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anonymous

  • 5 years ago

A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one wheel begins to stick. The train comes to a stop 7.7 m from the point at which it was released. What is the train's acceleration after its wheel begins to stick?

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  1. anonymous
    • 5 years ago
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    what is that Xi=4m for?

  2. anonymous
    • 5 years ago
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    are you there?

  3. anonymous
    • 5 years ago
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    yes 4 meter

  4. anonymous
    • 5 years ago
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    not sure thats how the problem is set up

  5. anonymous
    • 5 years ago
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    I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)

  6. anonymous
    • 5 years ago
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    very nice im gonna try to work it out. thanks! dont know how people remember this stuff though.. lol

  7. anonymous
    • 5 years ago
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    lol only that equation you need to remember. there are a few equations that's very helpful. You'll surely remember them all. it's just the matter of time.

  8. anonymous
    • 5 years ago
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    i think youre good with physics can you help me with this one? (ill post it) search for my name thanks! =)

  9. anonymous
    • 5 years ago
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    not quite, but i'll try

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