Two airplanes leave an airport at the same time. The velocity of the first airplane is 700 m/h at...

Question:

Two airplanes leave an airport at the same time. The velocity of the first airplane is 700 m/h at a heading of 52.2 degrees. The velocity of the second is 600 m/h at a heading of 97 degrees. How far apart are they after 3.1 h? Answer in units of m.

Distance

First, we need to know the distance traveled from the airport by each of the airplanes by using the formula {eq}d = vt{/eq}. After that, we can now find the distance between two airplanes using the cosine law.

Answer and Explanation:

Given:

  • Velocity of the first plane {eq}v_1 = 700 \rm\ mi/h = 312.93 \rm\ m/s{/eq}
  • Velocity of the second plane {eq}v_2 = 600 \rm\ mi/h = 268.22 \rm\ m/s{/eq}
  • Angle of the first plane {eq}\theta_1 = 52.2 ^{\circ}{/eq}
  • Angle of the second plane {eq}\theta_2 = 97 ^{\circ}{/eq}
  • Time {eq}t = 3.1 \rm\ hr = 11,160 \rm\ s{/eq}


The FBD is given below


The angle {eq}\theta{/eq} is given by

$$\begin{align} \theta &= \theta_2 - \theta_1\\ \theta &= 97 ^{\circ} - 52.2 ^{\circ}\\ \theta &= \boxed{44.8 ^{\circ}} \end{align} $$


Next, the distance traveled by each plane is

  • Plane 1

$$\begin{align} d_1 &= v_1 t\\ d_1 &= (312.93 \rm\ m/s)(11,160 \rm\ s)\\ d_1 &= 3492298.8 \rm\ m \end{align} $$


  • Plane 2

$$\begin{align} d_2 &= v_2 t\\ d_2 &= (268.22 \rm\ m/s)(11,160 \rm\ s)\\ d_2 &= 2993335.2 \rm\ m \end{align} $$


Now, applying the cosine law to obtain the distance between the two plane

$$\begin{align} d &= \sqrt{d_1^2 + d_2^2 - 2d_1d_2cos\theta}\\ d &= \sqrt{(3492298.8 \rm\ m)^2 + (2993335.2 \rm\ m)^2 - 2(3492298.8 \rm\ m)(2993335.2 \rm\ m)cos(44.8 ^{\circ})}\\ d &= \boxed{2514167.67 \rm\ m} \end{align} $$


Learn more about this topic:

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Distance, Time & Average Speed: Practice Problems

from Physical Science: Middle School

Chapter 1 / Lesson 3
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