Kinematics - identifying and describing types of motion

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The discussion focuses on calculating velocity and acceleration in kinematics using tangent slopes from a velocity-time graph. The user initially calculates acceleration but questions the reliability of their velocity data points, particularly those derived from drawing tangents by eye. It is noted that this method can introduce significant error, and a more reliable velocity can be obtained from known values at specific time points. The conversation emphasizes the importance of accuracy in drawing tangents and suggests revising the tangent at 0.5 seconds for improved results. Overall, the exchange highlights common challenges in kinematic calculations and the need for precise methods.
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Homework Statement


upload_2016-3-21_22-46-33.png


Homework Equations



acceleration equation:
a = v2 – v1 / t2 – t1

The Attempt at a Solution


Can someone please tell me if my answers make sense. Thanks in advance :)[/B]

21.
upload_2016-3-21_22-47-32.png


22.
to calculate the velocity, i determined the slope of the tangents, postioned at 0.50s, 0.75 s, and 1.00 s

upload_2016-3-21_22-48-12.png

Using this information, i created a velocity time graph for the orange
upload_2016-3-21_22-51-52.png


Acceleration:

a = v2 – v1 / t2 – t1

a = 8.5 m/s (down) – 6.7 m/s (down) / 1.00 s – 0.75 s

a = 7.2 m/s2 (down)
 

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Not bad, but which of all the available datapoints for velocity should be the most reliable?
 
what do you mean? are you referring to the velocity i calculated in my time-velocity data table? and what do you mean by most reliable? arent they all reliable?
 
alexandria said:
what do you mean? are you referring to the velocity i calculated in my time-velocity data table? and what do you mean by most reliable? arent they all reliable?
They're not wonderfully reliable because you are drawing tangents by eye and finding the velocities from their slopes. I can see, for example, that your tangent at 0.5s is a bit too steep.
There is one completely reliable velocity that you have not used. You do not even have to draw a tangent to find it!
 
i don't understand :P
 
alexandria said:
i don't understand :P
Which bit don't you understand, the inaccuracy inherent to the method you have been told to use to find velocities, or what data point you have not used?
 
both
i was just told to draw tangents and then calculate the slope of each tangent to determine the velocity
here's an example from my lesson:
upload_2016-3-22_2-42-30.png

upload_2016-3-22_2-41-55.png


upload_2016-3-22_2-43-24.png

upload_2016-3-22_2-43-44.png
 
alexandria said:
i was just told to draw tangents and then calculate the slope of each tangent to determine the velocity
Yes, I understand that, but you must realize that drawing tangents by eye like that is open to substantial error. The tangent you drew at 0.5s looks to me more like the tangent at 0.55s.
Since you are following a worked example, I suppose you cannot do much better than that. But had I prepared the example I would have taken the velocity at time 0 as an obvious known value. Note that using that value together with your t=1 tangent gives a rather more accurate value for g.
 
ok i understand, should i fix the tangent at 0.5 s, will that make it more accurate?
 
  • #10
alexandria said:
ok i understand, should i fix the tangent at 0.5 s, will that make it more accurate?
Yes, a bit.
 
  • #11
ok thank you.
 
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