Question
A robotic vehicle, or rover, is exploring the surface of Mars. The landing craft is the origin of the 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 as:
x(t) = 2.0 m – (0.25 m/s2)t2
y(t) = (1.0 m/s)t + (0.025 m/s3)t3
a) Find the rover’s displacement and average velocity between t = 0 and t = 2.0 s.
b) Find the instantaneous velocity at t = 2.0 s, both in vector form and in terms of magnitude and direction.
Solution
(a): Find the displacement and average velocity between t = 0 and t = 2.0 s.
- Displacement (d): Displacement is the vector difference between the rover’s position at t = 2.0s and t = 0,
Position at t = 0:
x(0) = 2.0m and y(0) = 0
So r(0) = (2.0m, 0)
Position at t = 2.0:
x(2) = 2.0 – (0.25)(2)2 = 2.0 – 1.0 = 1.0 m
y(2) = (1.0)(2) + (0.025)(2)3 = 2.0 + 0.2 = 2.2m
r(2) = (1.0m, 2.2m)
Therefore, displacement, d = r(2) – r(0) = (1.0 – 2.0, 2.2 – 0) = (-1, 2.2) m.
2. Average Velocity (Vaverage) = d / Δt = (-1, 2.2) / 2.0 = (-0.5, 1.1) m/s
(b) Instantaneous Velocity at t = 2.0s
- Velocity Components (vx and vy): Instantaneous velocity is the derivative of the position function with respect to time.
x(t) = 2.0 – 0.25t2:
v(x) = dx/dt = -0.5t
At t = 2.0:
v2(2) = -0.5 x 2 = -1.0 m/s
y(t) = (1.0)t + (0.025)t3:
vy = dy/dt = 1.0 + 0.075t2
At t = 2.0:
vy(2) = 1.0 + 0.075(2)2 = 1.0 + 0.3 = 1.3 m/s
Instantaneous velocity vector:
v = (vx, vy) = (-1.0, 1.3) m/s.
2. Magnitude of Velocity (|v|):
|v| = √(vx2 + vy2) = √(-1.0)2 + (1.3)2 = √1.0 + 1.69 = √2.69 = 1.64 m/s
2. Direction of Velocity (θ): The angle (θ) is given by:
θ = tan-1 (vx/vy) = tan-1 (1.3/-1.0) = tan-1 (-1.3) = 52.1°
This angle is measured clockwise from the positive x-axis.