Video Transcript
The given figure shows a force–time
graph. At time 𝑡 seconds, where 𝑡 is
greater than or equal to zero, the force is given by 𝐹 is equal to the quantity 𝑡
minus two squared newtons. Find the impulse over the first
four seconds.
Taking a look at this graph, we see
that it shows us force versus time and that this force starts out at four newtons,
decreases to zero newtons at a time of two seconds, and then increases at an
increasing rate until it reaches nine newtons at a time of five seconds. Focusing on the first four seconds,
we want to calculate the impulse of this force.
Graphically, this is equal to the
area under our curve from 𝑡 equals zero to 𝑡 equals four seconds. But note that we’re given the
algebraic equation for this force in terms of the time 𝑡. Knowing that in our case 𝐹 of 𝑡
is equal to the quantity 𝑡 minus two squared newtons, we can recall that when we
have a time-varying force, the impulse due to that force is equal to the integral of
the force over time. In this expression, 𝑡 one is the
initial time being considered and 𝑡 two is the final time. That is, we’re integrating 𝐹 of 𝑡
over the time interval of interest.
Applying this relationship to our
scenario, we’re working from a time interval of 𝑡 equals zero seconds up to 𝑡
equals four seconds. We’ll integrate 𝐹 of 𝑡 with
respect to time. And we’ll keep track of the units
involved.
Our first step is to multiply out
this expression. Quantity 𝑡 minus two squared is
equal to 𝑡 squared minus four 𝑡 plus four. And now we want to integrate each
of these three separate terms with respect to 𝑡. The integral with respect to 𝑡 of
𝑡 squared is 𝑡 cubed over three. The integral of negative four 𝑡 is
negative two 𝑡 squared. And that of four is four times
𝑡. And we’ll evaluate this expression
from 𝑡 equals four seconds to 𝑡 equals zero seconds.
Note that when we substitute in
zero seconds for 𝑡, all three of these terms will equal zero. So we only really need to
substitute in four seconds for the time 𝑡. Before we substitute in this time
though, let’s divide it up into its numerical value, that’s four, and its unit,
that’s seconds. We know that, in the end, our
impulse will have units of newton seconds. That’s what it means to multiply a
force by a time.
Therefore, when we substitute in
our time of four seconds, we won’t substitute in the unit. But we’ll instead pull that out and
group it with our other unit, the newton. So now, to finish evaluating this
integral, we’ll substitute in four for 𝑡 here, here, and here. Doing that gives us this
expression.
And now we just want to
simplify. Four cubed over three is sixty-four
thirds. Negative two times four squared is
negative 32. And four times four is 16. To combine these values, we’ll
first want to get a common denominator for all three. We’ll choose that denominator to be
three so that negative 32 becomes negative 96 over three and positive 16 becomes
positive 48 over three. Combining all these fractions, we
come up with a result of sixteen thirds newton seconds. This then is the impulse due to our
force over the first four seconds.
