A toy train is pushed forward and released at x0 = 1.6 m with a speed of 2.2 m/s. It rolls at a steady speed for 2.3 s, then one wheel begins to stick. The train comes to a stop 6.0 m from the point at which it was released. What is the magnitude of the train's acceleration after its wheel begins to stick?

Question

aum · Answer

Train starts at x = 0, is pushed until x = 1.6 m and then released.  
It travels for 2.3 s at a steady speed of 2.2 m/s.  
Distance traveled from the point of release = 2.3 * 2.2 = 5.06 m.  
The wheel sticks and then the train stops 6.0 m from where it was released. 
Distance traveled after the wheel sticks is: 6.0 - 5.06 m = 0.94 m.

Initial velocity at the time the wheel sticks u = 2.2 m/s
Final velocity v = 0
Distance traveled s = 0.94 m
Find the deceleration.  
Use the formula    $v^2 = u^2 + 2as$   where 'a' is the acceleration.
They only want the magnitude and so ignore the negative sign.
Express the answer in  m / s^2

anonymous · Answer

Thank you so much! I had entered 2.57 m/s^2 and gone through the whole procedure that you said, but the online assignment marked me wrong because I did not disregard the negative sign for magnitude.

aum · Answer

But did it accept 2.57 m/s^2 WITHOUT the negative sign?

anonymous · Answer

My physics teacher has barely taught us this. Can I ask how you know to use which formulas when and how I should know to use this formula in this situation?

anonymous · Answer

Yes it did accept 2.57 without the neg. sign just now. Thanks to you!!

aum · Answer

There are only a few formulae involving initial velocity u, final velocity v, distance traveled s, acceleration a and traveling time t.  Most of the formulae have time 't' in it.  There is only one formula that does not involve t and that is the one we used here.  We knew the train's initial velocity u, and since the train stops, its final velocity v is zero and we know the distance it traveled 's' before it stopped.  They want acceleration 'a'.  There is no time 't' involved and so we look for the formula that does not involve 't' which was $v^2=u^2+2as$.  Out of the 4 variables we know 3 and we can solve for the unknown 'a'.

anonymous · Answer

Thank you! :) I actually have a couple more questions I have no idea where to start with which I am about to post and I would love your help, but if you are busy that's understandable too.