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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?

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

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:

givens
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\begin

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[ x_

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= \mbox

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] [ x = \mbox

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] [v_

= \mbox

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] [ v = \mbox

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]\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:

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\begin

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[ v^

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= v_

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^2 + 2a(x - x_

) ]\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:

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\begin

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[ v^

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= 2ax ]\end

which greatly simplifies the algebra needed to isolate the acceleration. We find:

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\begin

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[ a = \frac{v^{2}}

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] \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.

Part B

What is the time required for takeoff?

Part C

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

Part D

What is the acceleration during landing?

Part E

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

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?

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