# Uniform acceleration

1. Sep 27, 2011

### sdoi

1. The problem statement, all variables and given/known data
A child pushes his friend in a wagon along a horizontal road and then lets go. The wagon rolls for 10.0 seconds before stopping. It travels a distance of 20.0cm during the last 1.00s of its motion. Assuming uniform acceleration,
(a) How fast was the wagon travelling at the instant the child released it?
(b) How fast was the wagon travelling when it had covered half of the distance?

2. Relevant equations

3. The attempt at a solution
I know this is a really basic question, but i've been over analyzing it and now i've gotten myself all confused. My first problem is working with uniform acceleration. Does that mean every 1.0s the wagon travels 0.20m? Working under that assumption, I came up with:

(a)Vi= 0.20m x 10.0s
Vi= 2m/s
Therefore, as the wagon was released it was travelling 2 m/s[ along the horizontal]
(b) Still undergoing uniform acceleration:
0.20m x 5.0s
=1m/s
Therefore, the wagon was travelling exactly half of the speed it was travelling initially.

2. Sep 27, 2011

### bacon

No, that would be uniform(or constant) velocity, not acceleration. Constant acceleration means the the velocity is changing at a constant rate.
Have you been introduced to the equations of motion with constant acceleration? You will need to use one or more of them. They will look like these:
http://en.wikipedia.org/wiki/Equations_of_motion
at the beginning of the article.

3. Sep 27, 2011

### sdoi

Ok, so then for (a) I am assuming my best bet is to work with v= vi + a(delta)t or,
v^2= vi^2+2a(s-si)

4. Sep 27, 2011

### sdoi

For question a i'm trying to find Vi
If I use the equation v= Vi+a(delta)t I have too many unknowns
v= d/t
a= vf^2/ 2(delta)d
vf= vi + at

For v= d/t:
v= 0.20m/1.0s
v= 0.2 m/s

(0.2m/s)= Vi + a(10.0s)
But, I still have too many unknowns.

5. Sep 27, 2011

### flyingpig

No you are going to have to use

$$\Delta x = -\frac{at^2}{2} + v_0 t$$

I will give you a hint. What is the speed of the wagon when it stops?

Also, what a naughty child, how could he push his friend like that?

6. Sep 27, 2011

### bacon

Yes, there are too many unknowns to solve in one step. The information given for the last second of travel will allow you to find the acceleration, and since it is constant you will then be able to apply it to the whole 10 sec roll. If you could find the velocity at the beginning of the last second of roll, do you see how you could use that to find the acceleration?

7. Sep 27, 2011

### sdoi

So, for the last second of travel:
t=1.0s
Δd=0.20m
vf=0m/s, Since the wagon comes to a complete stop
vi=?
would vi= d/t?

8. Sep 27, 2011

### flyingpig

$$\Delta x = -\frac{at^2}{2} + v_0 t$$

Here you go.

9. Sep 27, 2011

### bacon

No, that would give the average velocity during the last second of travel. Do you see an equation that has initial and final velocity, time and distance that you could use for the last one second of travel?

10. Sep 27, 2011

### sdoi

I feel like I may get deducted marks for using this equation, its no where in my textbook, but i'll give it a try.
To first solve for a:
v= d/t
v= (0.20m)/(1.0s)
v=0.2m/s

a=v/t
a=(0.2m/s)/ (0.1s)
a=2m/s^2

Then using the equation given:
(0.20m)=- (2m/s^2)(1.0s)/2 + Vi(1.0s)
(0.20m)=-1m/s + Vi
0.20m+1=Vi
Vi=1.2m/s

11. Sep 27, 2011

### sdoi

there is d= (vi+vf)/2 (t)
Is my previous calculation wrong then?

12. Sep 27, 2011

### sdoi

d= (vi+vf)/2 (t)
(0.20m)= (vi+0m/s)/2 (1s)
(0.20m)= Vi/2
Vi= (0.20m)2
Vi= 0.4m/s

13. Sep 27, 2011

### bacon

That looks right to me. Do you see how you can use this Vi to find acceleration?

I'm not following flyingpig(although I think we must be related), the equation he gave seems to me to be unusable at this point, only two of the four variables are known.

14. Sep 27, 2011

### sdoi

Haha, I noticed. And yes, I was quite lost.
Now, the Vi is only for the last one second, so at 9s the wagon is moving 0.4 m/s, but with that how do I find the actual initial velocity?
a= vf-vi/2
a=(0m/s-0.4m/s)/2
a=-0.2m/s^2

v= vi + a(delta)t
v=d/t
But, what is d now?

15. Sep 27, 2011

### sdoi

wait, to find how fast the wagon is travelling at the instant the child released it:
a= (Vf-Vi)/t
-2m/s^2= (0m/s-Vi)/ 10.0s
0m/s-Vi= -20m/s
Therefore Vi= 20m/s

16. Sep 27, 2011

### bacon

I'm not quite with you here.
You don't have to divide the initial velocity by 2. You know Vi,Vf and t, but don't know a(I am still referring to just the last second of travel, once we know a, we can move to the whole 10 second period). Is there an equation that relates these four variables?

17. Sep 27, 2011

### sdoi

For the acceleration of the last second I used the equation
a=(Vf-Vi)/2
a=(0m/s-0m/s)/2
a=-0.2m/s^2
Can I use this value for a when calculation the whole 10 second period?

18. Sep 27, 2011

### bacon

I was thinking of the equation
$V=V_{0} + at$
Does that look familiar?

19. Sep 27, 2011

### sdoi

0m/s= (0.4m/s) +a
so a=-0.4m/s^2

Vf= Vi +at
0m/s= vi + (-0.4m/s^2)(10.0s)
-Vi=(-0.4m/s^2)(10.0s)
Vi=40m/s

20. Sep 27, 2011

### bacon

This looks good to me except for a math error.
.4 x 10 =4