Sketch a diagram describing the given situation, and then recall that the velocity is distance divided by time.
≈ 2.8 mi/min
Practice makes perfect
We are given that an airplane at a constant altitude a flies a horizontal distance d toward us at velocity v, and observing for time t we measure angles of elevation ∠ E_1 and ∠ E_2 at the start and end of our observation. Let's sketch a diagram describing this situation.
Now let's substitute the values we are given in our exercise. Additionally, we will label the horizontal distance between us and the position of the airplane at the end of our observation by x.
To find the values of x and d, we will use one of the trigonometric ratios. Let's recall that the tangent of an angle is the ratio between the side opposite this angle and the side adjacent to this angle. Using this definition, we can write equations for tan 50^(∘) and tan 40^(∘).
tan 50^(∘)=2/x & (I) tan 40^(∘)=2/x+d & (II)
Now we will solve the system of equations using the Substitution Method. Our first step will be to solve the first equation for x.
The value of x is approximately 1.68 miles and the value of d is approximately 0.7 miles. Let's add this information to our diagram.
Finally, as we are asked to evaluate the velocity, let's recall that the velocity is the quotient of distance and time.
v=d/t
To find the velocity of this airplane, we will substitute 0.7 for d and 0.25 for t as we want our result to be in miles per minute, not seconds.
v=0.7/0.25=2.8
The velocity of this airplane is approximately 2.8 miles per minute. Notice that this is an approximation as we used an approximated value of d to evaluate it.
