Question

The forces on a small airplane (mass \(1160 \mathrm{kg}\) ) in horizontal flight heading eastward are as follows: gravity \(=16.000 \mathrm{kN}\) downward, lift \(=16.000 \mathrm{kN}\) upward, thrust \(=1.800 \mathrm{kN}\) eastward, and \(\mathrm{drag}=1.400 \mathrm{kN}\) westward. At \(t=0,\) the plane's speed is \(60.0 \mathrm{m} / \mathrm{s} .\) If the forces remain constant, how far does the plane travel in the next \(60.0 \mathrm{s} ?\)

Short answer

Answer: The airplane travels approximately 4221 meters eastward in 60.0 seconds.

Step-by-step solution

Sum up horizontal forces to find net force

In this problem, we only need to analyze horizontal forces, which are thrust (eastward) and drag (westward). To find the net horizontal force (\(F_{net}\)), subtract the drag force from the thrust force: \(F_{net} = F_{thrust} - F_{drag} = 1800 - 1400 = 400\,\text{N}\).

Calculate the airplane's acceleration

Now that we have the net force, we can find the airplane's acceleration using Newton's second law, which states \(F = ma\). Rearrange the formula to solve for acceleration (\(a\)): \(a = \frac{F_{net}}{m} = \frac{400\,\text{N}}{1160\,\text{kg}} \approx 0.345\,\text{m/s}^2\)

Use kinematic equations to find the distance traveled

We are given the initial speed (\(v_0 = 60.0\,\text{m/s}\)), time (\(t = 60.0\,\text{s}\)), and acceleration (\(a = 0.345\,\text{m/s}^2\)). We can use the kinematic equation for the distance traveled, which is: $$ d = v_0t + \frac{1}{2}at^2 $$ Plug in the given values and compute the distance: $$ d = (60.0\,\text{m/s})(60.0\,\text{s}) + \frac{1}{2} (0.345\,\text{m/s}^2)(60.0\,\text{s})^2 \approx 3600\,\text{m}+ 621\,\text{m} = 4221\,\text{m} $$

Report the final answer

In the next 60.0 seconds, the airplane travels a distance of approximately 4221 meters eastward.