Question

An airplane starts from rest and accelerates at 10.8 m/s^2. What is its speed at the end of a 400 m-long runway?

Answer 1

To calculate the speed of an airplane at the end of a 400 m-long runway given an acceleration of 10.8 m/s^2 from rest, use the kinematic equation v^2 = u^2 + 2as. This yields a final speed of approximately 92.9 m/s for the airplane.

Step-by-step explanation:

To find the speed of the airplane at the end of a 400 m-long runway, we can use the kinematic equation: v² = u² + 2as, where u is the initial speed, v is the final speed, a is the acceleration, and s is the distance. The airplane starts from rest, so u = 0, the acceleration a is given as 10.8 m/s², and the distance s is 400 m.

Substituting the known values into the equation, we have:

v² = 0² + 2 × 10.8 m/s² × 400 m

v² = 2 × 10.8 m/s² × 400 m

v² = 8640 m²/s²

Taking the square root of both sides to solve for v gives us:

v = √(8640 m²/s²)

v = 92.9 m/s (approx)

Therefore, the speed of the airplane at the end of the runway is approximately 92.9 m/s.

Answer 2

Final velocity will be 92.851 m/sec

Step-by-step explanation:

We have given length of runway that is s = 400 m

Acceleration
`a=10.8m/sec^2`

As the airplane starts from rest so initial velocity u = 0 m/sec

Now according to third equation of motion
`v^2=u^2+2as`

`v^2=0^2+2* 10.8* 400=8640`

v = 92.851m/sec

So final velocity will be 92.851 m/sec