Problem 5.7.1.
Since \(\dfdx{x}{t}=1\) we have \(x=t\text{.}\) Since \(\dfdx{y}{t}=-9.8t\) and we know that \(y=-4.9t^2.\) Use this to find a formula for \(y=y(x)\) and confirm that the arc is a parabola.
Section 5.7 Thinking Dynamically
If \(s(t)\) represents the position of an object moving in a straight line we have seen that its velocity is given by \(\dfdx{s}{t}\text{.}\) When the object is falling vertically we will use \(y(t)\) to represent its vertical position, so that \(\dfdx{y}{t}\) is its vertical velocity. Similarly if it is moving horizontally we will use \(x(t)\) and \(\dfdx{x}{t}\text{.}\)
Recall that in Problem 3.6.9 we imagined releasing an object from a helicopter which is flying horizontally horizontally with a speed of \(\dfdx{x}{t}=1\frac{\text{meter}}{\text{second}}\text{.}\) From Galileo’s work we now know that its vertical velocity will be \(\dfdx{y}{t}=9.8 t\, \frac{\text{meters}}{\text{second}}\text{.}\)
After its release the ball will fall neither horizontally, nor vertically. It won’t even fall along a straight line. If we ignore air resistance then the combination of the ball’s horizontal velocity (inherited from the helicopter) and its vertical velocity (due to the force of gravity) will cause the flight path to curve into an arc similar to that shown below. In our sketch the value of \(s(t)\) is the distance the ball has traveled along its flight path.
As the ball falls there are three distinct quantities we are interested in: the horizontal velocity of the ball, \(\dfdx{x}{t}\text{,}\) the vertical velocity of the ball, \(\dfdx{y}{t}\text{,}\) and its speed in the direction of travel, \(\abs{\dfdx{s}{t} }\text{.}\)
From the Pythagorean Theorem we see that
\begin{equation*}
\abs{\dfdx{s}{t}}
=\sqrt{\left(\dfdx{x}{t}\right)^2+{\left(\dfdx{y}{t}
\right)}^2}\text{.}
\end{equation*}
But what is not as clear is why the length of the diagonal (tangent) arrow in the sketch above should represent the (instantaneous) speed of the object. To see this, it will be helpful to look at a differential triangle and observe what is going on at an infinitely small scale.
From the Principle of Local Linearity, we know that during an infinitely small time change \(dt\text{,}\) the change in distance traveled, \(\dx{s}\text{,}\) along the curve is the same as the change along the tangent line. Even though the differential triangle above is infinitely small, we can still apply the Pythagorean theorem to get
\begin{equation*}
(\dx{s})^2=(\dx{x})^2+(\dx{y})^2.
\end{equation*}
Thus
\begin{equation*}
\abs{\dx{s}}=\sqrt{(\dx{x})^2+(\dx{y})^2}.
\end{equation*}
Dividing by \(\abs{\dx{t}}\) we see that the speed in the direction of travel at any time \(t\) is
\begin{align*}
\abs{\dfdx{s}{t}}\amp =\frac{\sqrt{(\dx{x})^2+(\dx{y})^2}}{\dx{t}}\\
\amp =\sqrt{\left(\dfdx{x}{t}\right)^2+\left(\dfdx{y}{t}\right)^2}\\
\amp = \sqrt{1+(9.8t)^2}.
\end{align*}
Problem 5.7.2.
Suppose \((x,y)\) are the coordinates of a ball moving along the given curves. If the ball is moving with a constant horizontal velocity of \(\dfdx{x}{t}=2\frac{\text{ units}}{\text{second}}\) for each curve find the following.
- The vertical velocity of the ball at the points indicated.
- The horizontal and vertical acceleration of the ball at the points indicated.
- The speed in the direction of travel of the ball at the points indicated.
(a)
\(y=x^2,\) at \((-1,1)\) and \((1,1)\)
(b)
\(y=x^3,\) at \((-1,-1)\) and \((1,1)\)
(c)
\(x^2-y^2=3^2\) at \((-5,4)\) and \((-5,-4)\)
Recall that we stopped analyzing in Problem 5.3.4 because we needed to be able to account for both the velocity and the acceleration of the craft. We now have everything we need to finish the analysis of the flight path of the commercial airliner which we started in Example 5.3.2.
Problem 5.7.3.
Recall that in Problem 5.3.4 we asked you find the equation of this flight path from the data in the figure at the right. You should have found that
\begin{equation*}
y=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2.
\end{equation*}
As we said when we suspended our earlier efforts, the problem with this picture is that it is static. It is a picture of the path the airliner has flown after the flight is finished. We now have the tools we need to address the characteristics of the flight as it occurs.
(a)
Use the differential triangle in our figure to show that airspeed of the airplane in its direction of motion, \(\abs{\dfdx{s}{t}}\text{,}\) its horizontal velocity, \(\dfdx{x}{t}\text{,}\) and slope of its flight path, \(\dfdx{y}{x}\text{,}\) are related by the formula:
\begin{equation}
\abs{\dfdx{s}{t}}=\abs{\dfdx{x}{t}}\sqrt{1+\left(\dfdx{y}{x}\right)^2}.\tag{5.10}
\end{equation}
Hint.
Recall that \(\dfdx{x}{t}\text{,}\) and \(\dfdx{y}{t}\) are the horizontal and vertical velocities respectively
(b)
Assume that the plane’s horizontal velocity is constant.
-
Use equation (5.10) to explain how we know that the plane is traveling fastest when it is at the point where the curve is steepest (about where the red dot is in our sketch).
- What is the slowest speed attained by the plane, and where on the flight path does this occcur?
The next problem will complete our analysis of the flight path of our commercial airliner.
Problem 5.7.4.
The flight path obtained Problem 5.3.4 was \(y=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2 \text{.}\) We need to put some limitations on the vertical acceleration, \(\dfdxn{y}{t}{2}\text{,}\) experienced by the passengers in a commercial airliner. To keep things simple we assume that the pilot must maintain a constant horizontal speed of \(v\frac{\text{
meters}}{\text{second}}\text{.}\) (That is, set \(\dfdx{x}{t}=-v\text{.}\))
(a)
Use the above equation to show that the vertical acceleration is given by
\begin{equation*}
\dfdxn{y}{t}{2}=\frac{6hv^2}{l^2}\left(1-\frac{2x}{l}\right).
\end{equation*}
(b)
On the interval \(0\le x\le l\text{,}\) what is the largest vertical acceleration and what is the smallest vertical acceleration and where do they occur? Does this make sense physically?
(c)
Suppose we put a restriction on the vertical acceleration so that
\begin{equation*}
-k\frac{\text{ meters}}{\text{second}^2}\le\dfdxn{y}{t}{2}\le k\frac{\text{ meters}}{\text{second}^2}
\end{equation*}
for some constant \(k\text{.}\) Show that with this restriction, \(l\ge\sqrt{\frac{6hv^2}{k}}
\text{.}\)
(d)
Suppose that initially \(h=10000\) meters, \(v=100\frac{\text{
meters}}{\text{second}}\text{,}\) \(k=0.1\frac{\text{ meters}}{\text{sec}^2}\) (which is approximately \(1\%\) of the acceleration due to gravity). Find what \(l\) must be (in kilometers).
In Section 5.3 we mentioned that NASA claims that the Vomit Comet can make passengers experience weightlessness for about \(25\) seconds. Let’s check on that claim.
To simulate weightlessness (neutral buoyancy) the pilot must execute a parabolic flight path \(y=Ax^2+Bx+C\text{.}\) In Problem 5.3.1 you should have found that \(B\) and \(C\) were \(1\) and \(7000\text{,}\) respectively, so the flight path is
\begin{equation}
y=Ax^2+x+7000\tag{5.11}
\end{equation}
with \(A\) yet to be determined.
The pilot is climbing at angle of elevations \(45^\circ\) to an altitude of about \(7000\) meters and then follows this parabolic path to produce a vertical acceleration of \(\dfdxn{y}{t}{2}=-9.8
\frac{\text{meters}}{\text{second}^2}\) (matching the acceleration due to gravity) and horizontal acceleration of \(\dfdxn{x}{t}{2}=0\text{.}\) This will provide neutral buoyancy inside the plane. On the way back down the pilot pulls out of this dive when the altitude returns to \(7000\) meters.
For training purposes this is repeated \(40\) times.
Problem 5.7.5.
(a)
To determine \(A\) we need one more fact. At the beginning of the maneuver, the initial airspeed is about \(180\frac{\text{ meters}}{\text{second}}\) (approximately \(400\) mph). Use the initial airspeed to determine \(\dfdx{x}{t}\text{.}\) In turn use this result and the fact that \(\dfdxn{y}{t}{2}=-9.8\) to show that \(A=-\frac{9.8}{180^2}\approx -0.0003\text{.}\)
Notice that you are not asked to find \(A\text{.}\) The problem is to show that the value we’ve given is correct.
(b)
Now that we’ve determined all of the coefficients, equation (5.11) describes the flight path of the Vomit Comet. Use equation (5.11) to determine the value of \(x\) when the pilot pulls out of the dive at \(7000\) meters.
(c)
Next find the value of \(t\) when the pilot pulls out of the dive. How does this compare with the \(25\) second claim?
Hint.
You may find it helpful to review, part (b) of Problem 5.5.2.
(d)
If the pilot wants to maintain a constant horizontal speed what must the airspeed of the plane be in terms of \(t\) (\(t=0\) representing the start of the maneuver)?
