is
measured in seconds, and
Since we know how to
compute the displacement when the velocity is constant, we will start by asuming
that the velocity is constant and equal to
\(v(0)\), the initial velocity, for the entire
four-second time interval,
\(\Delta t=4\).
This scenario will give us a very rough approximation of the displacement, but it
is a good starting point.
\[ \text {displacement}\approx v(0)\cdot \Delta t=1\cdot (4-0)=4 \unit {ft}/\unit {s} \]
This scenario is illustrated in the figure below.
From
the graph above, we see that the assumption of constant velocity is not quite
accurate. The velocity is changing, so we should allow our velocity to change as well.
Instead of assuming the velocity is constant over the entire four-second interval, we
will keep approximating the velocity with a constant over smaller and smaller
intervals of time.
In our next step we will assume that the velocity is constant during two-second time
intervals \([0,2]\) and \([2,4]\). For each of these two intervals, we assume that the velocity is equal
to the velocity at the beginning of the time interval.
In this case, the object would move by \(v(0)\cdot 2 \unit {ft}\) during the time interval \([0,2]\), and during the next
two seconds, the object would move by \(v(2)\cdot 2 \unit {ft}\).
If we let \(\Delta t\) denote the length of the time interval, we can approximate the displacement
and write
\[ \text {displacement}\approx v(0)\cdot \Delta t+v(2)\cdot \Delta t=1\cdot 2+2\cdot 2=\answer [given]{6} \unit {ft}/\unit {s} \]
Using sigma notation, we write
\[ \text {displacement}\approx \sum _{k=1}^2v((k-1)\cdot \Delta t)\Delta t \]
Since we evaluate the velocity at the sample points
\(t_{k}^*=(k-1)\cdot \Delta t\),
\(k=1,2\), we can also write
\[ \text {displacement}\approx \sum _{k=1}^2v(t_{k}^*)\Delta t. \]
This is a left Riemann sum for the function
\(v\) on the interval
\([0,4]\), when
\(n=2\). This scenario is illustrated in
the figure below.
Next, we approximate the displacement of the object assuming that the velocity is
constant during one-second time intervals, and that velocity is equal to the velocity
at the beginning of each time interval.
Under this assumption, during the first second, the object would move by \(v(0)\cdot 1 \unit {ft}\); during
the second time interval of one second, the object would move by \(v(1)\cdot 1 \unit {ft}\), during the third
second, the object would move by \(v(2)\cdot 1 \unit {ft}\), and during the last second, it would move by
\(v(3)\cdot 1 \unit {ft}\).
During the entire time interval, the object would move by
\begin{align*} \text {displacement} &\approx v(0)\cdot \Delta t+v(1)\cdot \Delta t+v(2)\cdot \Delta t+v(3)\cdot \Delta t\\ &=\frac {\answer [given]{30}}{4} \unit {ft}/\unit {s} \end{align*}
Using sigma notation and the fact that \(\Delta t=\frac {4-0}{4}=1\), we can write
\[ \text {displacement}\approx \sum _{k=1}^4v((k-1)\cdot \Delta t)\Delta t \]
Again, we denote the
sample points by
\(t_{k}^*=(k-1)\cdot \Delta t\),
\(k=1,2,3,4\), and write
\[ \text {displacement}\approx \sum _{k=1}^4v(t_{k}^*)\Delta t \]
This is a left Riemann sum for the function
\(v\)
on the interval
\([0,4]\), when
\(n=4\). This scenario is illustrated in the figure below.
Now,
we approximate the displacement of the object assuming that the velocity is constant
during half-second time intervals, and that velocity is equal to the velocity at the
beginning of each time interval.
Under this assumption, during the first half-second, the object would move by \(v(0)\cdot \frac {1}{2} \unit {ft}\);
during the second time interval of half a second, the object would move by \(v\left (\frac {1}{2}\right )\cdot \frac {1}{2} \unit {ft}\), during
the third half- second interval, the object would move by \(v(1)\cdot \frac {1}{2} \unit {ft}\),etc.
During the entire time interval, the object would move by
\begin{align*} \text {displacement} &\approx v(0)\Delta t+v\left (\frac {1}{2}\right )\Delta t+ \cdots +v\left (\frac {7}{2}\right )\Delta t\\ &=\answer [given]{8.375} \unit {ft} \end{align*}
Using sigma notation and the fact that \(\Delta t=\frac {4-0}{8}=\frac {1}{2}\), we can write
\[ \text {displacement}\approx \sum _{k=1}^8v((k-1)\cdot \Delta t)\Delta t \]
Using the standard notation for sample points, we can write
\[ \text {displacement}\approx \sum _{k=1}^8v(t_{k}^*)\Delta t \]
This is a left Riemann sum for the function \(v\) on the interval \([0,4]\),when \(n=8\). This scenario is
illustrated in the figure below.
We pause here to
illustrate the
\(k\)th rectangle when computing a left Riemann sum of the function
\(v\) on the interval
\([0,4]\).
We continue by approximating the displacement of the object assuming that the
velocity is constant during \(4/n\)-second time intervals, with velocity equal to the velocity
at the beginning of each time interval.
Under this assumption, during the first \(\frac {4}{n}\)-second interval, the object would move by \(v(0)\cdot \frac {4}{n} \unit {ft}\);
during the second such time interval, the object would move by \(v\left (\frac {4}{n}\right )\cdot \frac {4}{n} \unit {ft}\), during the third
time interval, the object would move by
\(v\left (2\frac {4}{n}\right )\cdot \frac {4}{n} \unit {ft}\), etc.
During the entire time interval, the object would move by
\begin{align*} \text {displacement} \approx v(0)\cdot \Delta t &+v\left (\frac {4}{n}\right )\cdot \Delta t+v\left (2\frac {4}{n}\right )\cdot \Delta t+\cdots \\ &+v\left ((n-1)\frac {4}{n}\right )\cdot \Delta t \end{align*}
Using sigma notation, notation for the length of each time interval, \(\Delta t=\frac {4-0}{n}=\frac {4}{n}\), and notation for
sample points, we write
\[ \text {displacement}\approx \sum _{k=1}^nv((k-1)\cdot \Delta t)\Delta t \]
and
\[ \text {displacement}\approx \sum _{k=1}^nv(t_{k}^*)\Delta t. \]
This is a left Riemann sum for the function
\(v\) on the interval
\([0,4]\).
What happens when we let \(n\to \infty \)?
Approximations get better and better, because we approximate the velocity with a
constant over smaller and smaller time intervals. Therefore, it seems reasonable to
write
\[ \text {displacement}=\lim _{n\to \infty }\sum _{k=1}^nv(t_k^*)\Delta t \]
By the definition of the definite integral, we can write
\[ \text {displacement}=\int _{0}^{4}v(t)\d t. \]
This gives us another interpretation of the definite integral. Namely,
“definite integral of the velocity over \([a,b]\)= the displacement over \([a,b]\).”
On the other hand, the displacement of an object during the time interval \([0,4]\) is given by
\[ \text {displacement}=\text {change in position} \]
The change in position can be written in terms of \(s\), the position function of the object
\[ \text {displacement}=s(4)-s(0). \]
But, we know that
\(s\) is an antiderivative of
\(v\). Therefore,
\[ s(t)=\int v(t)\d t=\int \left (\frac {t^2}{4}+1\right )\d t=\frac {t^3}{\answer [given]{12}}+\answer [given]{t}+C \]
If
\(t=0\), we get
\[ s(0)=C \]
So,
the position function is given by
\[ s(t)=\frac {t^3}{\answer [given]{12}}+\answer [given]{t}+s(0). \]
Now we can compute the displacement
\[ \text {displacement}=s(4)-s(0)=\frac {\answer [given]{28}}{3}. \]
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