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

In this exercise we have to use the knowledge of vertical acceleration to calculate the car's acceleration, so we have to:

The relationship is given as follows;

knowing that the values are:

Knowing that speed can be described by:

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

Replacing the known values, we find that:

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

Step-by-step explanation:

The relationship is given as follows;

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;

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

Which gives;

Hence the vertical component of the acceleration is given as follows;

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

351°f

step-by-step explanation:

it would open into a square depending on the type of envelope i’m not sure about the second part