A toy train is pushed forward and released at = 3.5 with a speed of 1.0 . It rolls at a steady speed for 1.2 , then one wheel begins to stick. The train comes to a stop 3.4 from the point at which it was released.What is the train's acceleration after its wheel begins to stick?

without units, none of this makes sense.

A toy train is pushed forward and released at = 3.5 m with a speed of 1.0m/s . It rolls at a steady speed for 1.2s , then one wheel begins to stick. The train comes to a stop 3.4m from the point at which it was released.What is the train's acceleration after its wheel begins to stick?

To find the train's acceleration after its wheel begins to stick, we can use the equation of motion:

vf^2 = vi^2 + 2ad

Where:
- vf is the final velocity of the train (which is 0 as it comes to a stop)
- vi is the initial velocity of the train
- a is the acceleration of the train
- d is the distance traveled by the train after the wheel begins to stick

First, let's calculate the initial velocity of the train. We know that it is released with a speed of 1.0 m/s, and it rolls at this steady speed for 1.2 m. Therefore, the initial velocity (vi) is 1.0 m/s.

The train comes to a stop at a distance of 3.4 m from the point at which it was released. This distance (d) is given as 3.4 m.

Plugging in the values into the equation of motion, we have:

0 = (1.0)^2 + 2a(3.4)

Simplifying the equation, we get:

0 = 1 + 6.8a

Rearranging the equation, we have:

6.8a = -1

Lastly, solving for the acceleration (a), we divide both sides of the equation by 6.8:

a = -1/6.8

Therefore, the train's acceleration after its wheel begins to stick is approximately -0.147 m/s^2 (rounded to three decimal places). Note that the negative sign indicates that the train is decelerating.

distance at steady speed: 1.0m/s*1.2s=1.2m

stopping distance=3.4-1.2=2.2m

Vf^2=Vi^2+2ad solve for a, knowing Vf=0, Vi=1m/s, d=2.2m