# Train wheels stick Problem| University Physics

### What is the magnitude of the train’s acceleration after its wheel begins to stick? Assume acceleration is constant after wheel begins to stick.

ANSWER: $a=1.0\:m/s^2$

The toy train has a constant speed of $v=2\:m/s$ from $x=2\:m\:\:to\:x=6\:m$.  Then, it began to decelerate at $x=6\:m$, and finally stop at $x=8\:m$. Considering the moment after the toy train’s wheel begin to stick up to the moment when it stopped.

$\left(v_f\right)^2=\left(v_i\right)^2+2a\left(\Delta x\right)$

$0^2=\left(2\:m/s\right)^2+2a\left(8\:m-6\:m\right)$

$4a=-4$

$a=-1\:m/s^2$

The negative sign indicates that the acceleration is moving opposing the motion (a deceleration). Basically, the negative sign is just an indication of the direction of the acceleration. The magnitude of the acceleration is simple $a=1.0\:m/s^2$.