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

Question

1 month 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:

tester [12K] · Answer 1 · 2026-02-17T19:53:16+03:00

Answer:

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

Step-by-step explanation:

The formula is presented 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

The velocity, denoted as v, can be determined using this 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, represented by a, is calculated by differentiating velocity concerning time as shown here;

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

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

Which results in;

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

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

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