Question

Two spaceships taxiing into starfleet drydock follow linear trajectories. Rocket 1 is initially at the position x₁= (100m, 0m, 200m) and has a constant velocity given by the vector v₁= (-100m/s, 200m/s, 0m/s). Rocket 2 is initially at the position x₂= (200m, -100m, 100m) and has a constant velocity given by the vector v2?= (100m/s, 100m/s, 100m/s).

a.) Write the equations for the linear trajectories followed by both the rockets.

b.) Find the distance of the closest approach to the station hub located at (0m, 0m, 0m) for rocket 2.

c.) Find the minimum distance between the rockets.

Answer 1

given two spaceships

Rocket 1:

x1 = (100,0,200)m

v1 = (-100, 200,0) m/s

Rocekt 2:

x2 = (200,-100,100) m

v2 = (100,100,100)m/s

a. for rocket 1

position as a funciton of time is given by

r1 = x1 + v1*t = (100 - 100t, 200t, 200)

similiarly

for rocekt 2

r2 = (200 + 100t , -100 + 100t, 100 + 100t)

hence in parametric form

r1 = (100 - 100t)i + 200t j + 200k

r2 = (200 + 100t)i + (-100 + 100t)j + (100 + 100t)k

where i , j and k are unit vectros along x, y and z directions

hence

for rocekt 1

x = 100 - 100t

y = 200t

z = 200

hence

x = 100 - 100*y/200

200x = 200,00 - 100y

2x = 200 - y, z = 200

for rocekt 2

x = 200 + 100t

y = -100 + 100t

z = 100 + 100t

y + z = 200t = (x - 200)2

y + z = 2x - 400

2x - y - z = 400

b. for rocekt 2, distance of closest approach to the dock located at (0,0,0)

is

d2 = 400/sqroot(4 + 1 + 1) = 163.29931618 m

c. distance betwwen two rockets at time t is given by

d = sqrt((100 - 100t - 200 - 100t)^2 + (200t + 100 - 100t)^2 + (200 - 100 - 100t)^2)

d = 100*sqrt(3 + 6t^2 + 4t)

d = 100*sqrt(6t^2 + 4t + 3)

the minimum happens when

6t^2 + 4t + 3 is minimum = M

dM/dt = 0

12t + 4 = 0

t< 0

hence not possibel

hence

hence for t = 0, the distnac eis minimum

d = 100*sqrt(3) = 173.2050807568 m