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View Full Version : Shopping cart down a hill


Magomad
Sep 14, 2009, 03:18 PM
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 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 2.0m/s^2.

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. :(

ebaines
Sep 15, 2009, 03:01 PM
You were fine right up until you said "cross off the t^2s." You can't just get rid of them, because there isn't a t^2 in every term. What you do is group the like terms together:

1/2*2*t^2 = 50 + 1/2*.51*t^2
t^2 = 50 +1/2*.51 t^2
(1 - .51)t^2 = 50
t^2 = 50/(1-.51) = 102
t = 10.1 s