Question

An airplane starts from rest on the runway. The engines exert a constant force of 78.0 kN on the body of the plane (mass 9.20 × 104 kg) during takeoff. How far down the runway does the plane reach its takeoff speed of 75.0 m/s?

Answer 1

The distance down the runway where the plane reaches its takeoff speed of 75.0 m/s is approximately 2640.9 meters.

Step-by-step explanation:

To determine the distance down the runway that the plane reaches its takeoff speed, we can use the equation:

d = (v2 - u2) / (2a)

Where d is the distance, v is the final velocity, u is the initial velocity, and a is the acceleration. Since the plane starts from rest, the initial velocity, u, is 0 m/s. The acceleration, a, can be calculated using Newton's second law of motion:

F = ma

Where F is the force exerted by the engines and m is the mass of the plane.

Substituting the given values, we have:

a = F / m = 78,000 N / (9.20 × 104 kg) = 0.847 m/s2

Now we can calculate the distance:

d = (75.02 - 02) / (2 * 0.847) = 2640.9 m

Therefore, the plane reaches its takeoff speed of 75.0 m/s when it has traveled approximately 2640.9 meters down the runway.

Answer 2

Given data

*The given initial speed of an airplane is u = 0 m/s

*The engine exerts a constant force is F = 78.0 kN = 78.0 × 10^3 N

*The given mass is m = 9.20 × 10^4 kg

*The given final speed of the plane is v = 75.0 m/s

The acceleration of the airplane is calculated by using the relation as

`\begin{gathered} F=ma \\ a=(F)/(m) \\ =(78.0*10^3)/(9.20*10^4) \\ =0.848m/s^2 \end{gathered}`

The formula for the distance of the runway is given by the kinematic equation of motion as

`\begin{gathered} v^2=u^2+2as \\ s=(v^2-u^2)/(2a) \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} s=((75.0)^2-(0)^2)/(2(0.848)) \\ =3316.62\text{ m} \end{gathered}`

Hence, the distance of the runway is s = 3316.62 m