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Deceleration with weight and a dent? Confused

  1. Aug 17, 2006 #1
    Ive been trying to fiure out how to get this answer but am just have some serious trouble.

    A 27 pound meteorite struck a car, leaving a dent 27 cm deep in the trunk. If the meteorite struck the car with a speed of 520 m/s, what was the magnitude of its deceleration, assuming it to be constant?

    I just cant figure out how this is done. To find the magnitude of deceleration i belive i would need to divide the 520 m/s by some other time in seconds yet...there is no other time in seconds. Also, I havent an idea what im supposed to do with the pounds and dent. Is there a way to find out seconds from those or something? Im really stuck here. If someone could just guide me along or give me a formula to figure it how to do this it would be greatly appreciated, as I just cannot find one!
     
  2. jcsd
  3. Aug 17, 2006 #2

    NateTG

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    What equations do you know?
     
  4. Aug 17, 2006 #3
    You can calculate the Kinetic Energy of the meteorite prior to striking the car. Use KE = F * d to get the average force that the dent acted on the
    meteorite during the impact. Once you have the average force you should
    be able to calculate the deceleration.

    A 27 lb object traveling at nearly the speed of a rifle bullet and a dent??
    You've got to be kidding!
     
  5. Aug 17, 2006 #4
    Don't forget to convert pounds to Newtons, and Newtons into kilograms.
     
  6. Aug 17, 2006 #5
    What does F, and d stand for? I know KE is the kinetic energy. I did the conversion of pounds to Kilograms. It would be 12.246994 kg.

    And the only equations i learned where the four kinetics equations, but i dont see how they could possibly help me yet as none of them have anything to do with mass.
     
    Last edited: Aug 17, 2006
  7. Aug 17, 2006 #6
    F is the force in Newtons in the metric system
    d is the distance in meters

    F * d is the work in Joules done in stopping the meteorite which is
    also equal to the Kinetic Energy of the meteorite at impact
    Also, there is no use in using more significant figures than 12.2 Kg
    for the mass since you are limited to 2 significant figures in the depth
    of the dent.
     
  8. Aug 17, 2006 #7

    Doc Al

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    You don't need the mass. You are given all the information needed to solve this as a kinematics problem.

    Hints:
    What distance does the object travel (during its deceleration)?
    What are its initial and final speeds?
     
    Last edited: Aug 17, 2006
  9. Aug 17, 2006 #8
    Thats what im missing. I dont see where it tells me the distance the object travels, or how i can figure that out. Also, I dont see its inital speed. I just know its final speed is 520 m/s
     
  10. Aug 17, 2006 #9

    NateTG

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    Well:
    1. Is the object decelerating before it hits the car?
    2. How far does the object travel while it's in contact with the car?
     
  11. Aug 17, 2006 #10
    Actually, Ive just made some progress. Ive found out i need to use the equation

    Vf^2=Vi^2+(2ad)

    but im not sure what i do next.
    Was i wrong before and is the INITIAL speed 520 m/s, the FINAL speed 0 m/s?

    That would have 0^2=520^2+2ad but...i dont know what the displacement is.

    Oh wait, Displacement=final v-inital v. so now i have

    0^2=520^2+2a(0-520)
    0=270400+2a(-520)
    0=270400-1040a
    1040a=270400
    a=260 m/s/s

    Am i right or did i place something wrong? Shouldnt the acceleration be negative? Since its deceleration im getting?

    Sorry about all the editing to this but i think i got it now. I just solved for A and got:
    A=(Vf^2-Vi^2)/2d
    A=(0-520^2)/2*520
    A=-260 m/s/s
     
    Last edited: Aug 17, 2006
  12. Aug 17, 2006 #11
    Ok. I dont wanna keep on posting my random ideas but this time im almost positive i did it right.
    A=(Vf^2-Vi^2)/2d
    A=(0-550^2)/(2*-.27) (Is the -.27 negative becuase its a dent so it went like..inwards kinda?)
    Then you have A=500740.7407 m/sec^2 as your final answer?
     
  13. Aug 17, 2006 #12
    Sorry, sorry guys. Ive just got one more thing, and i finished that question. Just the next one, its basically the same thing but im not sure about one part of it:

    Coasting due south on your bicycle at 8.0 m/s, you encounter a sandy patch of road 6.0 m across. When you leave the sandy patch your speed has been reduced to 6.6 m/s. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch? Give both magnitude and direction.

    So once again we use the same formula A=(Vf^2-Vi^2)/2d
    A=(6.6^2-8^2)/(0-6) A=3.406 m/sec^2 due north OR
    A=(6.6^2-8^2)/(6-0) A=-3.406 m/sec^2 due north

    im not sure which way the displacement part of it goes but im pretty sure my direction part of it is right, since its decelerating the direction would be the opposite. And i know, Displacement=final position-initial position...im just not sure what 6 is!
     
  14. Aug 17, 2006 #13

    Doc Al

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    That's OK, but please round it off to two significant figures.

    If you are worried about the sign of your answer, you must use a sign convention consistently. If you call up positive and down negative, then:
    Vi = -520 m/s
    d = change in position = -0.27 m - 0 m (calling the initial position at impact = 0)

    Using this convention the acceleration will end up positive (upwards), which should make sense. After all, the car pushes up on the meteorite, bringing it to rest. Since the acceleration is opposite to the velocity, the term "deceleration" is used (which just means it slows down).
     
  15. Aug 17, 2006 #14

    Doc Al

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    Realize that a negative acceleration due north means the same as saying that the acceleration is due south.

    Again, pick a sign convention like I explained in my last post. Call the initial position = 0.

    But you really should be able to write down the direction of the acceleration immediately, without fussing with sign conventions or equations. Hint: Is the bike slowing down or speeding up? Use the equation to get the magnitude of the acceleration, but use your understanding to assign the direction.

    Of course, if you are strict with the sign convention, the equation will automatically provide the correct sign.
     
  16. Aug 17, 2006 #15
    So then are you saying that 3.406 m/s^2 due north = -3.406m/s^2 due south, since its going south?
     
    Last edited: Aug 17, 2006
  17. Aug 18, 2006 #16

    Doc Al

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    No. I'm saying that taking negative two steps forward is the same as taking two steps backward. :wink:

    The magnitude of a vector is always positive (or zero). So when you specify a vector, such as acceleration, by giving its magnitude and direction, give a positive number and a direction (never a negative number).
     
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