### Video Transcript

A body moves in a straight
line. At time π‘ seconds, its
displacement from a fixed point is given by π equals two π‘ squared plus five π‘
plus four meters. Its mass varies with time such that
π is equal to six π‘ plus five kilograms. Determine the force acting upon the
body when π‘ equals three seconds.

In this question, weβve been given
information about the displacement of a body moving in a straight line with respect
to time. Weβve also been given information
about its mass with respect to time π‘. And so we have a body with variable
mass. This means we canβt use the
equation πΉ equals ππ that weβre so used to.

Instead, we use Newtonβs second
law, but we add a term to account for the momentum carried by mass entering the
system. We say πΉ is equal to π£ times dπ
by dπ‘ β thatβs the derivative of mass with respect to time β plus π by dπ£ by dπ‘,
which is the derivative of π£ with respect to time, in other words its
acceleration.

Now since π is equal to six π‘
plus five, weβre going to need to differentiate that with respect to time. And we can do that term by
term. The derivative of six π‘ is just
six, and the derivative of five is zero. So we so find that dπ by dπ‘ is
equal to six.

Now weβre told that π , the
displacement, is two π‘ squared plus five π‘ plus four. And so we need to recall that
velocity is the change in displacement with respect to time. Thatβs dπ by dπ‘. So we differentiate each term with
respect to time. The derivative of two π‘ squared is
two times two π‘, which is four π‘. And the derivative of five π‘ is
five. The derivative of four, however, is
zero. So we find when we differentiate π
with respect to π‘, we get four π‘ plus five. Similarly, we need dπ£ by dπ‘ for
the formula. So weβre going to differentiate
once again. And of course, in fact, that gives
us acceleration.

Now the derivative of four π‘ is
four, and the derivative of five is zero. Letβs substitute everything we have
into our formula. π£ times dπ by dπ‘ is four π‘ plus
five times six, and π times dπ£ by dπ‘, in other words ππ, is six π‘ plus five
times four. We distribute each of our
parentheses. Four π‘ times six is 24π‘, and five
times six is 30. Similarly, six π‘ times four is
24π‘, and five times four is 20. And so we collect like terms, and
we find that the force is 48π‘ plus 50. And since weβre working meters,
seconds, and kilograms, itβs actually 48π‘ plus 50 newtons.

Now, of course, weβre looking to
find the exact force at π‘ equals three. So πΉ equals 48 times three. We replace π‘ with three and then
we add 50. That gives us 194. And so the force acting on the body
when π‘ equals three seconds is 194 newtons.