Question

evaluate the numerical value of the vertical velocity of the car at time t=0.25s using the expression from part d, where y0=0.75m , α=0.951/s , and ω=6.3rad/s .

Answer 1

The numerical value of the vertical velocity of the car at time t=0.25s is 0.929 m/s.

Step-by-step explanation:

To evaluate the numerical value of the vertical velocity of the car at time t=0.25s, we need to use the expression provided in part d of the question. The expression is given as voy = vo*sin(theta), where voy is the initial vertical velocity and theta is the angle of the initial velocity. From the given information, we know that y0=0.75m, α=0.951/s, and ω=6.3rad/s.

Substituting these values into the expression, voy = (0.951/s) * sin(6.3rad/s) = 0.951 * 0.978 = 0.929 m/s.

Therefore, the numerical value of the vertical velocity of the car at time t=0.25s is 0.929 m/s.

Answer 2

The vertical velocity of the car at time t = 0.25s is 0m/s.

The vertical velocity (
`v_(y)`) of the car in simple harmonic motion can be given by the equation:

`\[ v_y(t) = \alpha \cdot \omega \cdot \cos(\omega t) \]`

Given:

-
`\( \alpha \)` = 0.951
`s^(-1)`

-
`\( \omega \)` = 6.3 rad/s

- t = 0.25 s

`v_(y)` (0.25) = 0.951 * 6.3 * cos (6.3 * 0.25)

= 0.951 * 6.3 * cos (1.575)

= 0.951 * 6.3 * cos (1.575)

= - 0.025 m/s

Since the value is very small, the numerical value of the vertical velocity of the car at time t = 0.25s using the given expression is 0m/s.