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Mariulka [41]
3 years ago
13

A design team for an electric car company finds that under some conditions the suspension system of the car performs in a way th

at produces unsatisfactory bouncing of the car. When they perform measurements of the vertical position of the car y as a function of time t under these conditions, they find that it is described by the relationship: y(t)=y0e−αtcos(ωt) where y0=0.75m, α=0.95s−1, and ω=6.3s−1. In order to find the vertical velocity of the car as a function of time we will need to evaluate the derivative of the vertical position with respect to time, or dydt. For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that acceleration is the derivative of velocity with respect to time.
Mathematics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

The vertical acceleration when t = 325 s is -2.76 × 10⁻¹³² m/s²

Step-by-step explanation:

The relationship is given as follows;

y(t) = y_0 \cdot e^{(-\alpha t)} \times cos (\omega \cdot t)

Where:

y₀ = 0.75 m

α = 0.95 s⁻¹

ω = 6.3 s⁻¹

Given that the velocity, v, is found by the following relation;

v = \dfrac{dy}{dt} = -\dfrac{\alpha \cdot y_0 \cdot cos(\omega \cdot t) + \omega\cdot y_0 \cdot sin(\omega \cdot t)  }{e^{\alpha \cdot t} }

The acceleration, a, can be found by differentiating the velocity with respect to time as follows;

a = \dfrac{d^2 y}{dt^2} =\dfrac{d\left (-\dfrac{\alpha \cdot y_0 \cdot cos(\omega \cdot t) + \omega\cdot y_0 \cdot sin(\omega \cdot t)  }{e^{\alpha \cdot t} } \right )}{dt}

a = {\dfrac{\left (\alpha ^2 - \omega ^2 \right )\cdot y_0 \cdot cos(\omega \cdot t) + 2 \cdot \omega\cdot \alpha \cdot y_0 \cdot sin(\omega \cdot t)  }{e^{\alpha \cdot t} } }

Which gives;

a = {\dfrac{\left (0.95 ^2 - 6.3 ^2 \right )\times 0.75 \times cos(6.3 \times 325) + 2 \times 6.3\times 0.95 \times 0.75 \times sin(6.3 \times 325)  }{e^{0.95 \times 325} } }Hence the vertical component of the acceleration is given as follows;

a_{vertical} = {\dfrac{ 2 \times 6.3\times 0.95 \times 0.75 \times sin(6.3 \times 325)  }{e^{0.95 \times 325} } } = -2.76 \times 10^{-132} m/s^2

The vertical acceleration when t = 325 s = -2.76 × 10⁻¹³² m/s².

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natita [175]

We can infer and logically deduce that the arc intersected by these chords are not congruent.

<h3>What is a circle?</h3>

A circle can be defined as a closed, two-dimensional curved geometric shape with no edges or corners. Also, a circle refers to the set of all points in a plane that are located at a fixed distance (radius) from a fixed point (central axis).

In Geometry, a circle is generally considered to be a conic section which is formed by a plane intersecting a double-napped cone that is perpendicular to a fixed point (central axis) because it forms an angle of 90° with the central axis.

<h3>What is an arc?</h3>

In Geometry, an arc can be defined as a trajectory that is generally formed when the distance from a given point has a fixed numerical value.

Let the radii of the two (2) different circles with center O and O' be r₁ and r₂ respectively. Also, two (2) isosceles triangles AOB and COD would be formed and as such we have:

AO = OB = r₁.

CO = O'D = r₂.

By critically observing the diagram (see attachment) of these circles, we can logically deduce that the arc intersected by chord AB is greater than the arc intersected by the chord CD.

This ultimately implies that, the arcs intersected by these chords are not congruent.

Read more on arcs here: brainly.com/question/11126174

#SPJ1

<u>Complete Question:</u>

Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chords also congruent? Explain.

Hint: It would be helpful to draw two circles and label them according to the given information, then evaluate possible arc measures.

Consider the type of triangle that may be drawn by connecting the endpoints of a chord to the center of a circle. Compare the triangles made by two circles with different radii.

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