Jill has just gotten out of her car in the grocery store parking lot.

Question

Jill has just gotten out of her car in the grocery store parking lot. The lot is on a hill and is tilted 3.0°. Twenty meters downhill from Jill, a little old lady lets 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 2.0 m/s2. How far has the cart rolled before Jill catches it?

 

Using x as the variable for position along the hill, and subscripts of “C” for the cart and “J” for Jill, arrive at expression for the positions of the cart and of Jill to arrive at the solution. 

Answer 1

Mia,

The first thing is to figure out the acceleration of the shopping cart.

Gravity, pulling straight down, can be split up into vector components which are perpendicular and parallel to the slope.

The parallel component provides the acceleration of the cart. The perpendicular would only be used if friction were involved.

 

a_c = 9.81sin(3) = 0.51 m/s²

 

Now we use the equation x = x₀ + v₀t + 1/2at² to find when there positions are equal.

Because v₀ = 0 for both, I'll drop that term from the computation.

 

x_j = -20 + 1/2(2)t²           x_c = 0 + 1/2(.51)t²  

 

Jill catches the cart when x_j = x_c. So set them equal:

 

1/2(2)t² -20 = 1/2(.51)t²

 

         t² = 20 + .255t²

 

   .745t² = 20

 

         t² = 20/.745

 

          t = 5.18 sec

 

x_c = 1/2(.51)(5.18)² = 6.8 meters   (rounded to two sig figs per the given data)