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?

Question

Physics

Timothy Mueller

19 October, 08:43

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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Answers (1)

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Finley · Answer 1

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