Question

A satellite in outer space is moving at a constant velocity of 21.0 m/s in the +y direction when one of its onboard thruster turns on, causing an acceleration of 0.330 m/s² in the +xdirection. The acceleration lasts for 45.0 s, at which point the thruster turns off.

a) what is the direction of the satelites velocity when the thruster turns off (anwser in meters per second)

B) what is the direction of the satelites velocity when the thruster turns off ( give your anwser as an angle measuerd counterclock wise from the + x axis) how many degrees counterclockwise from the + x axis ?

Answer 1

(a) Perpendicular acceleration doesent results in change in speed but only change its direction.

Speed will remain same as 21 m/s

(b)

Angular acceleration

α = a/r = 0.330 /r

here, a = v^2/r

r = v^2/a = 21^2 /0.330 =1336.36 m

α =0.330/1336.3 =2.469*10^-4 rad/s^2

w0 = v/r = 21/1336.36 = 0.0157

Now angle turned,

Theta = w0*t+1/2*α*t^2

=0.0157*45+1/2*2.469*10^-4*45^2

=0.9565

=54.83 degrees respect to +Y direction

Angle with respect to +x direction = 90-54.83 = 35.17 degrees.