Question

A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time: = 2.0 − (0.25 / 2 ) 2 = (1.0/) + (0.025/ 3 ) 3 () ℎ ′ and distance from the lander at t=2.0s. (B) Find the rover’s displacement and average velocity vectors for the interval t=0.0 s to t=2.0s. (c) Find a general expression for the rover’s instantaneous velocity vector ⃗ . Express ⃗ at in component form and in terms of magnitude and direction​

Answer 1

A) The rover’s coordinates and its distance from the lander at t = 2.0 s are; (1, 4) and 4.1 m

B) The rover’s displacement and average velocity vector during the interval are; s = (-1, 4) and v = (-0.5, 2) m/s

C) The magnitude and direction of the instantaneous velocity are; 2.24 m/s and 117°

What is the displacement and Velocity?

The rover's x and y coordinates are given as;

x = 2.0m − (0.25 m/s²)t²

y = (1.0m/s)t + (0.25 m/s³)t³

A) At t = 2 s, the rovers coordinates are;

x = 2.0m − (0.25 m/s²)2²

x = 1 m

y = (1.0m/s)2 + (0.25 m/s³)2³

y = 4 m

Distance from the lander is;

s = √[(1 - 2)² + 4²]

s = 4.1 m

B) Let us first find the distance coordinates for the interval t = 0.0 s to t = 2.0s. Thus;

s = r - r₀

s = (1 - 2), (4 - 0)

s = (-1, 4)

Thus, average velocity vector is;

v = ¹/₂s

v =  ¹/₂(-1, 4)

v = (-0.5, 2) m/s

C) A general expression for the instantaneous velocity components is;

v_x = -0.5t

v_y = 1 - 0.75t²

Thus, v(2) is;

v_x = -0.5(2) = -1

v_y = 1 - 0.75(2)²

v_y = -2

Instantaneous velocity vector is; v = (-1, -2)

Magnitude of instantaneous Velocity = √(-1² + -2²) = 2.24 m/s

Direction = 180° - tan⁻¹(-2/-1) ≈ 117°

Read more about displacement and velocity at; brainly.com/question/4931057