- #1

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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

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- Thread starter tgoot84
- Start date

- #1

- 7

- 0

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

- #2

Integral

Staff Emeritus

Science Advisor

Gold Member

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How are you finding the distance?

Please show us what you have done.

Please show us what you have done.

- #3

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

Just ignore this post.

cookiemonster

Just ignore this post.

cookiemonster

- #4

- 7

- 0

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

Science Advisor

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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

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

Science Advisor

Homework Helper

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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

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

- #7

ShawnD

Science Advisor

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

- #8

Integral

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[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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