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Sagot :
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v = [tex]\sqrt {2 \ 15.0 \ 645}[/tex]
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ = [tex]\frac{ 143.666^2 }{2 \ 18.2}[/tex]
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
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