Question

A toy train is pushed forward and released at xi = 1.0 m with a speed of 1.0 m/s. It rolls at a steady speed for 2.5 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?

Answer 1

So the train is moving initially at 1.0 m/s

If it rolls for 2.5 s at this speed, the product gives us the distance:
x = v * t = (1.0) * (2.5) = 2.5 m

The train moves 2.5 m before it's wheel gets stuck.

This means that there is 6 - 2.5 m or 3.5 m left until the train stops.

So we need to find how much the train decelerates over 3.5 m to go from 1.0 m/s to 0.0 m/s.

kinematic eq: Vf^2 = Vi^2 + 2a(Xf - Xi)
Vf = 0 m/s
Vi = 1.0 m/s
a = ? m/s^2 <-- we need to find this
Xf = 3.5 m
Xi = 0 m <-- we'll take the point where it starts to get stuck as the new 0 position

Now we solve:
0 = 1^2 + (2)(a)(3.5 - 0)
0 = 1 + 7a
-1 = 7a
a = -1/7 = -0.14 m/s^2

The negative sign indicates the train is slowing down, which is correct!

The train accelerates at -0.14 m/s^2 after it's wheel starts to stick.

Answer 2

If it is easier to understand, you could replace Xf and Xi with what you actually have:

Xf = 6 + Xi = 6 + 1 = 7 m
new Xi = old Xi + 2.5 = 1 + 2.5 = 3.5 m

Notice, 7 - 3.5 is the same as what we did above (3.5 - 0).