The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
To find the answer, we need to know about the Newton's equation of motion.
What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?
- The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
- V= final velocity, U= initial velocity, S= distance and a= acceleration
What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?
- Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
- So, 100²-0= 2×a×1800
=> 10000= 3600a
=> a= 10000/3600 = 2.8 m/s²
Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
Learn more about the Newton's equation of motion here:
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