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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.
<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
- 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
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
- 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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The answer would be false
Answer:
2.75 m/s^2
Explanation:
The airplane's acceleration on the runway was 2.75 m/s^2
We can find the acceleration by using the equation: a = (v-u)/t
where a is acceleration, v is final velocity, u is initial velocity, and t is time.
In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1
a = 2.75 m/s^2
