Question

A satellite in outer space is moving at a constant velocity of 21.5 m/s in the +y direction when one of its onboard thruster turns on, causing an acceleration of 0.330 m/s2 in the +x direction. The acceleration lasts for 42.0 s, at which point the thruster turns off.

Answer 1

Answer: 57.31 degrees

Step-by-step explanation:

v = u + at

u = initial velocity

v = final velocity

a = acceleration

t = time

Now,

vy= 21.5 m/s

vx = 0 + 42(0.33) = 13.8 m/s

v = sqrt ((vx)sqr + (vy)sqr)= sqrt((13.8)sqr+ (21.5)sqr)= 25.55m/s

at an angle θ = tan inv (vy/vx) =

tan inv (21.5/13.8) = 57.31 degrees (counterclockwise from the +x-axis)