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Photo by Craig Kinzer, courtesy Wikipedia. |
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.
Part A
Assuming constant acceleration, what is the Tomahawk's acceleration during the takeoff run?
Solution
System: In each of the parts, we will treat the Tomahawk as a point particle.
Interactions: 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.
Model: One-Dimensional Motion with 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 xi (= 0 feet in our coordinates).
We also know that v, the speed at liftoff is 55 knots. Finally, we assume that the starting speed, vi = 0 knots based upon the setup of the problem. Thus, our givens are:
\begin
[ x_
= \mbox
] [ x = \mbox
] [v_
= \mbox
] [ v = \mbox
]\end
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:
\begin
[ v^
= v_
) ]\end
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 xi and vi are each zero, we can write:
\begin
[ v^
= 2ax ]\end
which greatly simplifies the algebra needed to isolate the acceleration. We find:
\begin
[ a = \frac{v^{2}}
] \end
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]. The appropriate [conversions] in this case are: 1 foot = 3.281 m and 1 knot = 0.514 m/s. After conversions, our (nonzero) givens are:
\begin
[x = \mbox
] [ v = \mbox
]\end
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. 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.)
With our converted givens, we are ready to substitute, finding:
\begin
[ a = \frac{(\mbox
)^{2}}{2(\mbox
)} = \mbox
^
]\end
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.
Part B
What is the elapsed time from the instant the plane begins its takeoff run until it lifts off the ground?
Solution
System, Interactions and Model: The system, interactions and model are the same as in Part A.
Approach:
Using the work we did to convert in Part A, we have the givens:
\begin
[ x_
= \mbox
][x = \mbox
] [v_
= \mbox
] [v = \mbox
][ a = \mbox
^
]\end
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 ti = 0 seconds. (Just as we chose xi = 0 in setting up our coordinate system.)
Unless explicitly told to assume a certain ti, we will generally take ti = 0 to simplify the equations.
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:
\begin
[ x = x_
+ \frac
(v+v_
)(t-t_
) ] \end
becomes, after using the zeros:
\begin
[ x = \frac
v t ] \end
which is solved to give:
\begin
[ t = \frac
= \mbox
] \end
Method 2
If you did solve Part A, the simplest approach is to use the simplest of the Laws of Change:
\begin
[ v = v_
) = at ] \end
which gives:
\begin
[ t = \frac
= \mbox
] \end
Method 3
You could also choose the slightly more complicated:
\begin
[ x = x_
(t-t_
) + \frac
a(t-t_
)^
= \frac
a t^
]\end
giving:
\begin
[ t = \pm \sqrt{\frac
{a}} = \pm \mbox
] \end
We select the plus sign, since the plane began its run at ti = 0 s and it cannot possibly lift off before it has begun to move!
Whenever taking a square root in physics it is important to use your understanding of the problem to choose the appropriate sign.
Part C
How far does the plane travel in the first 9.0 seconds of the takeoff?
Solution
System, Interactions and Model: As in the previous parts.
Approach:
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:
\begin
[ x_
= \mbox
][ v_
= \mbox
][t_
= \mbox
] [t = \mbox
] [ a = \mbox
^
]\end
The unknown that we seek is x, the location at t = 9.0 s. The most direct approach is to use:
\begin
[ x = x_
(t-t_
) + \frac
a(t-t_
)^
= \frac
at^
] \end
No algebra is necessary, since x is already isolated. We can simply substitute to find:
\begin
[ x = \mbox
] \end
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 t2 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.
Part D
What is the acceleration during landing?
Solution
System, Interactions and Model: As in the previous parts.
Approach:
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:
and summarize our givens (this time already converted to SI units):
\begin
[ x_
= \mbox
][x = \mbox
][ v_
= \mbox
][ v = \mbox
]\end
Then, we choose the appropriate equation:
\begin
[ v^
= v_
^
+ 2a(x-x_
) ] \end
and substitute zeros:
\begin
[ 0 = v_
^
+ 2ax ] \end
Finally, we solve symbolically and substitute:
\begin
[ a = \frac{-v_
^{2}}
= -\mbox
^
]\end
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.
Part E
How far does the plane travel in the first 9.0 seconds of the landing?
Solution
System, Interactions and Model: As in previous parts.
Approach:
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:
\begin
[ x_
= \mbox
][ v_
= \mbox
][ a= -\mbox
^
]\end
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.
As in Part C, we will choose ti = 0 s, and we will use the equation:
\begin
[ x = x_
(t-t_
) + \frac
a(t-t_
)^
= v_
t + \frac
a t^
] \end
Note we cannot cancel the initial velocity for the case of a landing plane.
Again, no algebra is required, so we substitute:
\begin
[ x = (\mbox
)(\mbox
) + \frac
(-\mbox
^
)(\mbox
)^
= \mbox
] \end
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.
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