### Video Transcript

A ball of mass five grams was
moving in a straight line through a medium loaded with dust. The dust was accumulating on its
surface at a rate of one gram per second. Find the magnitude of the force
acting on the ball at time π‘ equals five seconds, given that the displacement of
the ball is expressed by the relation π of π‘ equals two-thirds π‘ cubed plus π‘
squared plus 7π‘ plus one π, where π is a unit vector in the direction of the
motion and the displacement is measured in centimeters.

We want to solve for the magnitude
of the force that acts on the ball at a particular time π‘ equals five seconds. Weβll call that force magnitude
capital πΉ. To solve for that force, it will
help us to recall Newtonβs second law of motion. Newtonβs second law of motion is
commonly written as force is equal to an objectβs mass times its acceleration
π. This is a perfectly valid
expression of the second law when the mass does not change with time and the speed
of the motion is nonrelativistic. A more general expression for the
second law is that the net force on an object is equal to the time rate of change of
its mass times its velocity.

In our scenario, we have a ball
that starts out initially with a mass of five grams. Weβll call that value π sub zero,
yet it takes on mass as it gathers dust every second. In fact, if we were to write the
mass of the ball as a function of time, it would be equal to its original mass five
grams plus the time in seconds since each second that passes has one gram of mass to
the ball.

So in our scenario, the mass of our
object does change in time. And therefore, we want to use this
more general formulation of the second law. In writing this equation for our
case, we see that π as a function of π‘ is something we know, but the velocity π£
is unknown currently.

We can recall though that our
velocity is equal to the time rate of change of position. And weβve been given a position
vector for our ball. This means we can take the time
rate of change of that position and substitute it in for π£ in our equation. When we take the time rate of
change or derivative with respect to time of our position function, we get a result
of two π‘ squared plus two π‘ plus seven in the π hat direction. This is the expression for π£ that
weβll put into our equation to solve for πΉ.

But before we do, we can take the
step of multiplying π£ by π as a function of π‘, the mass of our ball, so that we
can insert this expression entirely into our equation for πΉ. We discovered earlier that our mass
as a function of time is equal to five plus π‘ grams.

When we multiply this term through
with our velocity term, neglecting the units for the time being and grouping common
powers of π‘, we find that this product is two π‘ cubed plus 12π‘ squared plus 17π‘
plus 25. Thatβs π as a function of π‘ times
the velocity of the particle π£. So itβs this expression that weβll
plug in to the parentheses to take the time derivative of to solve for πΉ.

Weβre now set to take another time
derivative. And when we do, we find the result
of six π‘ squared plus 24π‘ plus 17. This is close, but not quite the
result that weβre looking for.

We want to solve for the force πΉ
magnitude at a particular π‘ value β that is when π‘ is equal to five seconds. So weβll now plug in that
particular value for π‘ to solve for the force at that instant in time.

With this value plugged in, when we
calculate πΉ, we find its magnitude is 287. And the unit is dynes. This is the magnitude of the force
that the ball experiences at time π‘ equals five seconds.