You’re flying your 10,000-kg Scooty Puff Pro spaceship at a constant speed of 20

Question

You’re flying your 10,000-kg Scooty Puff Pro spaceship at a constant speed of 20 km/s in the

positive x-direction. All of the sudden it malfunctions and many of the thrusters are stuck on. The main thruster is stuck on producing a 800 kN force in the positive x-direction. Two of the small maneuvering thrusters are also stuck on, one producing a 100 kN force in a direction 30° from the positive y-direction towards the negative x-direction, and the other is producing a 150 kN force in a direction 10° from the negative y-direction towards the negative x-direction. Af- ter 2 minutes of this the two maneuvering thrusters shut off and the main thruster decreases in magnitude to 200 kN, while still applying the force in the positive x-direction. It takes an addi- tional 6 minutes to fix the problem and turn all of them off. How far and in what direction would you have to travel to get to where you would have been if the malfunction had not oc- curred.

Explanation / Answer

We need to find the net force on the ship, in each situation.

We know that the force,F = ma

So acceleration a=F/m,

For distance, S = Vi t + 0.5at2

vi=initial velocity

It started with 20 km/s speed in positive x direction, so that will be added to the first calculation.
For each force given, there will be an x component and a y component.
First situation:
800kN in x direction: x component = 800kN, y component = 0

Net force F=800 kN
100 kN force in a direction 300 from the positive y-direction towards the negative x-direction:
x component = -100cos 300 kN=-86.6kN, y component = 100 sin 300 kN=50kN

F=150 kN force in a direction 100 from the negative y-direction towards the negative x-direction:
x component = -150 cos 100 kN=-147.7kN, y component = -150 sin 100 kN=26.04kN
Force on X direction=800 kN+-86.6kN+-147.7kN=565 kN

Force on Y direction=50kN+26kN=76KN

The net force for the two directions is found out.

Then accileration

mass=10,000 kg

Acceleration along x axis=565 kN/10,000 kg=0.0565 m/s2

Acceleration along y axis=76 kN/10,000 kg=0.0076 m/s2

By using the acceleration, we can apply into S = Vit + .5at2 equation

along Y axis, S=20000m/s*120s+0.5*0.0076*1202=2400.406 km in Y axis

along X axis, S=20000m/s*120s+0.5*0.0565*1202=2400.055 km in X axis

For the next section of thrust, you also need the speed in each direction. That is done using

Vf = Vi + at

t=720s

So Vf=20000+0.0565*720=20040.68 m/s in x axis

and in y axis

Vf=20000+0.0076*720=20005.472 m/s in Y axis

along Y axis, S=20005.472 m/s*720s+0.5*0.0076*7202=14405.9km in x axis

along X axis, S=20040.68m/s*720s+0.5*0.0565*7202=14443.4 km in y axis

direction=tan-1(Fy/Fx)=tan-1(76/565)=7.660

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