A plane is going in circles around an airport. The plane takes 3 minutes to complete one round. The angle of elevation of the plane from a point P on the ground at time t seconds is equal to that at time (t + 30) seconds. At time (t + x) seconds, the plane flies vertically above the point P. What is x equal to?

Question

Prepp Team · Accepted Answer

Understanding the Plane's Circular Motion

The problem describes a plane executing circular motion around an airport. We are given the time it takes to complete one full circle and information about its angle of elevation from a specific point P on the ground.

Time Period: The plane completes one round in 3 minutes. We convert this to seconds: $T = 3 \text{ minutes} \times 60 \text{ seconds/minute} = 180 \text{ seconds}$.
Angular Speed: Since the plane completes a full circle ($360^\circ$) in 180 seconds, its angular speed ($\omega$) is constant. $$ \omega = \frac{360^\circ}{180 \text{ s}} = 2^\circ/\text{s} $$ This means the plane covers 2 degrees of its circular path every second.

Analyzing the Angle of Elevation Condition

We are told that the angle of elevation ($\alpha$) from point P at time $t$ is the same as at time $t + 30$ seconds. Mathematically, this condition is $\alpha(t) = \alpha(t+30)$.

Let's consider the plane's angular position, denoted by $\theta(t)$, measured in degrees from a reference direction. Assuming P is at the center of the horizontal projection of the circular path, the angle of elevation might be constant. However, the condition $\alpha(t) = \alpha(t+30)$ suggests there's a dependency or symmetry involved. The angular position of the plane at time $t$ is:

$$ \theta(t) = (\omega \times t) \pmod{360} = (2t) \pmod{360} $$

The time difference given is 30 seconds. The angle covered during this time is:

$$ \Delta\theta = \omega \times 30 \text{ s} = (2^\circ/\text{s}) \times 30 \text{ s} = 60^\circ $$

So, the plane's angular position changes by $60^\circ$ between time $t$ and time $t+30$. The condition $\alpha(t) = \alpha(t+30)$ often implies a symmetry in the setup. For instance, if the angle of elevation depends on the angular position $\theta$ as $f(\theta)$, the condition $f(\theta(t)) = f(\theta(t+30))$ means $f(\theta) = f(\theta + 60^\circ)$. This implies the function $f$ is periodic with a $60^\circ$ period. A common interpretation leading to a solvable problem is that the positions $\theta$ and $\theta+60^\circ$ are related such that the elevation angle is the same. This can happen if $\theta = -(\theta + 60^\circ) \pmod{360}$, which simplifies to $2\theta = -60^\circ \pmod{360}$, or $2\theta = 300^\circ \pmod{360}$. Solving for $\theta$ gives:

$$ \theta(t) \equiv 150^\circ \pmod{180} $$

This means the condition $\alpha(t) = \alpha(t+30)$ holds true when the plane's angular position is $150^\circ$ or $330^\circ$ within its 180-second cycle.

Calculating the Specific Time

We can now calculate the specific times ($t$) when these angular positions occur during the flight cycle:

For $\theta(t) = 150^\circ$: $$ t_1 = \frac{150^\circ}{\omega} = \frac{150^\circ}{2^\circ/\text{s}} = 75 \text{ seconds} $$
For $\theta(t) = 330^\circ$: $$ t_2 = \frac{330^\circ}{\omega} = \frac{330^\circ}{2^\circ/\text{s}} = 165 \text{ seconds} $$

The problem asks for the value of $x$, where $t+x$ is the time when the plane flies vertically above point P. The statement "vertically above the point P" implies that point P is directly below the center of the circular path. Let's define the position $\theta = 0^\circ$ as the moment the plane is vertically above P. The cycle completes at $t=180$ seconds, which is the next time the plane is vertically above P.

The question implies we should consider a time $t$ that satisfies the angle condition $\alpha(t) = \alpha(t+30)$, and then find how much additional time ($x$) passes until the plane is vertically above P. If there are multiple such times $t$, we typically consider the earliest one. The first time the condition holds is $t_1 = 75$ seconds.

So, we start at $t = 75$ seconds. We need to find $x$ such that the plane is vertically above P at time $t+x = 75+x$. The next time the plane is vertically above P is at the end of the cycle, $t_{overhead} = 180$ seconds.

Setting up the equation:

$$ 75 \text{ s} + x = 180 \text{ s} $$

Solving for $x$:

$$ x = 180 \text{ s} - 75 \text{ s} $$ $$ x = 105 \text{ seconds} $$

Conclusion

By analyzing the plane's circular motion and the given condition on the angle of elevation, we find that the condition is met at $t=75$ seconds (and $t=165$ seconds). Starting from $t=75$ seconds, the time it takes for the plane to reach the position vertically above point P (at $t=180$ seconds) is $x=105$ seconds.

Prepp Team · Answer

75 seconds

Prepp Team · Answer

90 seconds

Prepp Team · Answer

135 seconds

Understanding the Plane's Circular Motion

Analyzing the Angle of Elevation Condition

Calculating the Specific Time

Conclusion

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