The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2xf=x

Question

The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2xf=xi+vit+12at2 where xfxf is the final position, xixi is the initial position, vivi is the initial velocity, aa is the acceleration, and tt is the time. Let’s say a car starts with an initial speed of 15 m/sm/s, and moves between the 1000 mm and 5000 mm marks on a roadway in a time of 60 ss. What is its acceleration?

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6 months 2021-07-13T17:29:38+00:00 1 Answers 24 views 0

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    2021-07-13T17:30:48+00:00

    Answer:

    a = -04978 m / s²

    Explanation:

    For this exercise we can use the kinematics equations, as the initial speed, distance and time indicate, we can use

            x = x₀ + v₀ t + 1 /2 a  t²

           ½ a t² = x-x₀ – v₀ t

           a = (x-x₀ – v₀ t) 2 / t²

    let’s reduce the magnitudes to the SI system

         x₀ = 1000 mm = 1 m

         x  = 5000mm = 5m

    let’s calculate

         a = (5 – 1 – 15 60) 2/60²

         a = -04978 m / s²

    negative sign indicates that braking is related

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )