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Grade 12th passModern Physics

A Boeing 707jet aircraft has a takeoff mass of 1.2 x 105kg. Each of its four engines has a thrust of 75kN. Calculate the acceleration and the length of the runway needed to become airborne if the takeoff speed is 73m/s.

Profile image of Karen espineda
4 Years agoGrade 12th pass
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Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To determine the acceleration of the Boeing 707 and the length of the runway required for it to become airborne, we can break the problem down into a few logical steps. First, we need to calculate the total thrust produced by the engines, then use Newton's second law to find the acceleration, and finally, apply kinematic equations to find the runway length needed to reach the takeoff speed.

Calculating Total Thrust

The Boeing 707 has four engines, each providing a thrust of 75 kN. To find the total thrust, we multiply the thrust of one engine by the number of engines:

  • Total thrust = Number of engines × Thrust per engine
  • Total thrust = 4 × 75 kN = 300 kN

Since 1 kN is equal to 1,000 N, we convert this to Newtons:

  • Total thrust = 300 kN = 300,000 N

Finding Acceleration

Next, we apply Newton's second law, which states that force equals mass times acceleration (F = ma). Rearranging this gives us acceleration (a) as:

  • a = F/m

Substituting the values we have:

  • F = 300,000 N
  • m = 1.2 × 105 kg

Now we can calculate the acceleration:

  • a = 300,000 N / (1.2 × 105 kg)
  • a = 2.5 m/s²

Determining Runway Length

To find the length of the runway needed for the aircraft to reach its takeoff speed of 73 m/s, we can use the kinematic equation:

  • v² = u² + 2as

Where:

  • v = final velocity (takeoff speed) = 73 m/s
  • u = initial velocity = 0 m/s (the aircraft starts from rest)
  • a = acceleration = 2.5 m/s²
  • s = distance (runway length)

Rearranging the equation to solve for s gives us:

  • s = (v² - u²) / (2a)

Substituting the known values:

  • s = (73 m/s)² / (2 × 2.5 m/s²)
  • s = 5329 m²/s² / 5 m/s²
  • s = 1065.8 m

Summary of Results

In conclusion, the Boeing 707 will experience an acceleration of 2.5 m/s², and it will require approximately 1065.8 meters of runway to reach its takeoff speed of 73 m/s. This calculation highlights the importance of thrust-to-weight ratio and acceleration in aviation, as they directly influence the performance and safety of aircraft during takeoff.