Question

A-777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the
braking force is 445,000 N, what is the speed of the airplane 10.0 s later?​

Answer 1

Given that, a aircraft of mass 300,000 kg and it's speed is 27 m/s. Also, the braking force is 445,000 N.

We have to find the speed of the airplane 10.0 s later.

From above data we have; mass (m) = 300000 kg, Initial speed (u) = 27 m/s, Force (F) = -445000N (as said in question braking force) and time (t) = 10 sec.

We know that force is product of mass and acceleration.

F = ma

Substitute the values,

-445000 = 300000 × a

-445000/300000 = a

-1.48 = a

Therefore, the retardation of the plane is -1.48 m/s².

Now, using the First Equation Of Motion:

v = u + at

Substitute the known values in the above formula,

v = 27 + (-1.48)(10)

v = 27 - 14.8

v = 12.2

Therefore, the speed of the plane is 12.2 m/s.

Answer 2

Answer:

Given:

• A-777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the

braking force is 445,000 N.

Find:

• What is the speed of the airplane 10.0 s later?

Know terms:

1. Initial speed (u)
2. Force = (F)
3. Mass = (m)
4. Time = (t)
5. Acceleration = (A)

Using formula:

★ ${\sf{\underline{\boxed{\orange{\sf{F = m \times a }}}}}}$

Calculations:

⇒ $\bold{-445000 = 300000 \times A}$

⇒ $\bold{\dfrac{-445000}{300000} = A}$

⇒ ${\sf{\underline{\boxed{\red{\sf{A = -1.48 \: m/s^2}}}}}}$

Therefore, -1.48 m/s² is the retardation of the plane.

Using formula:

★ ${\sf{\underline{\boxed{\orange{\sf{v = u + at}}}}}}$

Calculations:

⇒ $\bold{v = 27 + (-1.48 \times 10)}$

⇒ $\bold{v = 27 - 14.8}$

⇒ ${\sf{\underline{\boxed{\red{\sf{v = 12.2 \: m/s}}}}}}$

Therefore, 12.2 m/s is the speed of the plane.