Question

A satellite in outer space is moving at a constant velocity of 20.7 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.310 m/s2 in the x direction. The acceleration lasts for 49.0 s, at which point the thruster turns off.
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counter clockwise from the +x-axis.

Answer 1

Answer:

a)24.73 m/s

b)58.56°

Explanation:

given:

$U_{y}$=20.7 m/s

t=49 s

a=0.3

solution:

a)

$U_{y}$ = $V_{y}$ = 20.7 m/s

$U_{x}$ =0

$V_{x}$ = $U_{x}$  + at

   = 0 + 49 x 0.3

   = 14.7 m/s

magnitude:

   =$\sqrt{V_{x} ^{2}+V_{y} ^{2} }$        

   =24.73 m/s

b)

angle =tan-1 $(\frac{V_{y} }{V_{x} } )$

         = 58.56°