Question

A guy on an airplane allowed his pocket watch to hang freely, while the airplane accelerated on the runway. The airplane accelerated for 18 seconds before it lifted-off. During the acceleration of the airplane, the watch hung at an angle of 25o from vertical as shown below. What was the velocity of the airplane at take-off, if it started at rest?

Answer 1

The velocity of the airplane at take-off was approximately 82.6 m/s.

Identify the forces acting:

Gravity (mg) acting downwards
Tension in the string (T) at an angle of 25° from the vertical
Net force (Fnet) accelerating the plane forward
Apply Newton's second law:

Fnet = ma

where:

m = mass of the watch and

a = acceleration of the plane

Resolve forces into components:

Vertical component of tension (Tsin25°) balances the gravity (mg)

Horizontal component of tension (Tcos25°) provides the net force (Fnet)

Write down the equations of motion:

Tsin25° = mg (vertical equilibrium)

Tcos25° = ma (horizontal acceleration)

From the first equation, find T:

T = mg / sin25°

Substitute this value of T in the second equation:

mg/sin25° × cos25° = ma

Simplify the equation:

mg × cos25° / sin25° = ma

tan25° = a

Calculate the acceleration:

tan25° ≈ 0.466

a = 0.466 × g

≈ 4.59 m/s²

Calculate the take-off velocity:

v = at

v = 4.59 m/s² * 18 s

v ≈ 82.6 m/s

Answer 2

The velocity of the airplane at take-off was approximately 82.6 m/s.

How to find initial velocity?

Identify the forces acting:

• Gravity (mg) acting downwards
• Tension in the string (T) at an angle of 25° from the vertical
• Net force (Fnet) accelerating the plane forward

Apply Newton's second law:

Fnet = ma

where:

m = mass of the watch and

a = acceleration of the plane

Resolve forces into components:

Vertical component of tension (Tsin25°) balances the gravity (mg)

Horizontal component of tension (Tcos25°) provides the net force (Fnet)

Write down the equations of motion:

Tsin25° = mg (vertical equilibrium)

Tcos25° = ma (horizontal acceleration)

From the first equation, find T:

T = mg / sin25°

Substitute this value of T in the second equation:

mg/sin25° × cos25° = ma

Simplify the equation:

mg × cos25° / sin25° = ma

tan25° = a

Calculate the acceleration:

tan25° ≈ 0.466

a = 0.466 × g

≈ 4.59 m/s²

Calculate the take-off velocity:

v = at

v = 4.59 m/s² * 18 s

v ≈ 82.6 m/s

Therefore, the velocity of the airplane at take-off was approximately 82.6 m/s.