m. Actually in the question bank the answer they gave was 8m.
     
  16. Aug 1, 2015 #15

    I did put this in another forum, they said it was incorrect question. You can see in my original post, the choice answers they gave does not have 10m.
     
  17. Aug 1, 2015 #16
    But the answer isn't 10.It's 8..
     
  18. Aug 1, 2015 #17
    No.

    Clear it again. In 0 second, it is 4. That does not mean it traversed 4 meters in 0 sec.

    Remember, distance is the scalar and displacement is vector. So, in displacement you can use angles and positive negative signs for opposite directions. But for distance, it is sum of all separate displacements along all the directions neglecting the sign.

    So, look at the graph and think again.
     
  19. Aug 1, 2015 #18
    Yes, answer is 8. You can do it seeing graph. Add the displacements neglecting the signs
     
  20. Aug 1, 2015 #19
    Y-axis actually shows from 0 to 2 seconds the displacement is (-4) and from 2 to 4 seconds it is (+4).

    So, distance is |-4|+|+4|=4+4=8

    Best of luck for your niece anyway.
     
  21. Aug 1, 2015 #20
    Thank you for your time both of you.
    Now I am beginning to understand.
    How would explain this? things confusing me were
    1) how can a particle move in 0 seconds to 4 meter
    2) If you break the journey: at 1 second it comes down to 1 (meaning 3m) at 2 second to y=0 (meaning another 1m) and then back up again. Is that correct?
    3) If yes then if there is large number for example x is 20 how would we calculate.
    I am grateful for your input.
     
  22. Aug 1, 2015 #21
    thank you for your input and precious time. It is interesting to me.
     
  23. Aug 1, 2015 #22
    1) Suppose you are sitting and an ant is coming towards you. It is now 4m apart from you when you started to count the time. So, its initial distance from you is 4m. So, at time t=0, it didn't cross 4m, rather it is 4m away from the co-ordinate system with you.

    2) Now it is coming with a velocity of 2 m/s (though large for an ant). So after 2 minutes it is 0 m from you. So, it's position is 0 from origin (you)

    3) Let then it went 20 meters more after 10 mins. So, total distance by it, is 24 meters, and right now it is 20 meters from you.

    I think the motion of the particle for the function is unidirectional. It is just once left and another time right to the co-ordinate system.
     
  24. Aug 1, 2015 #23
    Sorry, it is double directional for your function. Something like, after the ant crossed you, it changed its direction and turned back. So, after 10 minutes it will go 20 m away from you again. Displacement for it is (4-20)m = -16 m. Here, negative sign means the displacement is opposite to its initial direction. And distance=|4|+|-20|=24m
     
  25. Aug 1, 2015 #24
    Thank you So much.
     
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