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°
<h3>What is the displacement and Velocity?</h3>
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
