Question

I tried this on my own but wrong help!
A toy train is pushed forward and released at xi = 2.2 m with a speed of 1.0 m/s. It rolls at a steady speed for 2.9 s, then one wheel begins to stick. The train comes to a stop 6.0 m from where it was released. What is the train's acceleration after its wheel begins to stick?
I actually tried this many ways on my own, but am not seeming to get the right answer? can someone explain what i'm doing wrong and what the right answer is?
to solve (using Vf^2 = Vi^2 + 2a(change in x)
0 = 1^2 + (2)(a)(.9)
-1=1.8a
a=-.56
Again, this is not right, but why?

Answer 1

i think since you're measuring acceleration after the wheel starts to stick so v(0) of 1m/s wouldn't be right as that was before the wheel started to stick. so i think the right v(0) would be to solve for v at t=2.9s and that would be you're v(0)

Answer 2

so what would the answer be then

Answer 3

you dont know enough of the variables to do that

Answer 4

ok looking through it again i see that i was wrong.

Answer 5

break the problem down, first off, ignore the xi info since you do not need it. The calculate the distance that the train travels before the wheel begins to stick:

(2.9s)(1m/s) = 2.9 meters

Now the train stops 6m after it is released so

6m-2.9m = 3.1m

Now we assume constant acceleration (negative of course), since if acceleration is not constant than this is a much more advanced problem. Thus we have a velocity of 1m/s and 3.1 meters to reach 0 velocity. From here you can just plug into the equation you wanted to use, let me know if you still need help.