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-   -   Shopping cart down a hill (https://www.askmehelpdesk.com/showthread.php?t=396442)

  • Sep 14, 2009, 03:18 PM
    Magomad
    Shopping cart down a hill
    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. :(
  • Sep 15, 2009, 03:01 PM
    ebaines

    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

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