Bo: Hey guys.
Billy: Hey Bo.
Bobby: Hi Bo.
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Mr. P: Ladies and gentlepeople
the bell has rung, therefore class has begun.
Therefore you should be seated in your seat,
ready and excited to do a graphical,
uniformly accelerated motion problem.
Billy: Uniformly.
Bo: I thought it was uniformally.
Bobby: Nope, it's uniformly.
Mr. P: (laughing) Thanks, I'm glad to see we can all
say "uniformly" correctly now.
Today we have a problem.
The problem specifically states:
Assuming an initial position of zero, complete the empty
graphs, you may assume two sig figs.
So we have a velocity as a function of time graph.
You can see we have one, two, and three seconds
on all of our time graphs.
You can see we have a blue line
that represents the motion.
And we have two, four, and six
meters per second, for our velocities.
And we have two empty graphs,
a position and an acceleration graph.
And our goal is to fill in those graphs.
Now we've done this in a broad sense, without numbers,
but today we're gonna have numbers.
Bobby, before we begin, could you
please tell me, how do we know
from looking at the graph, that this is
uniformly accelerated motion?
Bobby: The slope of the line is a constant number.
And the slope of a velocity vs. time graph is acceleration
and therefore the acceleration is constant.
Mr. P: Great, constant slope on a velocity vs. time graph
means we have a constant acceleration.
And how are we gonna find that acceleration?
Billy: We use the equation definition of acceleration.
Acceleration equals change in velocity over change in time.
Bo: Actually, we should use the UAM equation.
Velocity final equals velocity initial
plus acceleration times change in time.
Billy: No we should clearly use the definition.
Bo: Actually no, we should use the UAM equation.
Billy: Definition.
Bo: UAM equation.
Billy: You are so wrong.
Bo: Me? Yeah, right.
Billy: Yeah, you are wrong.
Bo: UAM equation.
Billy: I don't think so.
Bobby: Are you both delirious and daft?
They're the same thing!
Bo: No way. Billy: What?
Bo: Not possible.
Mr. P: Actually, Bobby you are correct.
If you start with the equation for acceleration,
acceleration equals change in velocity over change in time,
"change in" is always final minus initial
so we have velocity final minus velocity initial
Then if we multiply both sides by the change in time
you get the acceleration
times the change in time
is equal to velocity final minus the velocity initial.
You could then add the velocity initial
to both sides and switch sides here we get
velocity final equals the velocity initial
plus the acceleration times the change in time.
In other words, these two equations are equivalent.
Bobby, you're right.
Billy: Well, I'll be!
Bo: Okay, I'm down with that.
Very nice, Bobby.
Mr. P: Acceleration equals
change in velocity over change in time.
Delta change is always final minus initial,
so we get velocity final minus velocity initial on top
divided by the quantity time final
minus time initial on the bottom.
Billy, can you please figure out the numbers for me.
Billy: Well, we need to pick two points on the graph.
And because the slope is constant,
it doesn't really matter which two points we pick.
We will get the same answer.
So let's go from zero to three seconds.
Velocity final is then six meters per second
and velocity initial is zero.
Time final is three seconds and time initial is zero.
That gives us 2.0 meters per second squared
for our acceleration.
Mr. P: And how do we draw a constant acceleration
of 2.0 meters per second squared
on our acceleration vs. time graph, Bo?
Bo: Because the acceleration doesn't change,
draw a horizontal line at the
acceleration value of 2.0 meters per second squared.
Mr. P: Great, now all we need
is the position as a function of time graph.
We already have the initial position
from the problem statement.
The initial position is zero.
Bobby, what will the general shape
of the position vs. time graph be?
Bobby: Honestly, I have no idea where to start.
Billy: Remember that the slope of a
position vs. time graph is velocity.
Bo: And on the graph, the initial velocity
is zero and increases from there.
Bobby: Okay, thanks.
So the slope of the position vs. time graph
needs to start out at zero and then increase over time.
So it's an upward sloping curve.
(chuckles) You're gonna need your
magic tangent line finder to show us that.
Mr. P: (laughs)
Got it.
Magic tangent line finder.
So yeah, it's gonna be an upward sloping curve,
generally something like this.
So you can see the initial slope
of the tangent line needs to be equal to zero
because the initial velocity was equal to zero.
And as time goes by,
the slope of the position as a function of time graph
needs to increase because the velocity increases.
Therefore you can see, using our magic
tangent line finder, that the slope
of this curve increases as a function of time.
Now, this is just an approximation.
We actually need to go through now
and figure out the numbers.
Bobby: It's UAM.
We can use the UAM equations.
Billy: Use displacement equals one half
times the quantity of velocity final
minus velocity initial times change in time.
Oh and the change in time would then be
the time final minus the time initial.
Mr. P: Absolutely, all we need now is the numbers.
I'll go through and do this one.
So I chose three different displacements to solve for,
the displacement from zero to one seconds,
the displacement from zero to two seconds,
and from zero to three seconds.
I'll walk through the first one.
The displacement from zero to one seconds
is going to be one half times the velocity
at one second plus the velocity at zero seconds
multiplied by the time at one second
minus the time at zero seconds.
The numbers are then gonna be
one half times the quantity of two plus zero
then times the quantity of one minus zero
and we get 1.0 meters.
Doing the same thing for the displacement
from zero to two seconds,
we end up with four meters.
And the displacement from zero to three seconds
ends up being nine meters.
Bo, I'm now going to erase this graph
of position as a function of time
because this was just an approximation.
Tell me, how are we gonna take this information now
and draw our position as a function of time graph?
Bo: You plot the three points
and then add the upward sloping curve.
At one second, the position would be one meter.
At two seconds, the position is two meters.
And at three seconds, the position is nine meters.
Then you just draw on the upward sloping curve.
Mr. P: Very nice everybody.
That completes the problem.
You probably noticed one of the other things
that I do no excel at, is drawing curves.
Oh well.
Any questions on the problem we just went through?
Billy: It all makes sense to me.
Mr. P: Great!
Let's just go through and do a quick review then.
We started out by showing that because the slope
of the velocity as a function of time graph
is constant, we had a constant acceleration,
therefore we could use the UAM equations.
Then we showed that the equation for acceleration,
which is equal to change in velocity over change in time
is actually equivalent to one of the UAM equations,
velocity final equals velocity initial
plus acceleration times change in time.
We then figured out our acceleration.
Again, because it's constant, that means
it's going to have a horizontal line on the
acceleration as a function of time graph.
We then used one of our UAM equations
to figure out three different displacements,
plotted those on the position as a function of time graph,
then I did my best to draw the curve
to match those different points.
That's what we did.
Ladies and gentlepeople, I hope
you enjoyed learning with me today.
I enjoyed learning with you.
Voiceover: Lecture notes
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Please enjoy lecture notes responsibly.