Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.