Bo: Hey, guys.
Billy: Hey, Bo. Bobby: Hi, Bo.
♫ (lyrics) Flipping Physics ♫
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 another
uniformly accelerated motion problem.
Billy: Absoultely. Bo: Yes.
Bobby: Oh, boy.
Mr. P.: Good.
Last time we did an introductory
uniformly accelerated motion problem.
Today we're going to do a slightly
more complicated UAM problem.
Bobby, could you please read the problem?
Bobby: A toy car starts from rest
and experiences an acceleration
of 1.56 meters per second squared for 1.6 seconds,
and then brakes for 1.1 seconds
and experiences an acceleration
of -2.07 meters per second squared.
Part (a), How fast is the car going
at the end of the braking period,
and Part (b), how far has it moved?
Mr. P.: All right, now let's translate
the problem to physics.
Bobby, could you please read again
and Billy, could you please translate?
Bobby: A toy car starts from rest
and experiences an acceleration
of 1.56 meters per second squared for 1.6 seconds and then ...
Billy: Please stop! a equals 1.56 meters
per second squared,
and t equals 1.6 seconds.
Mr. P.: Please say what the letters mean
and not just the letters themselves.
The more you actually say
what the letters mean,
the more you'll understand the physics,
and that's good.
Billy: Sorry, I knew that.
Acceleration equals 1.56 meters
per second squared,
and time equals 1.6 seconds.
Mr. P: It's okay, Billy.
Bobby, please keep reading.
Bobby: And then brakes for 1.1 seconds
and experiences an acceleration
of -2.07 meters per second squared.
Billy: Time equals 1.1 seconds
and acceleration equals -2.07 meters
per second squared.
Bo: That doesn't look good.
Billy: No, it doesn't.
Mr. P.: True. We can't have the same variable
equal to 2 different numbers.
That just doesn't make any sense.
For example, if time is equal to 1.6 seconds
and time is equal to 1.1 seconds,
then 1.6 seconds equals 1.1 seconds
which doesn't make any sense.
So what are we going to do about this?
Bo: I think we need to split the problem
into 2 parts.
Billy: Yeah, so that 1.56 meters per second squared
is the acceleration for the first part,
and -2.07 meters per second squared
is the acceleration for the second part.
Likewise, the 1.6 seconds
is the time for the first part,
and the 1.1 seconds
is the time for the second part.
Mr. P.: And the way that we indicate the variables
are for 2 different parts of a problem
is by using subscripts.
So you can see we have the acceleration
for Part 1 and the time for Part 1,
and then the acceleration for Part 2
and the time for Part 2.
Bobby, please keep reading.
Bobby: Part (a) How fast is the car going
at the end of the braking period,
and Part (b) how far has it moved?
Billy: Part (a) velocity final equals question mark,
and part (b) displacement equals question mark.
Mr. P.: All right, we actually have some things
to fix in our known variables please.
Bobby: I think those times are actually
changes in time, not time.
Billy: Oh, yeah. That makes sense
because those aren't specific times
like 2:31 AM or something like that.
Rather each one is a time duration
so please change each one to a delta t
or a change in time.
Mr. P.: True, they're changes in time
not specific times.
Good. That fixes part of it.
We are still missing some subscripts.
Bo: We need a subscript on the displacement.
Bobby: Part (b) says we need to find
how far it moved.
That means for the whole thing.
Billy: So we're looking for the total displacement.
So add subscript of t for total
to the displacement for Part (b).
Mr. P.: And we're actually still not done
fixing our knowns.
Which velocity final are we trying to find?
Bo: We want the velocity at the end of Part 2
which means we need to add a subscript
of 2 to the final velocity we're looking for.
Mr. P.: Correct. We're trying to find
the final velocity for Part 2,
and please realize that Part 2 ends
before the car stops moving.
It's an important thing to realize.
Bo, how should we proceed from here?
Bo: Well, we know we can use the UAM equations
because they're part of the title of the lecture.
Mr. P.: Bo, that's not going to fly.
I need to know from the problem statement itself
how do we know we can use the
uniformly accelerated motion equations.
Bo: Oh, because the accelerations
are equal to numbers.
The acceleration 1 equals 1.56 meters
per second squared,
and the acceleration 2 equals -2.07 meters
per second squared,
and those are both constant numbers.
They don't change.
So we can use the UAM equations.
Mr. P.: That is correct, Bo.
That's what the uniformly in UAM means.
It means that the acceleration is constant.
At this point I usually draw a picture
and that helps us understand
what's going on.
You can see from our drawing
that the part where we're speeding up
I've labeled Part 1,
and the part where we're slowing down,
I've labeled Part 2.
I've listed the things we know for both parts.
We know the acceleration and the change in time
for both Part 1 and Part 2.
Let's review real quick.
Class, remind me.
There are how many UAM variables?
Class: 5.
Mr. P.: There are how many UAM equations?
Class: 4.
Mr. P.: If you know how many of the variables?
Class: 3.
Mr. P.: You could figure out the other ...
Class: 2.
Mr. P.: Which leaves you with 1 ...
Class: happy physics student.
Billy: Yeah.
Bo: Actually, that means we have a problem.
We only know 2 of the UAM variables in each part.
Mr. P.: True. We are actually missing a variable.
Why don't we watch the example 1 more time
and see if you can see what we're missing.
Billy: It starts out not moving.
Bobby: So the initial velocity is 0.
Bo: The initial velocity for Part 1 is 0.
Mr. P.: Yes. You can see it in the video,
and it stated in the problem
that the car starts from rest; therefore,
the velocity initial for Part 1 is equal to 0.
This is one of those slightly hidden variables,
one of those knowns that is sometimes hard to see.
So you need to make sure
that you're paying attention
and if it says something stops
or something starts from rest,
that means that the velocity
at that particular point is equal to 0.
On a side note, class, have we started
solving the problem?
Bobby: Nope.
Billy: No.
Bo: Sadly, no.
Mr. P.: I will bring this up repeatedly
because it's so important.
As students, you often just want to rush
your way through the problem,
but really the best way is to just slow down
and understand everything first
before you even start trying to solve the problem.
Please slow down.
Do you understand?
Bobby: Yes. Billy: Yup.
Bo: Sure.
Mr. P.: Good.
Bobby, what should we do next?
Bobby: I know we have to start with Part 1
because that's where we know our 3 UAM variables.
However, I don't even know
what to solve for in Part 1.
Mr. P.: You know what? That's fine.
Sometimes it works that way.
Let's just remind ourselves
of what we're trying to find
by adding them to our picture.
As you can see from our picture,
we're looking for the total displacement
from the beginning to the end,
and we're looking for the final velocity
for Part 2 and I just then added
that the initial velocity for Part 1
is equal to 0.
If you don't know what to do,
let's just start by listing what we know
for example for Part 1.
Bobby: Okay, we know the initial velocity
for Part 1 is 0,
the acceleration for Part 1 is 1.56 meters
per second squared,
and the change in time for Part 1 is 1.6 seconds.
Mr. P.: Okay. These are the variables
that we know for Part 1.
Bo, what could we use these 3 UAM variables
to find for Part 1?
Bo: Uh, the 2 remaining variables
that we don't know for Part 1;
velocity final for Part 1,
and the displacement for Part 1.
Bobby: We should solve for the final velocity
for Part 1 because it's the same
as the initial velocity for Part 2.
Mr. P.: Absolutely correct.
As you can see in the video,
the ending point for Part 1 is the same
as the starting point for Part 2.
In other words,
the final velocity for Part 1 is
the same as the initial velocity for Part 2.
Okay. Billy, please solve for the final velocity
for Part 1.
Bobby: Wow! Are we really just starting
to solve the problem?
Bo: Yeah.
I bet it takes less time to solve
than it does to understand.
Bobby: Yeah, you may be right.
Billy: Uh, okay, we just need to pick
the correct UAM equation.
Let's see.
Velocity final equals velocity initial
plus acceleration times the change in time.
Mr. P.: Very close.
Sadly we are missing ...
Billy: Subscripts.
Please add 1's to everything.
Mr. P.: Actually, now we have all of the givens
for this equation so we can just plug in numbers.
What do we get for the final velocity for Part 1?
2.496 meters per second.
Notice, now that we have
the final velocity for Part 1
which we know is the initial velocity for Part 2,
we actually know 3 UAM variables for Part 2.
We can now solve for the final velocity
for Part 2 or Part (a).
Bo, please go ahead.
Bo: I think we just use the same equation.
Mr. P.: Close. It's not exactly the same.
Bo: Okay. Different subscripts.
So then velocity final for Part 2 equals
velocity initial for Part 2
plus acceleration 2 times the change in time 2.
Again, we have all the numbers
so the final velocity for Part 2 equals
the initial velocity for Part 2
which we just figured out to be 2.496,
plus the acceleration from Part 2
which is -2.07 meters per second squared
times the change in time for Part 2
which is 1.1 seconds.
Mr. P.: What do we get for an answer?
Billy: 0.219 or 0.22 meters per second
with 2 sig figs.
Mr. P.: Great. We figured out the final velocity
for Part 2, 0.22 meters per second.
This is Part (a).
How about we work on Part (b) now,
the total distance traveled.
Billy: Oh, the total distance
will just be equal to the distance for Part 1
plus the distance for Part 2.
Mr. P.: Very nice.
The total displacement is the summation
of the displacement for Part 1
and the displacement for Part 2.
You can see that pretty well in our picture.
Again, I just want to be careful
to make sure that everybody understands
that while they've asked
for the total distance traveled,
and we're using the symbol for displacement
because we're moving in a straight line,
the magnitude of the displacement
and the distance traveled are the same thing.
Important to realize.
Bobby, how do we find the displacement for Part 1?
Bobby: We could use the UAM equation
displacement 1 equals 1/2 times the quantity
velocity 1 final plus velocity 1 initial.
That quantity times the change in time from Part 1.
We have all those numbers so then
the displacement for Part 1 equals
1/2 times 2.496 plus 0 times 1.6,
and we get ...
Billy: 1.9968 meters.
Mr. P.: Okay. We have figured out
the displacement for Part 1.
Now we need the displacement for Part 2.
Bo, how are we going to do that?
Bo: We can use the UAM equation
displacement for Part 2 equals
the initial velocity 2 times the change in time 2
plus 1/2 times the acceleration 2
times the change in time for Part 2 squared.
Plugging in the numbers we get 2.496
times 1.1 plus 1/2 times -2.07 times 1.1 squared
which give us 1.49325 meters.
Mr. P.: Great.
We are almost done with this problem.
We just figured out the displacement for Part 2.
Billy, please finish the problem.
Billy: Oh, okay.
The total displacement as I said equals
the displacement for Part 1 plus
the displacement for Part 2
which is going to be equal to the 1.9968
plus the 1.49325.
We get 3.49005 rounded 2 sig figs.
We get 3.5 meters.
Mr. P.: Excellent. We have now solved the problem.
We figured out the final speed
at the end of Part 2,
and the total distance traveled.
I have a question about finding
the total distance traveled.
Why couldn't we, once we had figured out
the final speed at the end of Part 2
have solved the problem this way?
In other words,
why couldn't we have looked at the whole problem
and used 1 UAM equation.
In other words,
the displacement total is equal to 1/2 times
the velocity for Part 2 final
plus the velocity for Part 1 initial
times the change in time total.
Where the velocity for Part 2 final is
the final velocity for the whole thing,
and the velocity for initial for Part 1 is
the initial velocity for the whole thing
and the change in time total is the addition
of the change in time for the first part
plus the change in time for the second part.
The question is why couldn't we have solved
the problem this way?
It clearly gives us the wrong answer, 0.30 meters
is clearly not far enough,
but what is wrong with solving the problem this way?
It actually all goes back to the very beginning
of the problem and why we could use
the UAM equations in the first place.
Think about that.
Why could we use the UAM equations
which makes it so that we can't do it this way?
Bobby: Oh, it's not uniformly accelerated motion
throughout the whole problem.
Billy: Yeah, the acceleration is only constant
during Part 1, and then Part 2.
Bo: And if you try to deal with the whole thing
at once the acceleration isn't uniform
because it changes from 1.56 meters
per second squared
to the -2.07 meters per second squared.
Mr. P.: Exactly.
Please be careful to only use the UAM equations
when the acceleration is uniform,
when the acceleration is constant,
when the acceleration equals a number
which was true for Part 1,
and also true for Part 2,
but not true for the whole thing
because the acceleration is only constant
for Part 1 and for Part 2 individually,
but not the whole thing
because it changes from Part 1 to Part 2.
So please be careful of that.
Any questions?
Billy: No.
Bobby: No.
Mr. P.: In that case, Bo,
I will point out that you were correct.
It did take us longer to understand the problem
than it did to solve the problem
and that's okay.
It actually isn't all that unusual.
Just to review our understanding the problem,
we listed everything that we knew.
We drew a picture to identify
those different pieces.
We split it up into 2 parts.
We listed what we knew based on those 2 parts.
We wrote out the equation we were going to use
before we used it.
We used subscripts.
That always help to understand what's going on.
We went through it slowly,
and again, slowly , methodically
to understand what's going on
before we tried to solve the problem.
Ladies and gentlepeople,
I hope you enjoyed learning with me today.
I enjoyed learning with you.
Voiceover: Lecture notes are available
at FlippingPhysics.com.
Please enjoy lecture notes responsibly.