| Wiki Markup |
|---|
{composition-setup}{composition-setup}
{table:alignborder=right1|cellspacingframe=0void|cellpaddingrules=1cols|bordercellpadding=18|frame=box|width=30%cellspacing=0}
{tr:valign=top}
{td:alignwidth=center350|bgcolor=#F2F2F2}*[Examples from Motion and Acceleration]*
{td}
{tr}
{tr}
{td}
{pagetree:root=Examples from Kinematics}
{search-box}
{td}
{tr}
{table}
{table}{tr}{td}live-template:Left Column}
{td}
{td}
|!tomahawk_craigKinzer.jpg!{td}{tr}{tr}{td}|
|Photo by Craig Kinzer, courtesy Wikipedia.{td}{tr}{table}|
The Piper Tomahawk, widely used for flying lessons, has a liftoff speed of about 55 knots and a landing speed of about 46 knots. The listed ground-roll for takeoff is 820 feet, and for landing is 635 feet. {excerpt}In this example we will calculate acceleration, time, speed, and distance{excerpt}
{composition-setup}{composition-setup}
{deck:id=partdeck}
{card:label=Part A}
h3. Part A
Assuming constant acceleration, what is the Tomahawk's acceleration during the takeoff run?
h4. Solution
{toggle-cloak:id=sysa} *System:* {cloak:id=sysa}In each of the parts, we will treat the Tomahawk as a point particle.{cloak}
{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}We are assuming that the thrust from the plane's engine, air resistance, rolling friction, and all other influences add up to give a constant acceleration.{cloak}
{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)].{cloak}
{toggle-cloak:id=appa} *Approach:*
{cloak:id=appa}
Once we have determined that we are using 1-D Motion with Constant Acceleration, we have essentially reduced the problem to math. The Constant Acceleration model is somewhat unique in that there is an overabundance of equations (Laws of Change) to choose from. In order to select the proper equation or equations, we have to clearly understand the information presented in the problem (often called the _givens_) and also what we are asked to find (the _unknowns_).
To understand our givens, it is a good idea to first develop a coordinate system. By drawing out the runway and marking where the plane starts its takeoff run (x = 0 feet in the picture) and where it lifts off the ground (820 feet in the picture) we can see that the 820 feet given in the problem statement is a _distance_ not a position. The difference between the plane's liftoff position and the starting point is 820 feet. Thus, in terms of the equations, we have *both* _x_ (= 820 feet in our coordinates) *and* _x_~i~ (= 0 feet in our coordinates).
!runway1.png|width=700!
We also know that _v_, the speed at liftoff is 55 knots. Finally, we assume that the starting speed, _v_~i~ = 0 knots based upon the setup of the problem. Thus, our givens are:
{panel:title=givens}{latex}\begin{large} \[ x_{\rm i} = \mbox{0 feet} \] \[ x = \mbox{820 feet} \] \[v_{\rm i} = \mbox{0 knots} \] \[ v = \mbox{55 knots} \]\end{large}{latex}{panel}
The unknown for this problem is the acceleration, since that is what we are asked to find. We can now look over the many possible Laws of Change to see which will allow us to use our givens to solve for our unknowns. In this case, it is pretty clear that the best choice is:
{latex}\begin{large} \[ v^{2} = v_{\rm i}^2 + 2a(x - x_{\rm i}) \]\end{large}{latex}
This equation uses all of our givens, and it contains our unknown. The other possibilities that contain the acceleration are ruled out because they also contain time. We have no information about the elapsed time for the takeoff.
Once we have the equation, we can solve symbolically for the unknown using algebra. It is good practice to solve algebraically with variables *before* substituting numbers. The only exception is that you are encouraged to substitute any *zeros* before doing the algebra. Substituting zeros simplifies the equation. In this case, because _x_~i~ and _v_~i~ are each zero, we can write:
{latex}\begin{large} \[ v^{2} = 2ax \]\end{large}{latex}
which greatly simplifies the algebra needed to isolate the acceleration. We find:
{latex}\begin{large} \[ a = \frac{v^{2}}{2x} \] \end{large}{latex}
Now, we would like to substitute in the given values, but we have an important problem. The units of our givens are mismatched. We have _x_ in units of feet, and _v_ in units of knots. If we substitute into the equation we have found, we will recover an acceleration in the units of square knots per foot. This is certainly not a standard unit of acceleration. Because the equations of motion with constant acceleration involve many quantities with different dimensions, it is good practice always to convert to [SI units|Introduction to Units]. The appropriate [conversions|Common Conversions] in this case are: 1 foot = 3.281 m and 1 knot = 0.514 m/s. After conversions, our (nonzero) givens are:
{panel:title=converted givens}{latex}\begin{large}\[x = \mbox{250 m} \] \[ v = \mbox{28.3 m/s} \]\end{large}{latex}{panel}
{note}We have retained three digits for the velocity (and also for the position, though it is not obvious) even though the original given, 55 knots, contained only 2 [significant figures|Significant Figures]. This is not a mistake. When performing conversions, you should treat them as _intermediate steps_, not final answers. Retain extra digits through all intermediate steps and round at the end. (This is one important way that solving with symbols before substituting numbers will help you. It reduces the number of intermediate calculations and hence limits the opportunities for rounding errors to enter the calculation.){note}
With our converted givens, we are ready to substitute, finding:
{latex}\begin{large}\[ a = \frac{(\mbox{28.3 m/s})^{2}}{2(\mbox{250 m})} = \mbox{1.6 m/s}^{2}\]\end{large}{latex}
This is basically the answer, though the problem is slightly ambiguous. The word "acceleration" could stand for the vector or simply the magnitude. The magnitude is 1.6 m/s ^2^. Reporting the full vector is tricky, since we are not told which way the plane is moving as it takes off. The best we could do is to say that the vector acceleration is 1.6 m/s ^2^ _in the direction of the plane's movement_.
{cloak}
{card:Part A}
{card:label=Part B}
h3. Part B
What is the elapsed time from the instant the plane begins its takeoff run until it lifts off the ground?
h4. Solution
*System, Interactions and Model:* The system, interactions and model are the same as in Part A.
{toggle-cloak:id=appb} *Approach:*
{cloak:id=appb}
Using the work we did to convert in Part A, we have the givens:
{panel:title=givens}{latex}\begin{large}\[ x_{\rm i} = \mbox{0 m}\]\[x = \mbox{250 m}\] \[v_{\rm i} = \mbox{0 m/s}\] \[v = \mbox{28.3 m/s} \]\[ a = \mbox{1.6 m/s}^{2}\]\end{large}{latex}{panel}
Note that we have used the acceleration of Part A as a given for Part B. This extra given makes an enormous difference in the next step of the problem. Instead of one path to the solution, we can now find at least three different equations among the model's Laws of Change that will work. This kind of redundancy is common in this particular model. To illustrate, we will solve the problem using different methods. For simplicity, in using each of the methods, we will *choose* to make _t_~i~ = 0 seconds. (Just as we chose _x_~i~ = 0 in setting up our coordinate system.)
{note}Unless explicitly told to assume a certain _t_~i~, we will generally take _t_~i~ = 0 to simplify the equations.{note}
{deck:id=methdeck}
{card:label=Method 1}
h4. Method 1
Suppose you were unable to solve Part A. It is still possible to solve Part B without knowing that the acceleration is 1.6 m/s ^2^. The equation:
{latex}\begin{large} \[ x = x_{\rm i} + \frac{1}{2}(v+v_{\rm i})(t-t_{\rm i}) \] \end{large}{latex}
becomes, after using the zeros:
{latex}\begin{large} \[ x = \frac{1}{2} v t \] \end{large}{latex}
which is solved to give:
{latex}\begin{large} \[ t = \frac{2 x}{v} = \mbox{18 s} \] \end{large}{latex}
{card:Method 1}
{card:label=Method 2}
h4. Method 2
If you did solve Part A, the simplest approach is to use the simplest of the Laws of Change:
{latex}\begin{large} \[ v = v_{\rm i} + a(t-t_{\rm i}) = at \] \end{large}{latex}
which gives:
{latex}\begin{large} \[ t = \frac{v}{a} = \mbox{18 s} \] \end{large}{latex}
{card:Method 2}
{card:label=Method 3}
h4. Method 3
You could also choose the slightly more complicated:
{latex}\begin{large} \[ x = x_{\rm i} + v_{\rm i}(t-t_{\rm i}) + \frac{1}{2}a(t-t_{\rm i})^{2} = \frac{1}{2} a t^{2} \]\end{large}{latex}
giving:
{latex}\begin{large}\[ t = \pm \sqrt{\frac{2x}{a}} = \pm \mbox{18 s} \] \end{large}{latex}
We select the plus sign, since the plane began its run at _t_~i~ = 0 s and it cannot possibly lift off before it has begun to move!
{note}Whenever taking a square root in physics it is important to use your understanding of the problem to choose the appropriate sign.{note}
{card:Method 3}
{deck:methdeck}
{cloak}
{card:Part B}
{card:label=Part C}
h3. Part C
How far does the plane travel in the first 9.0 seconds of the takeoff?
h4. Solution
*System, Interactions and Model:* As in the previous parts.
{toggle-cloak:id=appc} *Approach:*
{cloak:id=appc}
Based upon the results of Part B, we know that the plane has not reached its takeoff speed within 9.0 seconds, and also that it has not traveled its full takeoff distance. We can still use the result of Part A that the acceleration is 1.6 m/s ^2^, however, since we have been assuming the acceleration is constant throughout the run. Thus, we are left with the truncated set:
{panel:title=givens}{latex}\begin{large}\[ x_{\rm i } = \mbox{0 m} \]\[ v_{\rm i} = \mbox{0 m/s}\]\[t_{\rm i} = \mbox{0 s} \] \[t = \mbox{9.0 s}\] \[ a = \mbox{1.6 m/s}^{2}\]\end{large}{latex}{panel}
The unknown that we seek is _x_, the location at _t_ = 9.0 s. The most direct approach is to use:
{latex}\begin{large}\[ x = x_{\rm i} + v_{\rm i} (t-t_{\rm i}) + \frac{1}{2} a(t-t_{\rm i})^{2} = \frac{1}{2}at^{2} \] \end{large}{latex}
No algebra is necessary, since _x_ is already isolated. We can simply substitute to find:
{latex} \begin{large} \[ x = \mbox{65 m} \] \end{large}{latex}
{note}It is important to note that the plane travels much less than 1/2 the takeoff distance in about half the takeoff time. In fact, based upon the _t_^2^ dependence of the formula we just employed, we expect that the plane should cover only 1/4 of the takeoff distance in the first half of the run. That leaves it to cover the remaining 3/4 in the second half. The reason for this difference is simple: the plane is constantly accelerating, and so it moves faster during the second half of the run.{note}
{cloak}
{card:Part C}
{card:label=Part D}
h3. Part D
What is the acceleration during landing?
h4. Solution
*System, Interactions and Model:* As in the previous parts.
{toggle-cloak:id=appd} *Approach:*
{cloak:id=appd}
This problem is essentially identical in structure to Part A, except that when landing, the plane's _initial_ speed is nonzero and its _final_ speed is zero. Thus, we proceed in the same fashion as in Part A. We first define a coordinate system:
!runway2.png|width=700!
and summarize our givens (this time already converted to SI units):
{panel:title=converted givens}{latex}\begin{large}\[ x_{\rm i} = \mbox{0 m} \]\[x = \mbox{194 m} \]\[ v_{\rm i} = \mbox{23.6 m/s} \]\[ v = \mbox{0 m/s} \]\end{large}{latex}{panel}
Then, we choose the appropriate equation:
{latex}\begin{large} \[ v^{2} = v_{\rm i}^{2} + 2a(x-x_{\rm i}) \] \end{large}{latex}
and substitute zeros:
{latex}\begin{large} \[ 0 = v_{\rm i}^{2} + 2ax \] \end{large}{latex}
Finally, we solve symbolically and substitute:
{latex}\begin{large} \[ a = \frac{-v_{\rm i}^{2}}{2 x} = -\mbox{1.4 m/s}^{2} \]\end{large}{latex}
Again, the question is ambiguous about what is expected. The magnitude of this acceleration is simply 1.4 m/s ^2^. To report it as a vector, the best option is probably to say 1.4 m/s ^2^ in the direction opposite the plane's motion.
{cloak}
{card:Part D}
{card:label=Part E}
h3. Part E
How far does the plane travel in the first 9.0 seconds of the landing?
h4. Solution
*System, Interactions and Model:* As in previous parts.
{toggle-cloak:id=appe} *Approach:*
{cloak:id=appe}
This problem is similar to Part C. Again, 9.0 seconds is not the full time over which the plane accelerates (can you find the total time needed to bring the plane to rest?) and so we cannot assume that the final speed is zero or the final position is 194 m. The givens that we have, including the result of Part D, are:
{panel:title=converted givens}{latex}\begin{large}\[ x_{\rm i} = \mbox{0 m} \]\[ v_{\rm i} = \mbox{23.6 m/s} \]\[ a= -\mbox{1.4 m/s}^{2}\]\end{large}{latex}{panel}
{note}It is extremely important to retain the negative sign on the acceleration when using it in the equations. We will see why in a moment.{note}
As in Part C, we will _choose_ _t_~i~ = 0 s, and we will use the equation:
{latex}\begin{large} \[ x = x_{\rm i} + v_{\rm i}(t-t_{\rm i}) + \frac{1}{2}a(t-t_{\rm i})^{2} = v_{\rm i}t + \frac{1}{2} a t^{2}\] \end{large}{latex}
{note}Note we cannot cancel the initial velocity for the case of a landing plane.{note}
Again, no algebra is required, so we substitute:
{latex}\begin{large} \[ x = (\mbox{23.6 m/s})(\mbox{9.0 s}) + \frac{1}{2}(-\mbox{1.4 m/s}^{2})(\mbox{9.0 s})^{2} = \mbox{156 m} \] \end{large}{latex}
{note}To see the importance of the negative sign, calculate the position after 9.0 s using a _positive_ acceleration. What do you find? The negative ensures that the plane's deceleration causes it to cover _less_ distance than it would if it were simply coasting at its initial speed. Similarly, a positive acceleration will cause the plane to cover _extra_ distance.{note}
{cloak}
{card:Part E}
{deck:partdeck}
{td}
{tr}
{table}
{live-template:RELATE license}
|
Page History
Overview
Content Tools
Activity