1. Feb 23, 2004

### tgoot84

In a grocery store, you push a 14.5 kg shopping cart with a force of 11.0 N. If the cart starts at rest, how far does it move in 3.00 s?

i've got that acceleration = force/mass

so the acceleration i got was .7586
but for some reason i keep getting the answer to the problem wrong

_____m

2. Feb 23, 2004

### Integral

Staff Emeritus
How are you finding the distance?

Please show us what you have done.

3. Feb 23, 2004

Think I'll just let Integral handle it...

Just ignore this post.

4. Feb 23, 2004

### tgoot84

my work

as i got .7586 m/s2 as the acceleration,

i multiplied by 3 seconds to get 2.28 m/s

then i multiplied 2.28x 3 seconds to get 6.83 meters

5. Feb 24, 2004

### ShawnD

But it wasn't going 2.28m/s the whole time. That's the problem.

You know the velocity formula obviously. Just integrate that to get a distance formula.

$$d = \int V dt$$

$$d = \int (V_i + at) dt$$

$$d = V_it + \frac{1}{2}at^2 + d_o$$

Sorry for making it sound complicated, I just wanted to write that

Last edited: Feb 24, 2004
6. Feb 24, 2004

### HallsofIvy

Staff Emeritus
Re: my work

ShawnD pointed out your error: the final speed, at the end of the 3 seconds is 2.28 m/s. It was not going at that speed all the time. If you have not taken calculus, you can use this "short cut": as long as the acceleration is a constant, you can find the "average" speed by averaging the first and last speeds.

The shopping carts initial speed was, of course, 0. After 3 seconds, it had a speed of 2.28 m/s. The average speed was (2.28+0)/2= 1.14 m/s. How far did the shopping cart go at an average speed of 1.14 m/s for 3 seconds?

7. Feb 24, 2004

### ShawnD

Oh cool. I didn't know you could do that.

8. Feb 24, 2004

### Integral

Staff Emeritus
One of the given equations in a non calculus class is something like

$$x= \frac 1 2 a t^2 + v_0t + x_0$$

For this problem

$$v_0 =0$$ and $$x_0 = 0$$

you have a and t, simply do the computation to get the same result obtained above.