Calulate the acceleration of a 300,000kg jet on take when the thrust is equal to 120,000N. I used a = f/m to get 0.4m/s/s
Tricky part. How long a runway is required for the plane to take off at a speed of 300km/h?
Help?
To calculate the acceleration of a jet during takeoff, we can use the formula a = f / m, where a is the acceleration, f is the thrust, and m is the mass of the jet.
Given:
- Thrust (f) = 120,000 N
- Mass (m) = 300,000 kg
Using the formula a = f / m, we can substitute the values to find the acceleration:
a = 120,000 N / 300,000 kg
a = 0.4 m/s²
So, you were correct in calculating the acceleration as 0.4 m/s².
Now, let's move on to the second part of your question, which involves calculating the runway length required for the plane to take off at a speed of 300 km/h.
To determine the runway length required, there are a few additional factors we need to consider, such as the deceleration during takeoff, braking efficiency, and other environmental factors. However, I can provide you with a rough estimation assuming constant acceleration during takeoff.
First, we need to convert the takeoff speed from km/h to m/s:
Takeoff speed = 300 km/h = 300,000 m/3,600 s = 83.33 m/s
Next, we can use the kinematic equation v² = u² + 2as, where
- v is the final velocity (takeoff speed)
- u is the initial velocity (0 m/s)
- a is the acceleration (0.4 m/s², as calculated earlier)
- s is the distance (runway length) required
Rearranging the equation to solve for s:
s = (v² - u²) / (2a)
Substituting the values:
s = (83.33 m/s)² / (2 * 0.4 m/s²)
s = 693.096 m
Therefore, the rough estimation for the runway length required for the jet to take off at a speed of 300 km/h is approximately 693.1 meters.
Keep in mind that this estimation assumes ideal conditions with a constant acceleration and neglects other factors that may affect the actual runway length needed.