Solve Mass of Dog Food Given Force, Acceleration, Time

[Pencil]

At the local grocery store, you push a 15.2 kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you accelerate the cart from rest through a distance of 2.24 m in 3.00 s. What was the mass of the dog food?
F=ma m=f/a
I rearranged x(f)=x(i)+v(i)+1/2at^2 to be a=(1/2t^2)/x
so,
1/2(3)^2/2.24= 2.01 m/s^2.

m=f/a
m=12.0/2.01=5.97kg.
This defenately looks wrong. How can the cart's mass of decreased? I'm confused what did I do wrong?
I tried 15.2-5.97 and that was also wrong.

 

[blackwizard]

I rearranged x(f)=x(i)+v(i)+1/2at^2 to be a=(1/2t^2)/x

The only thing i regognise there is 1/2at^2.
This wudv worked:

I rearranged x = vt +1/2at^2 to be a= 2x/t^2

 

[Pencil]

awsome, got it. 8.80kg. I don't know how i mess these up. What are the steps you use to convert x = vt +1/2at^2 to a= 2x/t^2? thanks for your help.

 

[BobG]

awsome, got it. 8.80kg. I don't know how i mess these up. What are the steps you use to convert x = vt +1/2at^2 to a= 2x/t^2? thanks for your help.

Initial velocity was 0.

Multiply both sides by 2 to get rid of the 1/2.

Divide both sides by t^2 to isolate your a.