April 7th, 2018, 06:54 PM  #1 
Newbie Joined: Apr 2018 From: Canada Posts: 9 Thanks: 0  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) 110/40=2.75m/s d) 4m Last edited by mayak201; April 7th, 2018 at 06:58 PM. 
April 7th, 2018, 07:14 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,159 Thanks: 878 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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 upside 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)40/110=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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,159 Thanks: 878 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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) 40/110 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 
Newbie Joined: Nov 2013 Posts: 28 Thanks: 8  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  
April 8th, 2018, 02:37 PM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521  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: 2,159 Thanks: 878 Math Focus: Wibbly wobbly timeywimey 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 