A spacecraft traveling at a velocity of 215 967 430 ms is ob

A spacecraft traveling at a velocity of (-21.5, -96.7, 43.0) m/s is observed to be at a location (200, 300, -500) m relative to an origin located on a nearby asteroid. At a later time the spacecraft is at location (-380, -2310, 660) m. How long did it take the spacecraft to travel between these locations? How far did the spacecraft travel? What is the speed of the spacecraft? What is the unit vector in the direction of the spacecraft\'s velocity? (Express your answer in vector form.)

Solution

a) displacement = final position - initial position

r = ( -390, -2310, 660) - ( 200, 300,-500) = (-590, -2610, 1160) m

and v = (-21.5, -96.7, 43) m/s

time = r / v = -590/-21.5 = 27.44 sec

Or -2610/-96.7 = 26.99 s

Or 1160/43 = 26.97 s

these all three must be same but this is due to some calculation error.

so we can assume t = 27 sec

b) distance = magnitude of r = sqrt(590^2 + 2610^2 + 1160^2) =2916.47 m

c) speed = magnitude of v = sqrt(21.5^2 + 96.7^2 + 43^3) = 108 m/s

for a) t = d/v = 2916.47/108 = 27 sec

d) unit vector v = vector v / magnitude v

= (-21.5, -96.7, 43) / 108 = (-0.20, -0.90, 0.40)