easy problem but i'm getting something wrong please help

[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

theanswer is in meters

_____m

[Integral]

How are you finding the distance?

Please show us what you have done.

[cookiemonster]

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

Just ignore this post.

cookiemonster

[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]

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 [:D]

The answer should be 3.4137m

[HallsofIvy]

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]

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.