### Video Transcript

An object has a mass of nine
grams. The mass of this object is reduced
with a rate of one gram per second. The velocity of this object is
given by vector π is equal to three π‘ plus 10 π’ centimeters per second. Find the magnitude of the force
acting on this object at π‘ equals two, where π‘ is the time in seconds.

Now, initially weβre given
information about the mass of the object and the fact that the mass reduces over
time. Now, we normally look to find the
magnitude of a force given its mass and acceleration, where force is mass times
acceleration. However, when given an object with
variable mass, we use an alternative formula. That is, for an object with mass π
and velocity π£, the force is given by π times dπ£ by dπ‘ plus π£ times dπ by
dπ‘.

The question gives us an expression
for π at time π‘. So we can quite easily calculate
dπ£ by dπ‘. But how do we calculate π and dπ
by dπ‘? Well, since the mass of the object
reduces at a rate of one gram per second, we can say that the change in mass with
respect to time, that is, dπ by dπ‘, is equal to negative one. We can then integrate this
expression with respect to π‘ and solve to find the constant of integration to find
an expression for π. In other words, π is equal to the
indefinite integral of negative one with respect to π‘.

Now, the integral of a constant is
just the value of that constant multiplied by the variable. So here the integral of negative
one is negative π‘. So π is equal to negative π‘ plus
some constant of integration C. But in fact weβre told information
about the initial mass of the object. At π‘ equals zero, the mass of the
object is nine grams. So we can rewrite our equation
substituting these values in as shown. And as such, C is equal to
nine. Our expression for the mass of the
object at time π‘ is π is equal to negative π‘ plus nine.

We might notice at this stage that
we do need to restrict the domain of this function. π‘ will be able to take values less
than or equal to nine. This is because if we substitute
anything greater than nine into the expression, weβll end up with a negative value
for the mass. Now, in fact, this doesnβt really
matter as weβre finding the magnitude of the force at π‘ equals two. However, it is something to bear in
mind. We now have our expression for π
and dπ by dπ‘. And then we have a vector equation
for π. π is equal to three π‘ plus 10
π’.

Now, at this stage, itβs useful to
spot that since three π‘ plus 10 is the π’-component of this vector, the velocity of
the object is only acting in a straight line. And so we can actually identify
that the magnitude of the velocity will be three π‘ plus 10 for our value of π‘. Next, we need to calculate an
expression for dπ£ by dπ‘. Well, the derivative of three π‘
with respect to π‘ is three and the derivative of 10 is zero. So dπ£ by dπ‘ is equal to
three. This essentially tells us that the
acceleration of the object is constant. It accelerates at a rate of three
centimeters per second squared.

We are now in a position where we
can substitute everything we know about our object into the given formula. We have π
is equal to negative π‘
plus nine multiplied by three plus three π‘ plus 10 multiplied by negative one. Distributing our parentheses gives
us negative three π‘ plus 27 minus three π‘ minus 10, which in turn simplifies to
negative six π‘ plus 17. We now have an expression for the
magnitude of the force at a given time π‘.

Now, the question asks us to find
the magnitude of the force at π‘ equals two. So thatβs π
is equal to negative
six multiplied by two plus 17. Negative six multiplied by two is
negative 12. And adding 17 gives us five. Now, since weβre working in grams,
centimeters, and seconds, our unit for the magnitude of the force is dynes. The magnitude of the force acting
on the object at π‘ equals two is five dynes.

Now, it is worth noting that we
calculated the velocity in terms of its magnitude. And we were able to do this because
the object is traveling in a straight line. We couldβve alternatively used the
vector form of our velocity and differentiated this with respect to π‘. This wouldβve given us the
derivative of vector π with respect to π‘ is three π’ and, as a result, a vector
expression for π
, from which we couldβve then calculated the magnitude of the
force. Either way, we get a final answer
of π
is equal to five dynes.