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marissa [1.9K]
2 years ago
6

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

it+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?
Physics
1 answer:
Mademuasel [1]2 years ago
7 0

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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<em>F</em> = 153 N

\theta = 11.3°

Explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components F_{x} and F_{y}:

F_{x} = 350 N - 200 N = 150 N

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