Hello, I am Francesc Soriguera
and I will present this video about the physics of movement.
I will introduce the topic,
I will present the time-space diagram,
then I will show a trajectory representation in the diagram,
I will also define speed and acceleration,
then we will interpret physically the trajectories on the diagram, and finally
we will derive the equations for the uniformly accelerated motion.
We are going to end video with references for further materials.
So, why we study the movement in this course?
Transportation or mobility deals with the movement of persons and goods.
Movement is defined by the change of the physical position in time.
This means that, Time and Space, are key variables to physically understand transportation and mobility.
Then in this video I will explain how the physics of movement
can be analyzed in a two dimensional plot: this is the time-space diagram,
and how we can derive the properties of this movement,
like the speed or acceleration, from this representation.
We are going to end the lecture by illustrating these definitions
with two particular types of of movement:
the movement with constant speed and the movement with constant acceleration,
the so called uniformly accelerated motion.
So, let's start with
the time – space diagram. If we consider "x" as
a one dimensional space variable,
so that "x" has units of space like meters or Kilometers,
and we consider time "t" with units of time
like seconds, minutes or hours.
So, the movement is defined by the change of "x" with respect to "t",
so that it is useful to represent these couple of variables in a coordinate
plain, in the (t,x) plain.
So, (t,x) defines the time – space diagram.
Typically, we represent "t" in the abscissa (i.e. horizontal axis)
and "x" in the ordinate (i.e. vertical axis),
as you can see in the figure bellow the slide.
A trajectory is the representation of the movement of one object
in this time – space diagram.
Mathematically speaking, this is the representation of the function x(t)
in the (t,x) plane.
One trajectory defines all what can be measured from the motion of one vehicle.
Here, in the figure, you can see the trajectory of vehicle "i",
in the time-space diagram. So, you can see the position
of the vehicle x(t) in the vertical axis, for any given
instant of time in the horizontal axis.
Let's define speed and acceleration.
Speed is defined as the rate at which an object covers distance.
So, imagine that the position of vehicle at t1 was x1,
and some time later, at t2, the position is x2.
Then, the average speed, this "v"
with an over bar, is the average
speed of the vehicle between t1 and t2.
So, this is simply computed as the space distance
between x1 and x2 divided by the time period between t1 and t2,
as you have in the equation in the slide, and obviously this have units
of space over time, meters per second, kilometers per hour.
We can see the representation of a trajectory in the time-space
diagram, below in this slide, where we can derive
the average speed of this vehicle just by the slope
of the straight line joining the positions x1 and x2,
at times t1 and t2, so you can see this slope of this line "v".
Acceleration, in turn, is defined as
the rate at which speed changes with time.
So, if the speed of a vehicle at t1 is v1,
and some time later at time t2 is v2, then the average acceleration,
"a" with the over bar, of the vehicle between t1 and t2, is simply
the increase or decrease of the speed in the period, v2 minus v1,
divided by the duration of the time period, t2 minus t1.
In units of acceleration, space over time square, for instance meter per squared second.
if this time interval, between t1 and t2,
is reduced to an infinitesimal size, that we call "dt",
then we obtain the instantaneous speed "v" and the instantaneous acceleration "a", at a given time "t".
Mathematically, the instantaneous speed is defined as the derivative,
which means the variation, of space with respect to time.
In turn, instantaneous acceleration is the derivative of speed with respect to time,
or equivalently the second derivative of space with respect to time
As you can see in the two equations below the slide.
We can interpret physically this instantaneous speed and instantaneous acceleration
in a trajectories diagram.
The instantaneous speed is the slope of the trajectory at a given time,
while instantaneous acceleration is the curvature of the trajectory at that time.
So, here, in this time-space diagram, you have the trajectory of one vehicle,
where you can see that, at the start of the movement, the vehicles is accelerating.
This can be seen because the curvature is positive.
So, the second derivative of space with respecto a time is positive. Then, at some point, the vehicle
starts decelerating, which means that the curvature turns negative.
Still, the vehicle moves forward. We can see that the vehicles is moving
forward because the slope of the trajectory, so the speed, is positive.
The derivative of space with respect to time is positive in this period. You can see that because
the slope of the tangent at this point is positive.
At some time, the vehicles stops, this can be see because the trajectory
is horizontal, so there is no movement in this time period.
Then, later on, the vehicle starts travelling backwards.
You can see that because the slope turns to be negative, so the speed of the vehicle is negative in this period,
which means that the vehicle is traveling in the negative direction
of "x", of the "x" axis. Note that a vehicle cannot travel backwards in time.
This would involve that a vehicle is at two points
at the same time, and this is not physically posible.
Now, we are going to change and
start talking of a particular type of movement, which is
the uniformly accelerated motion. This is a movement where the acceleration is constant.
We can obtain the speed and the position of this vehicle, with this type movement, if
we integrate this constant acceleration with respect to time,
then we obtain the speed, and if we integrate a second time, then we obtain the acceleration.
So, if we perform these integrations,
we can see that the speed changes linearly with time.
So, the speed is equal to the acceleration times the time pus v0,
which is the initial speed, one constant integration.
With respect to the position, if we integrate the speed with respect
to time, we have that the position
evolves quadratically with time, so that the position is equal to the acceleration
times the time squared, divided by two, plus the initial speed
times the time, plus the initial position x0, which is the second constant integration.
So, v0 and x0,
are the initial speed and initial position of this vehicle.
Here we have the representation of this particular
type of movement in the time space-diagram.
On the left hand side, you can see the position, this is the quadratic trajectory of the uniformly accelerated
movement, according to the equation that we have seen in the previous slide.
On the right hand side, you can see
one particular case when the acceleration is cero.
So, this yields a constant speed and the constant the speed can
be seen in the trajectories diagram with the linear trajectory,
where the slope of this linear trajectory is simply the speed.
One particular question that
usually appears when talking about the uniformly accelerated motion
is the determination of the stopping distance and the stopping time.
So, imagine the vehicle
traveling at speed v0 and decelerating uniformly at an acceleration -a_b
Note: this a_b is a deceleration,
which means that a_b is negative,
is a deceleration.
So, if the final speed needs to be 0,
because we want the stopping distance, the vehicle to be stopped,
we substitute this into the previous equations and we find that 0 needs to be equal to a_b times
the time stopping plus de initial speed. By solving this equation for the
stopping time, t_stop, we obtain that this is equal to minus v0 divided by a_b.
Note this negative sign, because then
the time of stopping will be positive, because recall that a_b is negative.
To obtain the stopping distance, we only need to substitute this time of stop
at the equation of the position,
so, we obtain x_stop.
By solving this equation, we can obtain that x_stop
is equal to minus the square of the initial speed divided by two times a_b plus the initial position.
Again, this negative sign
turns to be a positive stopping distance, because a_b is negative.
These equations are very useful in the problems dealing with this type of motion.
I finish this video with the references, where
you can find further materials about
trajectories, the time-space diagram, and the uniformly accelerated motion.
Thank you very much for watching and enjoy, the course.