grocery cart physics problem?

Question

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.
How far has the cart rolled before Jill catches it?

Answer 1

The acceleration of the cart is ac = g sine 3 degrees = 0.51 m/s^2. The initial position of the cart is 50 m more than jill's. Jill's acceleration is 2 m/s^2 whether gravity helps or not. The initial velocity of both is 0. Setting their two position equations equal,

yc = yj

yc = yoc + 0.5gct^2

yj = y0j + 0.5gjt^2

when yc = yj,


yoc + 0.5gct^2 = yoj + 0.5gt^2

50m + 0.5*0.510m/s^2 t^2 = 0 + 0.5*2m/s^2* t^2.

solve for t;

t = sqrt [ (100m/(1.49 m/s^2))] = 8.19 s

Now it takes jilll 8.2 s to catch up with the cart. In that time the cart has rolled y = 0.5 * 0.51* 8 19^2 m = 17 m.

Answer 2

Assume Gravity is 10 meters per second (if this is your homework, you can redo it with 9.8) then the lateral acceleration is sine(3) times 10. Since we know sine (3) is in radians sine (2pi/360*3) or sine (.1) which is about .1 since for small angles in radians the value of sine is equal to the number, the acceleration is about 1 m/s^2. Therefore the difference in accleration between Jill and the cart is 1 meter/s^2. Using x=1/2 a t^2 and substituting the seperation in our moving reverence frame of 50 = 1/2 *1 *t^2 we get t =10. So it takes Jill ten seconds to catch the cart. In that 10 seconds the cart moved an additional fifty feet x = 1/2 *1*10^2.
And to check Jill moved x = 1/2*2*10^2 = 100 feet.

Answer 3

The cart would only move 3 inches. Frictionless wheels? No such thing in reality. There isn't a shopping cart on this planet that doesn't have at least one busted wheel. Not to mention the parking lot is NOT a flat, featureless surface. It's got cracks, dips, hills, "islands" with trees, cars, other people and so on.

50 meters is a long distance. Us non metric folks would know it to be ~ 150 feet away (More or less.) That's halfway across the parking lot. The old lady is closer, she should go running after it.

The problem list's Jill's top speed, but doesn't mention if she asthma. "Top Speed" is NOT a consistent speed. And what if Jill is in a bad mood and decides not to go running after the cart? She's not required to do so.

Answer 4

8.20 seconds and 67.2 meters for her and 17.2 meters for the cart


The acceleration of the cart is 9.80*sine (3degrees), or about 0.51289...
The girl and cart must travel the same distance, therefore
d girl = (1/2) * 2.0 m/s^2*t^2 while
d cart = (1/2) * (9.80 * sine (3)) * t^2 + 50

Set the two equations equal and you get
t^2 = 0.256t^2 + 50
solve agebraically and you get...
8.20 seconds...tada!

I would appreciate the best answer vote, and sorry I took so long, but my dog needed to potty.

Answer 5

Where is this store with the frictionless wheeled carts? I want to go there. They would be so much fun.

Answer 6

Not far. The $100,000 motorhome just downhill from the old lady takes the brunt of the assault.

Answer 7

find acceleration of the cart
a=sin(3)g
a=.513m/s^2

since Jill and the cart start from rest, their intial velocity is 0m/s
1/2at^2+Vt+Xi=1/2at^2+vt+Xi
1/2(2m/s^2)t^2-50m=1/2(.513m/s^2)t^2
(1m/s^2)t^2-50= (0.2565m/s^2)t^2
(0.7435m/s^2)t^2=50m
t= 8.2s

X=1/2at^2
X=(1/2)(0.513m/s^2)(8.2s)^2
x=17.24m

Answer 8

(1/2)*2.2*t² = 50 + (1/2)*9.9*sin(3)*t² so t = 7.2793s
s = (1/2)*9.8*sin(3)*(7.2793)² = 13.58m


Doug