Ask Question
23 October, 00:41

Chegg A late passenger, sprinting at 8.00 m/s, is 30.0 m away from the rear end of a train when the train starts from rest with a constant acceleration of 1.00 m/s2. Will the passenger catch the train, and if so, how far must he run to do so? How long will he have to run?

+3
Answers (1)
  1. 23 October, 02:20
    0
    He will catch the train.

    He as to run for 6 seconds.

    Distance he travels = 48 m

    Explanation:o

    Speed of Chegg = 8 m/s

    Distance to train = 30 m

    Acceleration of train = 1 m/s²

    We have s = ut + 0.5 at²

    Distance traveled by Chegg at a time t

    s₁ = ut + 0.5 act² = 8t + 0.5 x 0 x t² = 8t

    Distance traveled by train at a time t

    s₂ = ut + 0.5 at² = 0 x t + 0.5 x 1 x t² = 0.5t²

    For Chegg to catcseh the train we have

    s₁ ≥ s₂ + 30

    8t ≥ 0.5t² + 300

    t² - 16t + 60 ≤ 01

    (t-6) (t-10) = 0

    t = 6 seconds or t = 10 seconds.

    So he will catch the train at time t = 6 seconds.

    So he as to run for 6 seconds.

    Distance he travels in 6 seconds = 8 x 6 = 48 m
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Chegg A late passenger, sprinting at 8.00 m/s, is 30.0 m away from the rear end of a train when the train starts from rest with a constant ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers