Finding acceleration from a distance vs time graph of a puck on a slope

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saad1109
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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


I WOULD DEEPLY BE THK YOU TO U IF U SOLVE THIS!
 
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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.
 


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
 


saad1109 said:
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.
 


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