WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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To find the velocity of
this jet car at 20 seconds
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we need to take a tangent line
of the graph at 20 seconds.
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So here's the 20
second mark here
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and I've drawn this
tangent line in red here
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and the tangent touches the curve
at one point and only one point.
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We find the slope of
this tangent line.
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This is also the instantaneous
velocity at 20 seconds.
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So the slope is the rise
divided by the run,
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so the amount by which
this line goes up,
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divided by the amount by
which it goes sideways.
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We have the change in position
seems to go from about here
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at about 250 meters up
to about 2250 meters
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for a change in position
of 2000 meters.
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For the run,
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it goes from about
10 seconds over to 30 seconds
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which make the change in time
of 20 seconds, 30 minus 10.
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So the velocity is change of
position divided by change in time,
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2000 meters divided by 20 which
is about 100 meters per second.
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Then to find its acceleration, we take the
slope of the velocity versus time graph.
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The rise here -- now this is
a bit of a technical point,
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but I've chosen to
take my slope triangle
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to not use any of the data points
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because the reason this
the line is being drawn,
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is the line is a kind of average
of all the data points.
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So your analysis of your graph
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is better if you're
using this average.
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Otherwise, if you take your slope
triangle from particular points
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to another particular point,
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then you may as well not have even
taken these other data points
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because they're not
being used in any way.
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But if you take
your slope between
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the points that are on just the
line and not actual data points,
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then you're kind of -- this line is
an average containing information
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from all of the data points. So all
of the data points were useful.
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This is kind of a
contrived graph obviously
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because you know,
there is no real data
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that would have a line going
perfectly through each point
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and so you don't really see any
difference in this particular case.
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But it's a general best practice
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to take your slope triangle ending on
the line itself and not on data points.
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Okay. So the slope
is rise over run,
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the rise being the change in velocity
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and it looks like this goes from
about 150 or so down to 30.
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That makes the change in velocity
of 120 meters per second
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and the time over which that
change of velocity occurs
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seems to end at about 27 seconds
there and starting at about two,
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for a total of 25
seconds time interval.
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So acceleration is change of
velocity over change of time,
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120 divided by 25 giving an answer
of 4.8 meters per second squared.
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So we've basically
confirmed that the jet car
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is accelerating at 5 meters
per second squared, '
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cause that's pretty close to 5.