Acceleration word problem A shuffleboard disk is accelerated at a constant rate from rest to a speed of 4.8 m/s over a 1.9 m distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of 2.2 m/s$^2$ until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel? So, I have vo = 0, vf = 4.8 = (vo + vf) *t/2 = 9.6. Then, x= 1/2 (vf+vo)*t(2.2) = 10.56. It says the answers are wrong, though.

You have two intervals of constant acceleration: the acceleration part, and the deceleration part.

The questions are asking you about both parts taken together. So,

* Calculate the time over which the first part happens.
* Calculate the time and distance over which the second part happens.
* Add the times and distances for the two parts, and you're done.



Can you take it from here?

Some more hints:

$$x = x_0 + v_0t + at^2/2$$ $$v = v_0 + at$$

To calculate the time for the acceleration part, start with what you know. You know $a$, and you know $x$. (Both $x_0$ and $v_0$ are zero.) The first equation then becomes $t = \sqrt{2x/a}$.