Graph of Motion Pls help

  • Thread starter saad1109
  • Start date
  • #1
5
0
Graph of Motion Pls help urgent

Case: We are performing a lab in our class in which on a slop (made from books) is made. Then we graph the motion of a puck via a machine which places a parabola type curve on the graph lick a ticker tape timer. The points were plotted under the motion of the puck.
the question i would like for u to solve would be something like this:

In a

Distance vs Distance Graph, or Displacement vs Displacement graph,
given interval between each point is .05 seconds, how would you find the average acceleration or the acceleration of the object, if the graph looks like this :

ONLY LOOK AT HALF OF THE GRAPH, THAT IS THE RIGHT HALF & IGNORE ANY LABELLING.

http://xamplified.com/wp-content/uploads/2009/03/image0042.jpg [Broken]


I WOULD DEEPLY BE THK YOU TO U IF U SOLVE THIS!!!!
 
Last edited by a moderator:

Answers and Replies

  • #2
phinds
Science Advisor
Insights Author
Gold Member
16,738
7,425


You should read the forum rules regarding posting homework. You are supposed to show what you know and how far you can get, and then you'll get help to take you further. Asking for someone to just solve your problem for you isn't the way this forum works.
 
  • #3
5
0


ohh sorry I'm new...i just didn't know...i tried it by getting the velocity between each points....given the distance and time between the points...but i am unable to go any further and find the acceleration
 
  • #4
DaveC426913
Gold Member
19,195
2,682


ohh sorry I'm new...i just didn't know...i tried it by getting the velocity between each points....given the distance and time between the points...but i am unable to go any further and find the acceleration
Show. Us. Your. Work.
 
  • #5
5
0


ok would this work fine:


*Velocity of each interval:

V1 = 0.02m /0.05s = 0.4 m/s

V2 = 0.025m /0.05s = 0.5 m/s

V3 = 0.015m /0.05s = 0.3 m/s

V4 = 0.018m /0.05s = 0.36 m/s

V5 = 0.016m /0.05s = 0.32 m/s

V6 = 0.019m /0.05s = 0.38 m/s

V7 = 0.022m /0.05s = 0.44 m/s

V8 = 0.021m /0.05s = 0.42 m/s

V9 = 0.025m /0.05s = 0.5 m/s

*Vav =

0.4 m/s + 0.5 m/s + 0.3 m/s + 0.36 m/s + 0.32 m/s + 0.38 m/s + 0.44 m/s + 0.42 m/s + 0.5 m/s
9

= .402 m/s

* Total Distance = 0.02m + 0.025m + 0.015m + 0.018m + 0.016m + 0.019m + 0.022m + 0.025m
= .181m

* Total Time = .05 x 9
= .45 sec

Total Time Total Distance Average Velocity Acceleration
.45 sec .181m .402 m/s ?

Formulae: d = (v)(t) + ½(a)(t^2) or a = 2(d – (v)(t) / t^2)

a = 2(0.181 – (.402)(.45) / .45^2)
= 9.9 x (10^-3) m/s^2= or 0.00099 m/s^2
 

Related Threads on Graph of Motion Pls help

  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
12
Views
2K
Replies
13
Views
3K
  • Last Post
Replies
4
Views
3K
Top