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31 March, 01:21

A speeder traveling at 50 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 1.64 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

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Answers (2)
  1. 31 March, 02:02
    0
    61s

    Explanation:

    the speedr travels at constant speed therefore it is the following equation

    X=VT

    the police travel in a uniformly accelerated movement, therefore the following equation is valid considering the initial speed is zero

    X = (1/2) AT^2

    as the displacement of the speeder and the policeman is the same we could match the previous equations

    VT = (1/2) AT^2

    Solving for T

    T=2V/A

    T=2 (50) / 1.64=61s
  2. 31 March, 05:58
    0
    t = 61s

    Explanation:

    Analysis

    When the policeman catches the speeder they will both have traveled the same distance d and the same time.

    d₁ = d₂ = d

    d₁: displacement of policeman

    d₂: displacement of speede r

    t₂ = t₁ = t

    t₂: time of the speeder

    t₁: time of the Policeman

    Policeman kinematics

    The policeman moves with uniformly accelerated movement:

    d = v₀₁ t + (1/2) (a) t² (Formula 1)

    d: displacement in meters (m)

    v₀₁: initial speed = 0

    a: acceleration = 1.64m/s²

    t: time in s

    We replace data in formula (1)

    d = (1/2) (1.64) * t²

    d = 0.82t² equation (1)

    Speeder kinematics

    The speeder moves with constant speed:

    d = v * t Formula (2)

    d: displacement in meters (m)

    v: speed in m/s = 50 m/s

    t: time in s

    We replace data in formula (2)

    d = 50t equation (2)

    Time calculation

    equation (1) = equation (2)

    d = 0.82t² = 50t

    0.82t = 50 (dividing by t both sides of the equation)

    t = 50/0.82

    t = 60.9756s = 61s
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