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- 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?

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- anonymous

- jamiebookeater

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- anonymous

what is that Xi=4m for?

- anonymous

are you there?

- anonymous

yes 4 meter

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- anonymous

not sure thats how the problem is set up

- 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

very nice im gonna try to work it out. thanks! dont know how people remember this stuff though.. lol

- 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

i think youre good with physics can you help me with this one? (ill post it) search for my name thanks! =)

- anonymous

not quite, but i'll try

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