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April 7th, 2018, 06:54 PM   #1
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I need help

I need help with question 2..please help me...

These are my answers..please check and help me...

2a) its 0s to 6s
b) 1.8/2=0.9m/s or 1m/s
c) 11-0/-4-0=-2.75m/s
d) 4m
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Last edited by mayak201; April 7th, 2018 at 06:58 PM.
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April 7th, 2018, 07:14 PM   #2
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Originally Posted by mayak201 View Post
I need help with question 2..please help me...

These are my answers..please check and help me...

2a) its 0s to 6s
b) 1.8/2=0.9m/s or 1m/s
c) 11-0/-4-0=-2.75m/s
d) 4m
a) You missed an interval. The velocity is increasing when the slope the graph at that point is positive. Where else does this happen?

b) What you wrote is actually the average velocity of the object between t = 0 s and t = 2 s. Unless the give you a ruler I don't see how you could accurately figure this out, though you might come close.

c) Your slope formula is up-side down. It's $\displaystyle \frac{\Delta d}{ \Delta t}$.

d) Is that at t = 8 s? (The image is a bit fuzzy.) At t = 8 s it looks to me that the object is at d = -2 m. That means the distance from the starting point is 2 m.

-Dan
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April 7th, 2018, 08:41 PM   #3
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2a) 9 to 11s
b) will 1.9 work?
1.9/2=0.95m=1m(round off)
c)-4-0/11-0=-0.36=-0.4m/s
d)actaully its at -2m

I am not sure if I am right and the picture is below

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April 8th, 2018, 04:00 AM   #4
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Originally Posted by mayak201 View Post
2a) 9 to 11s
b) will 1.9 work?
1.9/2=0.95m=1m(round off)
c)-4-0/11-0=-0.36=-0.4m/s
d)actaully its at -2m

I am not sure if I am right and the picture is below

Hmmmm.. Your picture doesn't show for me. Anyway
a) Yes

b) The average slope formula is $\displaystyle \frac{d_f - d_i}{t_f - t_i}$. You are still doing the average velocity from the origin (0, 0) to the final point (2, 1.9) (or so.) You need to pick two points close together on the graph that are close to t = 2 s. If you had access to a ruler of some kind you could simply sketch the tangent to the curve near t = 2 s and measure hte tangent to the curve at that point.

c) -4-0/11-0 needs to be (-4 - 0)/(11 - 0). Parenthesis are needed!

d) Sounds good to me, then.

-Dan
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April 8th, 2018, 11:23 AM   #5
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Thank you
For b) I am actually confused for this one...

Try this link for the graph

file:///C:/Users/Pc/Downloads/kinematics_asmt_graph.png

Last edited by mayak201; April 8th, 2018 at 11:27 AM.
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April 8th, 2018, 01:59 PM   #6
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Quote:
Originally Posted by mayak201 View Post
Thank you
For b) I am actually confused for this one...

Try this link for the graph

file:///C:/Users/Pc/Downloads/kinematics_asmt_graph.png
That seems to be a link to something on your computer which no one (that I no of) have access to. Please try to post a larger image.
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April 8th, 2018, 02:05 PM   #7
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https://ibb.co/ijuUUx

Try this link
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April 8th, 2018, 02:37 PM   #8
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ok ... it's a position vs time graph. what about it? looks like topsquark provided the requisite help in posts #2 and #4

Last edited by skeeter; April 8th, 2018 at 02:58 PM.
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April 8th, 2018, 03:50 PM   #9
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I do not know how to solve for b...
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April 8th, 2018, 03:59 PM   #10
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The equation you are using (the slope formula) is what you want to use, but as the graph has a slight curvature there we can't make $\displaystyle v = \frac{d_f - d_i}{t_f - t_1}$ come from the origin. What you are doing is finding the average v from t = 0 s to t = 2 s. This is not correct.

What you need to find is the instantaneous velocity at t = 2 s. Take a look at the graph. What you need to do is find the slope of the blue line, that is tangent to the curve at t = 2 s.

-Dan
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