Math Problem Statement

Cities P, Q and A are in different time zones. P and Q are located 4,500 km east and 4,500 km west of A respectively. below describes the schedule of an airline operating non-stop flights between A and P, A and Q. All the times indicated are local times and on the same day. Departure Time (1) : 7 am from City A to city P at 3 pm Departure Time (2) : 9 am from City A to city Q at 12 pm All flights travel at the same speed, but effective relative speed is influenced by a steady wind blowing from east to west at 75 kmph. What is the plane's cruising speed (in kmph)?

Solution

The problem gives us details about the flight schedule between three cities (A, P, and Q), taking into account time zones and the influence of wind. We are tasked with determining the plane's cruising speed, vcruisev_{\text{cruise}}, based on the given flight times and wind conditions.

Step 1: Define Variables

  • Let the cruising speed of the plane (without wind) be vcruisev_{\text{cruise}}.
  • The wind speed is given as 75 km/h, blowing from east to west.
  • The distance from city A to city P is 4,500 km, and from A to Q is also 4,500 km.

Step 2: Account for Wind Effect

The wind affects the plane’s effective speed differently depending on the direction of travel:

  • From A to P (eastward): The wind is helping the plane, so the effective speed will be: veast=vcruise+75km/hv_{\text{east}} = v_{\text{cruise}} + 75 \, \text{km/h}
  • From A to Q (westward): The wind opposes the plane, so the effective speed will be: vwest=vcruise75km/hv_{\text{west}} = v_{\text{cruise}} - 75 \, \text{km/h}

Step 3: Time Calculations

Using the flight schedules:

  • Flight from A to P:
    • Departure time: 7:00 AM (A's local time)
    • Arrival time: 3:00 PM (P's local time)
    • Duration: The time difference between 7 AM and 3 PM is 8 hours.

Thus, the time taken for the flight from A to P is 8 hours. Therefore, the effective speed from A to P is: veast=4500km8hours=562.5km/hv_{\text{east}} = \frac{4500 \, \text{km}}{8 \, \text{hours}} = 562.5 \, \text{km/h} This gives us the equation: vcruise+75=562.5v_{\text{cruise}} + 75 = 562.5 Solving for vcruisev_{\text{cruise}}: vcruise=562.575=487.5km/hv_{\text{cruise}} = 562.5 - 75 = 487.5 \, \text{km/h}

  • Flight from A to Q:
    • Departure time: 9:00 AM (A's local time)
    • Arrival time: 12:00 PM (Q's local time)
    • Duration: The time difference between 9 AM and 12 PM is 3 hours.

Thus, the time taken for the flight from A to Q is 3 hours. Therefore, the effective speed from A to Q is: vwest=4500km3hours=1500km/hv_{\text{west}} = \frac{4500 \, \text{km}}{3 \, \text{hours}} = 1500 \, \text{km/h} This gives us the equation: vcruise75=1500v_{\text{cruise}} - 75 = 1500 Solving for vcruisev_{\text{cruise}}: vcruise=1500+75=1575km/hv_{\text{cruise}} = 1500 + 75 = 1575 \, \text{km/h}

However, this speed is much higher than the cruising speed calculated

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Math Problem Analysis

Mathematical Concepts

Algebra
Time-Speed-Distance
Wind Effect on Speed

Formulas

v_{effective} = v_{cruise} ± wind speed
v = distance / time

Theorems

Basic Speed-Time-Distance Relationship

Suitable Grade Level

Grades 9-12