# Shuffleboard problem

1. Jan 19, 2008

### Cathartics

A shuffleboard disk is accelerated at a constant rate from rest to a speed of 4.2 m/s over a 1.4 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 2.7 m/s2 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?

I dont even know where to begin on this one, Please give me any hints, concept or formula's to begin with. Thanks in advance.

2. Jan 19, 2008

### Tom Mattson

Staff Emeritus
Break the problem into 2 pieces. First determine how long the cue is in contact with the disk, then determine how much more time it takes for the disk to come to a stop after the cue loses contact with it.

Again, break it up into 2 pieces. Determine the distance traveled while the cue is in contact with the disc, then determine the additional distance traveled after the cue loses contact.

3. Jan 20, 2008

### Cathartics

Firstly thanks tom for that info. Here is what i did.

first i found t1 that's when it was in contact with the cue where
vo= om/s (initial velocity), vf = 4.2 m/s (final velocity) and x= total distant
using the formula x = ((vo + vf)*t) / 2 then t = .66s this is the contact time.

After it loses contact i found how long did it travelled by using
vf = 0 (final velocity) and vo = 4.2 (initial velocity) and a = 2.7 m/s^2 using the formula
x = vf^2 - (4.2)^2 / - 2.7 which would yield 6.53 m

then by using the above value i found t for the first part of the problem by formula
x = 1/2 (vf+vo) *t where x =6.53 and vo = 4.2

But when i plug in the numbers it say's that i'm doing something wrong and answers dosent match. Please tell me what went wrong am i working on with a wrong concept or am i junt punching numbers to a formula, thanks in advance..

4. Jan 21, 2008

### Cathartics

5. Jan 22, 2008

### Cathartics

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Last edited: Jan 23, 2008
6. Jan 23, 2008

### Cathartics

7. Jan 26, 2008

### Tom Mattson

Staff Emeritus
Sorry, I was away for a few days.

That's correct.

That's the right formula, but I get 3.27m not 6.53m.

That error will propagate through the rest of your calculation.

There's one other misconception you have.

That doesn't make any sense. You already found t for the first part of the problem: t=0.66s. Also, why would you use x for the second part of the problem to find t for the first part?

This doesn't make any sense either. You're saying that the x you found for the second part of the problem is the same as x for the entire trip of the puck from start to finish. That would imply that the puck travels a zero distance for the first part of the problem, contrary to the given info.