in this video we extend the idea of position velocity and acceleration to higher dimensions
everything we do in this video is given in two dimensions but all the formulas could just be
amended with a z coordinate or k hat term to extend to three dimensions so in the picture
we're looking at a trajectory or path of an object in two dimensions and we've labeled two moments in
time on the curve t1 and t2 we start with how to define the position of the object and it
turns out the best way to do it is to label the position of an object by using a vector pointing
from the origin to the location of the object like this so there's r1 the position at time one
in r2 the position at time two and anywhere in between we could write a position vector
as a function of time for some arbitrary moment in time t and so in this view as a
particle travels from t1 to t2 we're actually watching a position vector rotate and grow
as it continues to point at the location of the object for each moment in time
and after defining positions as vectors the change in position or displacement keeps its
old definition from one dimension and we can write delta r is r2 minus r1 and then we recall
that a vector difference has a nice geometric interpretation when i write down r2 minus r1
i can visualize that by putting the tails of the two vectors in
the same place well that's already done because they're coming from the origin
and that vector difference points from the head of the second vector in the difference to the head of
the first in other words it's a vector pointing from the head of r1 to the head of r2 and this
makes a lot of sense for a displacement vector it simply points from the initial position to the
final position so it tells us the total direction the object moved and exactly how far it went
now we can put each of these position vectors into component form we could write r one equals x one i
hat plus y one j hat where the coordinates of this original point are x one y one
and we can write r two equals x two i hat plus y two j hat where those are the final coordinates
x2 and y2 and then we expand the displacement into components and we combine like terms
and i get it x2 minus x1 i hat plus a y2 minus y1
j-hat and finally we can put this back into delta notation using the fact that delta always means
final minus initial or the second coordinate minus the first i can write this as delta x
i-hat plus delta y j hat so our displacement vector in two dimensions is really just given
by the x displacement in the i hat direction added to the y displacement in the j hat direction now
that we have displacement we consider how to define average velocity in a two-dimensional
space and the definition is the same as it was in one dimension it's the displacement
during some time interval divided by how long it takes so in the picture at the left we have
a picture of our standard displacement vector pointing from the initial to final position
and that has a length measured in meters and when i divide that by delta t i get meters per second
and the average velocity is a vector let's put the vector hat on there real quick and it points in
exactly the same direction as the displacement vector because delta t is just a scalar so it
can change the length of the original vector of delta r but it can't change the direction so the
length of the velocity vector really depends on how fast the object got from point a to point b
but we'll just put it into the picture like this pointing in the same direction
as the original displacement vector and we can expand the displacement vector
into its components and distribute that one over delta t to both terms individually and when we do
that we discover that the coefficient of i hat in the average velocity vector is just the average x
velocity that's given by delta x divided by delta t a change in the x coordinate divided by how long
it took so we could call that v x bar i hat and then a v y bar j hat now just like we did in one
dimension we can obtain the instantaneous velocity vector by finding the average velocity over a very
small time interval and the idea here is that if we let the time interval become small enough the
velocity vector doesn't have time to change so for example the average velocity over one nanosecond
would then become a very good definition of instantaneous velocity so we're going to
compute the instantaneous velocity in this case at the time t1 and geometrically this
limiting process that i'm talking about is it will compute the average velocity
on t1 to t2 by taking the displacement vector and then dividing by the amount of time it
took to get between those two points and i get an average velocity vector same as we've already done
and then i consider a displacement vector that's a little bit closer so we're moving t2 in closer
to t1 and we compute the displacement vector and we divide by the amount of time it took
to get between those points again the length of the displacement vector is getting shorter as the
time interval also becomes shorter and i've got a new average velocity and i continue like this
letting t2 become closer and closer and closer to t1 and i keep computing that displacement vector
dividing by the time interval separating those two points so one thing to notice here is that
when that time interval becomes sufficiently small the velocity vector that i compute out
of this becomes tangential to the trajectory and that's an important thing to remember
instantaneous velocity is always tangential to the trajectory but this thing can be expressed
as a limit and i'm going to write it this way first it's the limit as delta t goes to zero so
the time interval size goes to zero of the average velocity but we can put the average velocity in
in its component form and you might be comfortable getting to the conclusion from this point but just
to make it extra clear what we're looking at i'm going to change the limit instead
of it being in terms of a delta t i'm going to write it as the limit as t2 goes to t1 so that
second time we're looking at t2 is getting really really close to the original time t1
and then we'll expand our delta x and our delta t as a t2 minus t1 and similar for the y coordinate
and this is the moment that i realized the coefficients of these unit vectors are actually
derivatives so that coefficient of i-hat that's the time derivative of x evaluated at t one and
the coefficient of j-hat is the time derivative of y evaluated at t one so this is a really cool
result all we have to do to get the velocity vector is take that original position vector
and take the time derivative of each of its components and instead of writing it specifically
in terms of t1 i'd like to write it in general that the velocity vector at any moment in time
is simply the time derivative of the components of the position
vector and sometimes it's useful to write this in terms of the velocity components
so i could write it as v x i hat plus v y j hat where v x is just the instantaneous
velocity in the x direction v y is the instantaneous velocity in the y direction
so as a quick check on our understanding we'll work a short example in this example we're given
the position function for a particle explicitly in terms of t so the position is given by t squared
minus three i hat plus a point five t cubed plus two t multiplied by j hat and we wanna get the
average velocity on the time interval zero to two and average velocity is the change in the position
vector divided by the change in time and so that'll be given by r of two minus r of zero
over two minus zero and then we need to get r of two and r of zero by subbing in those t values
so r of two and i get for my first component 2 squared minus 3 which is 1 so i get an i hat there
and then in my second component i would have half of 2 cubed that's 4 plus a 2 times 2 that gives me
a total of 8 j hat and then i subtract r of zero when i sub in t equals zero i get a negative three
i hat plus a zero j hat and that's all divided by the width of the time interval which was two
i distribute that minus sign in the numerator and combine like terms and i get 4 i hat
plus 8 j hat divided by 2 and finally i divide the 2 out of this and i get 2i hat
plus 4 j hat and i really should finish by putting units on my answer
that's 2 meters per second i hat plus 4 meters per second j hat and what this means is that
this particle was on average moving 2 meters per second to the right and 4 meters per second up
on the time interval 0 to 2. in part b i want the instantaneous velocity at t equals 1 second
so i need to find the velocity function and that's the time derivative of the position vector
so i just differentiate each of the coefficients of the unit vectors with respect to time
this gives me a 2t i hat for the x component and then for the y component when i differentiate t
cubed to bring the 3 in front subtract 1 from the exponent i get 1.5 t squared and then the
derivative of 2t is just 2 and i've got a 1.5 t squared plus 2 times j hat now i just evaluate
the instantaneous velocity function at t equals 1 and i've got 2i hat plus 3 and a half for 3.5
j hat and again i really should express this with units on it so 2 meters per second i hat
plus 3.5 meters per second j hat so at that moment in time the particle is moving exactly 2 meters
per second to the right and simultaneously three and a half meters per second upward
so naturally the next thing to do is to define average acceleration and learn how to compute
it and then we'll take a small time limit and come up with a definition for instantaneous
acceleration and average acceleration retains the same definition from the one-dimensional case
it's a change in velocity divided by a change in time in other words an average rate of change in
a velocity but this time it's a velocity vector so what do we do geometrically we would have to look
at what's the velocity at t1 and i'll just call that v1 and then we'd look at what's the velocity
at t2 i'll call that v2 and of course delta v in this expression for average acceleration that's
v2 minus v1 so that's a vector difference and the way i can compute that geometrically again is by
moving the tails of the two vectors into the same position so i slide v2 over here and put the tail
in the same place as the tail of v1 and the vector difference then would point from the
head of v1 to the head of v2 there's my delta v for the given time interval then i divide by
the length of the time interval which rescales that vector but does not change its direction
and i've got my average acceleration vector and then if we want to extend this to instantaneous
acceleration we just take a small time limit again so it's the same idea where if i compute the
average acceleration over a smaller and smaller and smaller time interval eventually the average
acceleration becomes a really good definition of instantaneous acceleration at a moment in
time so again we imagine moving t2 closer and closer and closer to t1 and we keep repeating
this process where we compute the velocity vector at each of those points in time we take the vector
difference and then we divide by delta t and i'll go ahead and write this in the delta notation
first but things become more recognizable if i write it as a limit as t2 goes to t1
and then we can split v2 and v1 into their components and then we gather like terms
and again we recognize time derivatives in these coefficients so my end result is that the
instantaneous acceleration vector at a time t 1 is given by the time derivative of the x component of
the instantaneous velocity vector multiplied by i hat and then a time derivative of the y
component of the instantaneous velocity vector evaluated at t one multiplied by j hat and then
we generalize to any time t not just t one and our acceleration vector is given by again taking the
time derivatives of the instantaneous velocity vector components this could also be written as
a second time derivative of the position vector components so i could write x double prime of t
i hat plus a y double prime of t j hat so we'll practice these concepts by working a short example
and in this one again i have the same position function given to me but this
time i want the average acceleration on the time interval t equals zero to t equals one
so remember to get average acceleration i've got to take the change in the velocity vector
and divide by the amount of time that went by this means i need to find the instantaneous
velocity vector as a first step and v of t is the time derivative of the
position vector so i just differentiate each of the components and this gives me 2t i hat
plus a 1.5
t squared plus 2 all multiplied by j hat and then i can evaluate what's the velocity at t equals 1
second so i sub in t equals 1 and i get 2i hat plus 3.5 j hat so there's my instantaneous
velocity when t is equal to 1. then i have to compute that when t is equal to 0 and the x
component for v of 0 is 0 because i sub n t equals 0 the y component is just a 2 j hat so my average
acceleration vector again that's a difference between my final and initial velocities and
that's divided by the width of the time interval which happens to be one second in this case
and combining like terms i get 2i hat plus 1.5 j hat and i should express this with units on it
and if i go back to my original definition of average acceleration i have a delta v in the
numerators that has units of meters per second on each of the components and i'm dividing by a
delta t measured in seconds and that means i get the correct units of meters per second squared
on each of my components in the final answer so that's two meters per second squared i hat
plus 1.5 meters per second squared j hat
so on average over this time interval the particle was accelerating at 2 meters per second per second
in the plus x direction and 1.5 meters per second per second in the plus y direction now i want my
instantaneous acceleration at t equals zero so i start with my instantaneous velocity function
and i just take the next derivative again the acceleration function is given by differentiating
each of the components of the velocity function with respect to time and the derivative of 2t is
just 2 so i end up with a 2i hat in the first term and then in the second term i get a 3t
j hat and the derivative of the 2 there vanishes so this acceleration is a non-trivial function
of time it's not a constant with respect to time there's still a t dependence in there and when i
evaluate the acceleration at t equals zero i just sub n t equals zero there and the j
hat term vanishes and i end up with 2i hats again i should probably write units on this and again
that's meters per second squared so at t equals zero the particle happens to be accelerating
exactly in the i hat direction with a magnitude of 2 meters per second squared so here's a quick
summary of what we've accomplished in this video first we started writing positions as vectors
pointing from the origin to the location of the object second we discovered that the instantaneous
velocity of an object at any moment in time is given by simply taking the time derivative
of the components of that position vector and we could also write that as an x velocity i hat and
a y velocity j hat and then finally we discovered that the acceleration vector at any moment in time
is given by just taking another time derivative of those same components so i got x double prime and
y double prime as my components or i could write it as an x acceleration i hat y acceleration j-hat
our next order of business is to apply these ideas to solving the problem of projectile motion and
i'll post a link to that video in the in screen if you enjoyed this video or at least found it useful
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