Calculating Acceleration of a Toy Train with a Sticking Wheel

[Steve Halt]

Homework Statement

A toy train is pushed forward and released at x_0 = 2.0 m with
a speed of 2.0 m/s. It rolls at a steady speed for 2.0 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?

Homework Equations

x_f = x_i +v*t +(1/2)*a*t^2

The Attempt at a Solution

O, well, I guess my attempt is this:

at x_0 = 2 m
v_0 = 2m/s
a_0 = 0 m/s^2

at x_1 = 6 m
v_1 = 2 m/s
a_0 = 0 m/s^2

and finally, through the relevant equation above,

8 = 6 + (2)(t) + (1/2)(a*t^2)

I have no idea how to complete this from here. Help?

 

[learningphysics]

Use a different kinematics equation.

 

[Steve Halt]

Well, okay, how about

v_f^2 = v_i^2 + 2*a*x

0 = (4) +2*a*(2)

-4 = 4*a

a = -1 m/s^2

so the magnitude of the acceleration is 1 m/s^2. But it can't be this simple now can it?

 

[learningphysics]

That looks right to me.