Suppose the mass of a fully loaded module in which astronauts take off from the moon is 1.0 x 10^4 kg. the thrust of its engines is 3.0 x 10^4n. calculate the magnitude of acceleration in a vertical takeoff from the moon with two significant figures. (a) 6.8 m/s^2 (b) 4.7 m/s^2 (c) 1.7 m/s^2 (d) 1.3 m/s^2

The right option is (d) 1.3 m/s^2

Explanation:

From the mass of the lunar module and the force printed by the engines I can obtain the acceleration that they produce:

F= m × a ⇒ a= (30000 kg × (m/s^{2}) ) / (10000kg) = 3 (m/s^{2})

This acceleration has an upward direction, and the gravitational acceleration of the moon has a downward direction, opposite to the first, so they are subtracted and the resulting acceleration will be:

Amoon= 1.6249  (m/s^{2})

Atotal=  3 (m/s^{2}) - 1.6249  (m/s^{2}) = 1.3751 (m/s^{2})

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