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Determine the car's displacement

  1. Oct 7, 2007 #1
    Hi, I am new to the fourm. If it is not any touble, please help me with a kinematic problem.

    1. The problem statement, all variables and given/known data

    A car travels east along a straight road at a constant velocity of 18 m/s. After 5.0s, it accelerates uniformly for 4.0s, reaching a velocity of 24 m/s. For the next 6.0s, the car proceeds with uniform motion. Determine the car's displacement for the 15.0s trip.

    2. Relevant equations

    Δd = v1 x Δt + 1/2 x acceleration x Δt(squared)
    Δd = v2 x Δt - 1/2 x acceleration x Δt(squared)
    Δd = 1/2 (v1 + v2) Δt
    2aΔd = v2^2 - v1^2
    a = v2 - v2 / Δt

    3. The attempt at a solution

    This is the only question I have not experienced before. Please guide me through this question.
     
    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 7, 2007 #2

    Kurdt

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    You will have to split the problem into 3 parts. There are two where the car is travelling with a constant velocity and one where it is accelerating. The equation you have stated is almost correct. For constant velocity you will need:

    [tex] d = vt [/tex]

    For constant acceleration you will need one of the constant acceleration kinematic equations.

    https://www.physicsforums.com/showpost.php?p=905663&postcount=2
     
  4. Oct 7, 2007 #3
    so do i do:

    18 m/s x 5.0s then 24 x 4.0s then add them up?

    What do i do with the 6.0s and the uniform motion? When adding do i find the displacement?
     
  5. Oct 7, 2007 #4

    Kurdt

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    No. The two time intervals when the car is travelling with constant velocity are the first 5 seconds and the last 6 seconds after the acceleration. To work out the distance travelled while accelerating you will need one of the kinematic equations in the link I gave you above. To save you the time here it is:

    [tex] d = ut +\frac{1}{2} at^2[/tex]
     
  6. Oct 7, 2007 #5
    How am i able to find the acceleration and v1 by using all of these variables (specifically time and velocity)? Would i use [tex] d = vt [/tex] at all?
     
    Last edited: Oct 7, 2007
  7. Oct 7, 2007 #6

    HallsofIvy

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    Which question are you talking about? You have posted on question and were told to break it into 3 parts.
    Okay, how far will it go at 18 m/s for 5 s?

    Hint: at uniform acceleration, the "average speed" is just the average of the lowest and highest speeds: (18+24)/2 m/s. How far will the car go in 4.0s at that average speed?

    At 24 m/s how far will the car go in 6.0 s? NOW add them all up.
     
  8. Oct 7, 2007 #7
    Ohh i see. If i were to add them, I would get the displacement correct? or should i use [tex] d = ut +\frac{1}{2} at^2[/tex] after adding.
     
  9. Oct 7, 2007 #8

    Kurdt

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    You can ignore that one, I made a mistake. You will need to use the average velocity as HallsofIvy suggested in d = vt.
     
  10. Oct 7, 2007 #9
    so, bastically i do [tex] d = vt [/tex] with the time and velocities, add them up and thats my displacement?
     
    Last edited: Oct 7, 2007
  11. Oct 7, 2007 #10
    Please help me with this proble. I have tried everything.
    A rifle is aimed horizontally at a target located 44.0 m away. The bullet hits the target 3.0 cm below the aim point. What was the bullet's time of flight?
     
  12. Oct 7, 2007 #11

    Kurdt

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    Yes but remember the middle time stint will require you work out the average speed.

    Please do not hijack other peoples threads. You can start a new thread of your own if you have a question.
     
  13. Oct 7, 2007 #12
    I see, so after adding 24+18 and dividing by 2, multply the time and that will be my distance ? Once done, do the same for the other 2 (but without finding the average acceleration) and add, thus giving my displacement?
     
    Last edited: Oct 7, 2007
  14. Oct 7, 2007 #13

    Kurdt

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    Yes.
     
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