I already sketched the slope of lab result to find a value. But I need to do it in Excel which I will include in the lab report. But the slope value is different.

The one I got from sketching in the lab is y = 3333.333x which lead to a 12% error.

But when I use Excel trendline with intercept 0,0, the value is very different as shown in the picture. It's y = 0.696x which seems wrong.

The value that I put is the same as the paper:

enter image description here


I think you need to look into a few points here:

1) Based on your graph I would not use linear interpolation in Excel. Your data resembles a polynomial shape as opposed to a linear one. If you're looking for a trendline that matches your curve, try different polynomial trendlines (quadratic, cubic, etc.) until you get something that closely matches.

2) How did you sketch the slope of the curve in your lab? Since the slope of your curve is constantly changing, you could get a lot of error based on what two points you used to measure the slope.

  • 2) In the lab I got 2 type of value to plot in the graph which is D^2 and Vt. I sample the value 5 times and plot the graph 5 spot. After that, I draw the linear trendline intercept with 0,0 and try to close with those spots. Lastly, I find slope from the trendline by delta(y)/delta(x). – C. Nititorn Jul 7 '18 at 14:43
  • I have to use Linear equation y = mx+c which when the trendline intercepts 0,0 the c is 0. The value from hand sketching slope which has some error. But the value I got from excel is the worst, the value not even close with the theory. So maybe I've done something wrong. Dunno why tho. – C. Nititorn Jul 7 '18 at 14:48
  • Also sorry for my bad English. – C. Nititorn Jul 7 '18 at 14:52
  • If you have to use linear interpolation, try changing the y-intercept to the first y data point on your graph. Then draw a trendline in excel. The y-intercept of your curve is not zero, it is some other value. You can notice this by observing that your curve will not hit y = 0 at x = 0. – Grady F. Mathews Iv Jul 7 '18 at 15:12
  • @C.Nititorn, I assume Vt is velocity at time t and D is distance at time t. If the action is starting from "rest" (zero velocity), the curve should pass through zero. If motion is already underway at time zero (you're graphing a window on action that is already underway), the velocity will be greater than zero at your starting point (which your graph seems to show). Otherwise, the graph might look like that if your values are inaccurate. So the intercept of your data won't be zero. (cont'd) – fixer1234 Jul 7 '18 at 21:52

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