Find the area of the shaded region. Mean 0 and standard deviation 1

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redraider4915

24.11.2021

Mathematics

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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