# Need some help

1. Feb 12, 2010

### barthayn

1. The problem statement, all variables and given/known data
An object is pushed along a rough horizontal surface and released. it slides for 10.0s before coming to rest and travels a distance of 0.2m during the last 1.0s of its slide. Assuming the acceleration to be uniform throughout
a) How fast was the object traveling upon release?
b) How fast was the object traveling when it reached the half way position in its slide?

vf = 0m/s
vi = ?
d1 = ?
d2 = 0.2m
a = ?
t = 10s

2. Relevant equations
a = Δv/Δt
Δd = vΔt = ((Vi+Vf)/2)Δt
vf = Δt-1/2a(Δt)2
Δd = vfΔt-1/2a(Δt)2

3. The attempt at a solution

a = -0.4/1
a = -0.4m/s2

Δd = (0.4(100))/2
Δd = 40/2
Δd = 20 m

vi2 = 0-(-0.2)(20.2)
vi2 = 4.04m2/s2
vi = 2.009m/s

Therefore the object was traveling at 2.009 m/s when first released.

For part b, I do not know where to begin. The acceleration is -0.4m/s2, as founded in the first part. But I do not know how to apply the information because I get a negative answer for the final velocity for the half way point.

2. Feb 12, 2010

### Redbelly98

Staff Emeritus
Your notation is a little on the sloppy side, but you are basically okay up to here.

Where does this come from? It seems wrong to me.

I got a different answer. Can you explain the stuff beginning with "vi2=0-(-0.2)(20.2)" please?

A negative answer is a problem. Can you show your work, then I can help figure out what is wrong.

3. Feb 12, 2010

### barthayn

vi2=0-(-0.2)(20.2) is from the acceleration found, and the total distance traveled.

The final answer I got 2.009m/s. For b) I got 2.04m/s. To get that answer I did this:

Divided the total distance and time by two.

vf = (2d+a(t)2)/t
vf = (20.2-10)/5
vf = 10.2=5
vf = 2.04 m/s

4. Feb 12, 2010

### Redbelly98

Staff Emeritus
The acceleration is -0.4 m/s2, but that value does not appear anywhere in your equation.
The distance travelled is 20.0 m, but you have 20.2 in your equation.
If you're using the equation I think you're using, check whether it has a factor of 1/2 or 2. OR you could use the simpler a=Δv/Δt that you listed as one of the Relevant Equations.

Actually, it is just the distance that is divided by two, not the time:
"half way position" refers to the distance.

5. Feb 12, 2010

### barthayn

You get the 0.2m plus from the question given. Also, in my notes, I got -0.4 m/s2. So I mistyped the number. The rest is correct.

How would I get the time to get it to reach the half way point. It states that the acceleration is uniform.

6. Feb 12, 2010

### Redbelly98

Staff Emeritus
According to the problem statement, the object comes to rest in 10.0 s.
You correctly calculated that it moves 20 m in those 10 seconds. At that point it is at rest, so it cannot travel an additional 0.2 m. Instead, those 0.2 m comprise the final part of the 20 m total distance (and the final 1.0 s of the 10.0 s total).

It is not correct, since you (1) used the mistyped -0.2 acceleration value to do the calculation, and (2) used a formula that I think is incorrect.

What is the general formula you used, before you substituted values into it? I mean the formula that has vi2=______? My further assistance is contingent upon your providing this information.

You have acceleration and distance. Once part (a) is done correctly, you'll have initial velocity. Three quantities are sufficient to solve any 1-d uniform acceleration problem, you just need to find and use an equation that has a, Δd, and vi.

7. Feb 12, 2010

### barthayn

The new first velocity I got was 4 m/s. Is this correct? For part b I got 2 m/s as well

8. Feb 12, 2010

### Redbelly98

Staff Emeritus
I agree with the 4 m/s. But it would be 2 m/s at half the total time, not half of the distance.

p.s. logging off for the night, will be back tomorrow.

9. Feb 12, 2010

### barthayn

How would I find out the time from this then? I am completely confused because the negative acceleration is uniform, meaning it is constant?

10. Feb 13, 2010

### Redbelly98

Staff Emeritus
Yes, the acceleration is constant.

You have Δd, a, and vi. Look through your list of kinematic equations for an equation that has those three terms in it.

11. Feb 20, 2010

### barthayn

How would I find the time it took to reach the half way distance then? I do not understand how you can get it mathematically. Wouldn't it just be total time divided by two because it is consent negative acceleration.

12. Feb 20, 2010

### Redbelly98

Staff Emeritus
Not, it would have to be constant velocity for it to go half the total distance in half the total time. Hopefully that is common sense. This is not a constant velocity situation, since acceleration is not zero.

But -- guess what? -- you do not even need to know what that time is anyway. Use the kinematic equation that has Δd, a, vi (all of which you know) and vf (which you are trying to find).

Do you, or do you not, have the kinematic equations in your text book or class notes?

13. Feb 20, 2010

### barthayn

The final answer I got was 2.8 m/s. Is this correct?

14. Feb 21, 2010

### Redbelly98

Staff Emeritus
Yes, looks good to me