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

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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 2018-04-07 (1).jpg (14.2 KB, 14 views)

Last edited by mayak201; April 7th, 2018 at 06:58 PM.

April 7th, 2018, 07:14 PM   #2
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by mayak201 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

 April 7th, 2018, 08:41 PM #3 Newbie   Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0 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
April 8th, 2018, 04:00 AM   #4
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Quote:
 Originally Posted by mayak201 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

 April 8th, 2018, 11:23 AM #5 Newbie   Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0 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.
April 8th, 2018, 01:59 PM   #6
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Quote:
 Originally Posted by mayak201 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.

 April 8th, 2018, 02:05 PM #7 Newbie   Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0 https://ibb.co/ijuUUx Try this link
 April 8th, 2018, 02:37 PM #8 Math Team   Joined: Jul 2011 From: Texas Posts: 2,771 Thanks: 1426 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.
 April 8th, 2018, 03:50 PM #9 Newbie   Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0 I do not know how to solve for b...
 April 8th, 2018, 03:59 PM #10 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,889 Thanks: 769 Math Focus: Wibbly wobbly timey-wimey stuff. 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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