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Mathematics 19 Online
OpenStudy (anonymous):

A toy train is pushed forward and released. What is the train's acceleration after its wheel begins to stick? at xi = 2.5m with a speed of 2.0m/s . It rolls at a steady speed for 2.6s , then one wheel begins to stick. The train comes to a stop 7.3m from the point at which it was released.

OpenStudy (anonymous):

What does xi = 2.5m mean here?

OpenStudy (anonymous):

xi???

OpenStudy (anonymous):

it means the distance between the train pushed forward and it starts to roll

OpenStudy (anonymous):

Vf^2 = 2ad + Vi^2 Vf = 0 d = displacement Vi = 2 solve

OpenStudy (anonymous):

\(v^2_f-v^2_o = 2ad\) so \(a = \frac{v^2_f-v^2_o}{2d}\) = \(\frac{\left(0\frac{m}{s}\right)^2 - \left(2\frac{m}{s}\right)^2}{2\left(7.3m\right)^2}\) = \(\frac{-4\frac{m^2}{s^2}}{14.6m}\) = \(-0.274\frac{m}{s^2}\)

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