Question

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.

Answer 1

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.

Answer 2

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

$a= -2.76 * 10^{-132} m/s^2$

The relationship is given as follows;

$y(t)=y_0e^{-\alpha t}cos(wt)$

knowing that the values ​​are:

• $y_0 = 0.75 m$

• $a = 0.95 s^{-1$

• $w = 6.3 s^{-1$

Knowing that speed can be described by:

$v= \frac{dy}{dt} = -\frac{\alpha y_0 cos(wt) +wy_0 sin(wt)}{e^{\alpha t}}$

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

$a = \frac{d^2y}{dt^2}= \frac{(\alpha ^2-w^2) y_0 cos(wt) + 2w\alpha y_0 sin(wt)}{e^{\alpha t}}$

Replacing the known values, we find that:

$a= -2.76 * 10^{-132} m/s^2$

See more about vertical acceleration at  brainly.com/question/11776718