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marissa [1.9K]
3 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]3 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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If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

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Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared eas
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A spatially challenged goldfish swims along the x-axis only. Its initial position is 7.8 m. After swimming back and forth a whil
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Answer:

<em>The fish displacement was -3.4 m</em>

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<u>Distance and Displacement</u>

A moving object constantly travels some distance at defined periods of time. The total moved distance is the sum of each individual distance the object traveled. It can be written as:

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The displacement only takes into consideration the initial and final points of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.

The displacement, when the object moves in one axis only is given by:

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We know the fist started at 7.8 m from a given reference along the x-axis.

After some undisclosed movements, it ends up at the position 4.4 m. Thus, the displacement is:

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This means the fish ended up to the left of the position it started from.

The fish displacement was -3.4 m

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