Uniform acceleration of moving wagon

[sdoi]

Homework Statement

A child pushes his friend in a wagon along a horizontal road and then let's 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 traveling at the instant the child released it?
(b) How fast was the wagon traveling when it had covered half of the distance?

Homework Equations

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

 

[bacon]

My first problem is working with uniform acceleration. Does that mean every 1.0s the wagon travels 0.20m?

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.

 

[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)

 

[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.

 

[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?

 

[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?

 

[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?

 

[flyingpig]

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?

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

Here you go.

 

[bacon]

...
would vi= d/t?

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?

 

[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

 

[sdoi]

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

 

[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

 

[bacon]

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

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.

 

[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^2v= vi + a(delta)t
v=d/t
But, what is d now?

 

[sdoi]

wait, to find how fast the wagon is traveling 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

 

[bacon]

a= vf-vi/2
a=(0m/s-0.4m/s)/2
a=-0.2m/s^2

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?

 

[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?

 

[bacon]

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

 

[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

 

[bacon]

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

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

 

[sdoi]

Ok thank you very much for all of your patience.
Just for question b does this look right?

d= [(Vi+Vf)/2] t
d= [(4.0m/s)/2](10s)
d=20m
Therefore for half the total distance:

Vf^2=Vi^2 +2aΔd
Vf^2= (4.0m/s)^2 + 2(-0.4m/s^2)(5m)^2
Vf^2=8m/s
vf=2.82 m/s

 

[bacon]

That's the answer I got.