anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
what is that Xi=4m for?
anonymous
  • anonymous
are you there?
anonymous
  • anonymous
yes 4 meter

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anonymous
  • anonymous
not sure thats how the problem is set up
anonymous
  • anonymous
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)
anonymous
  • anonymous
very nice im gonna try to work it out. thanks! dont know how people remember this stuff though.. lol
anonymous
  • anonymous
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.
anonymous
  • anonymous
i think youre good with physics can you help me with this one? (ill post it) search for my name thanks! =)
anonymous
  • anonymous
not quite, but i'll try

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