Easy problem but i'm getting something wrong please help

  • Thread starter tgoot84
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  • #1
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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
 

Answers and Replies

  • #2
Integral
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How are you finding the distance?

Please show us what you have done.
 
  • #3
Think I'll just let Integral handle it...

Just ignore this post.

cookiemonster
 
  • #4
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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
ShawnD
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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.

[tex]d = \int V dt[/tex]

[tex]d = \int (V_i + at) dt[/tex]

[tex]d = V_it + \frac{1}{2}at^2 + d_o[/tex]

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

The answer should be 3.4137m
 
Last edited:
  • #6
HallsofIvy
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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?
 
  • #7
ShawnD
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Oh cool. I didn't know you could do that.
 
  • #8
Integral
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One of the given equations in a non calculus class is something like

[tex] x= \frac 1 2 a t^2 + v_0t + x_0 [/tex]

For this problem

[tex] v_0 =0 [/tex] and [tex] x_0 = 0 [/tex]

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

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