The plane 5 seconds after acceleration reached 100 km / h, the speed of separation from the ground 200 km / h. what is the takeoff time?
Data: t1 (considered time) = 5 s; V1 (first speed) = 100 km / h (in SI V1 = 27.78 m / s); V2 (speed at the moment of separation from the ground) = 200 km / h (in SI V2 = 55.56 m / s).
1 way) a) Acceleration during takeoff: a = (V1 – 0) / t1 = 27.78 / 5 = 5.56 m / s2.
b) Time taken for takeoff: t2 = (V2 – 0) / a = 55.56 / 5.56 = 10 s.
Method 2) For uniformly accelerated movement, the proportion is correct: V1 / t1 = a = V2 / t2, whence t2 = V2 * t1 / V1 = 200 * 5/100 = 10 s.
Answer: The plane took off 10 seconds.
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