When overtaking, the car began to move with an acceleration of 0.6 m / s and after 5 seconds it reached a speed of 23 m / s. what way did the car go during this time?
These tasks: a (acceleration when overtaking) = 0.6 m / s2; t (overtaking time) = 5 s; V2 (speed reached by a given vehicle) = 23 m / s.
1) Speed before overtaking: V2 = V1 + a * t and V1 = V2 – a * t = 23 – 0.6 * 5 = 20 m / s.
2) Distance covered during overtaking: S = (V2 ^ 2 – V1 ^ 2) / 2a = (23 ^ 2 – 20 ^ 2) / (2 * 0.6) = 107.5 m / s.
Check: S = (V1 + V2) * t / 2 = (23 + 20) * 5/2 = 107.5 m.
Answer: In 6 seconds, the given car covered a path of 107.5 m.
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