Finding Acceleration in a Shopping Cart Rolling Downhill

[Mirole]

Homework Statement

Jill has just gotten out of her car in the grocery store parking lot. The parking lot is on a hill and is tilted 3 degrees. Fifty meters downhill from Jill, a little old lady let's go of a fully loaded shopping cart. The cart, with frictionless wheels, starts to roll straight downhill. Jill immediately starts to sprint after the cart with her top acceleration of 2.0m/s^2.

Homework Equations

The Attempt at a Solution

a(cart)=g*sin3=.51m/s^2

I know I need to use two equations to find the solution, but I am utterly confused on which two to use.

 

[tiny-tim]

Welcome to PF!

Hi Mirole! Welcome to PF!

Jill has just gotten out of her car in the grocery store parking lot. The parking lot is on a hill and is tilted 3 degrees. Fifty meters downhill from Jill, a little old lady let's go of a fully loaded shopping cart. The cart, with frictionless wheels, starts to roll straight downhill. Jill immediately starts to sprint after the cart with her top acceleration of 2.0m/s^2.

Use the usual https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, but use s in one and s + 50 in the other.

 

[Mirole]

Hi Mirole! Welcome to PF!


Use the usual https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, but use s in one and s + 50 in the other.

Alright, thinking about it.

I went:

S(1) = 1/2*a(j)*t^2
S(2) = 50 + 1/2*a(cart)*t^2

I set them equal to each other.

1/2*a(j)*t^2 = 50 + 1/2*a(cart)*t^2

1/2*2*t^2 = 50 + 1/2*.51*t^2
Cross off the t^2s.

1=50+1/2*.51
1=50.255

That's obviously not right, and this problem has given me a mental block. :(

 

[tiny-tim]

Hi Mirole!

(try using the X² tag just above the Reply box )

1/2*2*t^2 = 50 + 1/2*.51*t^2
Cross off the t^2s.

1=50+1/2*.51

"Cross off the t^2s" ??

get some sleep! :zzz:​