Training Flight

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).

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:

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:

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:

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

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:

With our converted givens, we are ready to substitute, finding:

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.

Using the work we did to convert in Part A, we have the givens:

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.)