Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80-kg astronaut in orbit moving at 27,500 km/h.

for the truck :

M = mass of the truck = 20000 kg

V = speed of truck = 110 km/h = 110 (km/h) (1000 m/1 km) (1 h/3600 sec) = 30.6 m/s

kinetic energy of truck is given as

KE₁ = (0.5) M V²

inserting the values

KE₁ = (0.5) (20000) (30.6)² = 9363600 J

for the astronaut :

m = mass of astronaut = 80 kg

v = speed of astronaut = 27500 km/h = 27500 (km/h) (1000 m/1 km) (1 h/3600 sec) = 7638.9 m/s

kinetic energy of astronaut is given as

KE₂ = (0.5) m v²

inserting the values

KE₂ = (0.5) (80) (7638.9)²

KE₂ = 2.33 x 10⁹ J

Ratio of kinetic energy of truck with kinetic energy of astronaut is given as

KE₁ /KE₂ = (9363600)/(2.33 x 10⁹)

KE₁ /KE₂ = 0.00402

kinetic energy of truck is  0.00402 time the kinetic energy of astronaut.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west