in this video we're given a time varying force vector that's acting on a mass of 0.23 kilograms
and what we're trying to get out of this in part a is the position vector as a function of time
for this object and then in part b we want to use the position vector to get the average velocity
on a given time interval now two of the key pieces of information are written into the problem
up here the object starts from rest and that means the initial velocity components are both
zero and we're told it starts at the origin and that means the initial position components are
both zero and then we get into finding the position vector and really what we want to
work with here is the acceleration vector and then we can find its anti-derivative twice in
order to arrive at the position vector so to get the acceleration vector from
the force vector we're going to have to use newton's second law so the acceleration as a
function of time will just be that force vector divided by the mass of the object which was
0.23 kilograms so all we have to do is distribute that factor of 1 over 0.23 to all the terms
and i'll just round everything to 4 significant digits for now because i'm planning to round to
3 significant digits in my final answer and this gives me 1.087 times t squared i hat plus a 4.348
t minus 2.174
j hat now that i have that acceleration vector i recall that to get the velocity vector as a
function of time all i have to do is find the anti-derivative of each component of
this acceleration vector and all we have involved here is constants and powers of t so we're going
to use the power rule for all the terms and so in my first term i get a 1 3 times 1.087 times t
cubed and don't forget to put an additive constant in this remember the antiderivative
has the flexibility of an additive constant and that additive constant depends on the initial
conditions that we were given for the problem and we'll apply those in just a second and in
our j-hat term we end up with one-half times 4.348 times t squared minus the 2.174 t plus a second
arbitrary constant c2 and at this point we want to apply our initial conditions in order to nail down
the c's and i already know that the initial velocity in the x direction is zero the initial
velocity in the y direction is zero and i can quickly see the only way this is going
to happen if i replace t with zero the only way i end up with a zero for my x component is if c one
is equal to zero and the same is true for the coefficient of j hat the only way that initial
value when i plug in t equals 0 is going to give me 0 is if c 2 is equal to 0. so in this problem
we have the most friendly initial conditions possible because our arbitrary constants are
coming out to 0. just keep in mind that in general you may not have such friendly initial conditions
so it's important to keep those c's in your antiderivatives so we start cleaning things up
and in the first term i have 0.362 t cubed hat and the coefficient of j hat i have 2.174 t squared
minus 2.174 times t all multiplying j hat and now i have to remember that to get the position
vector from the velocity vector i take the antiderivative of the components for a second time
and again we're just using the power rule here so i get the antiderivative of the x component of
the velocity and i get 1 4 times .362 times t to the fourth plus an arbitrary constant times i hat
then for my j hat term i get a 1 3 2.714 t cubed minus a one half 2.714
t squared plus another arbitrary constant and you probably see what's coming here
my initial condition for the position was also the simplest possible one and if i plug in t
equals zero the only way these two components are going to be equal to zero is if these c's are
equal to zero and i'll clean things up and write down my position vector as a function of time
so now we know exactly where this particle is at every moment in time moving on to part b we
want the average velocity of the particle from t equals zero to t equals two seconds and the
average velocity vector is just the change in the position vector divided by the change in the time
so i take the position at the final moment in time that's two seconds r of two minus the position
at the initial time which was zero seconds r of zero all divided by how much time went by that's
two minus zero and things are pretty simple here because our initial position vector components are
both zero in other words the initial position is the zero vector so it drops out of our calculation
i'll go ahead and move the factor of two in the denominator out in front
and just call it a factor of one half and i have one half times r of two so all i have to do is
plug in t equals two into the position vector that i computed so now that i've replaced t with
two all i have to do is crunch the numbers i'm going to distribute the factor of one-half as i go
and i end up with an average velocity vector given by 0.724 i-hat plus 0.720
j-hat now each of those coefficients should carry units of meters per second because we
were told from the very beginning that our force vector had all the correct coefficients in there
to carry units of newtons and perhaps the most appropriate answer to the question is to
actually show those units and the coefficients and we're done if you enjoyed this video or at
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