in this video we're given a velocity time graph and then we're asked to find the instantaneous
acceleration at a particular moment in time and in part b we're asked to find the average velocity on
a time interval we're told the grid spaces can be trusted in this graph and i'll just adjust
my title real quick that's velocity in meters per second on the y-axis and then i have time
in seconds on the x-axis so you can trust the numbers you see in the graph this is one two
three four meters per second for the velocities and then one through six seconds for the times
now to get the instantaneous acceleration at t equals four i go to that point on the graph when
t is four i'm right here and v looks like it's about two and a half well the acceleration at that
moment in time is just the slope of the graph and the easiest way to get the slope is to focus on
integer points here and i have a couple integer points that are connected by that line segment
then i look at the rise between these two points as t goes from three to five and the rise is
going to be a negative number that's a negative three meters per second and the run is a positive
two seconds i could be more formal about it and say that a is going to be delta v over delta t
the slope of the graph and my final velocity here is 1 meter per second my initial velocity
was 4 meters per second my final time 5 seconds my initial time is three seconds and this gives
me the same thing i have a negative three meters per second divided by two seconds and i can see
this simplifies to negative one and a half meters per second per second or meters per second squared
in part b i'm asked to compute the average velocity from t equals zero to t equals six
and average velocity is the change in position divided by the change in time and i have to
remember here that change in position is given by the area under the velocity time graph so
i've got to find the area under this thing and i'm going to think of it as a rectangle
and then a trapezoid the big trapezoid and then a little trapezoid and i have to remember that
the area under a trapezoid is the average of the two heights on the left and right side multiplied
by the width of the base so delta x starts with the area of this rectangle and that has a base
of three seconds and a height of four meters per second and then the area of the large trapezoid
that has a base of two seconds it has one height of four meters per second and then
another height of one meter per second and i take the arithmetic mean of those two heights
so that's one half times their sum one half of four meters per second plus one meter per second
and then times the width of the base which was two seconds and then finally i have the
small trapezoid and i already have one of its heights written in the diagram
and the height on the right hand side there is two meters per second so i add in a one-half
and then at one meter per second plus two meters per second
times the width of the base there which is just one second you can see in all the units
that seconds are going to cancel leaving me with meters which is good because we're talking about
displacement and this gives me a grand total of 18.5 meters displacement then just a little
reminder of what the total change in time is on this interval t went from zero to six so delta t
is equal to 6 seconds and i can finally write down my average velocity displacement was 18.5 meters
time 6 seconds and i get 3.08 meters per second and we're done if you enjoyed this video or at
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