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

Integral
Staff Emeritus
Gold Member
How are you finding the distance?

Please show us what you have done.

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

Just ignore this post.

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

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:
HallsofIvy
Homework Helper

Originally posted by tgoot84
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

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?

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

Integral
Staff Emeritus
$$x= \frac 1 2 a t^2 + v_0t + x_0$$
$$v_0 =0$$ and $$x_0 = 0$$