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{table:align=right}{tr}{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.  

h3. Part A

Assuming constant acceleration, what is the Tomahawk's acceleration during the takeoff run?

System:  In each of the parts, we will treat the Tomahawk as a point particle.

Model:  [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)].

Approach:  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!

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_.

h3. Part B

What is the elapsed time required for takeoff? from the instant the plane begins its takeoff run until it lifts off the ground?

Approach: The system, model and methods are the same as in Part A.  This time, however, we have an extra given.  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}



h3. Part C

How far does the plane travel in the first 1.0 seconds of the takeoff?


h3. Part D

What is the acceleration during landing?

h3. Part E

How far does the plane travel in the first 1.0 seconds of the landing?

h3. Part F

Suppose that the Tomahawk was landing in a tailwind of 5 knots, so that it actually approached the runway moving at 51 knots along the ground instead of 46 knots.  How much distance would the plane require to stop?