Question

A package of relief supplies is dropped and moves according to the set of parametric equations given below. Find the velocity and acceleration (magnitude and direction) when t=3.0 s. Draw both vectors on xy plane.

a) x=45t,y=−4.9t²
b) x=35t,y=−4.5t²

Answer 1

The velocity of the package at t = 3.0s has a magnitude of approximately 53.75 m/s and points down and to the right, while the acceleration is -9.8 m/s² in the vertical direction only.

Step-by-step explanation:

Finding the Velocity and Acceleration

To find the velocity and acceleration of a package of relief supplies at t = 3.0 s, given the parametric equations x = 45t and y = -4.9t², we first need to find the derivatives of the position functions with respect to time to get the velocity components. For the equation x = 45t, the velocity in the x-direction is constant at 45 m/s. For the equation y = -4.9t², the velocity in the y-direction is the derivative -9.8t; at t = 3.0 s, this gives -29.4 m/s. The velocity vector at t = 3.0 s has components Vx = 45 m/s and Vy = -29.4 m/s.

The acceleration in the x-direction is 0 since the velocity is constant, and the acceleration in the y-direction is the second derivative of y with respect to time, which is a constant -9.8 m/s². Therefore, the acceleration vector at any time, including t = 3.0 s, is 0 m/s² i - 9.8 m/s² j.

The magnitudes of these vectors are found using the Pythagorean theorem. For velocity, |v| = √(45² + (-29.4)²) = √(2025 + 864.36) = √(2889.36) ≈ 53.75 m/s, and the direction is given by θ = arctan(|Vy|/|Vx|) = arctan(29.4/45). The acceleration's magnitude is just the absolute value of the y-component, 9.8 m/s², and it points in the negative y-direction.