A space probe has two engines. Each generates the same amount of force when fire

Question

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 30.3 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular? Number Units the tolerance is + /-2%

Explanation / Answer

let the mass of the probe be m and force applied by single thruster be T
in the case when the thrusters are applying force in one diorection
from newtons second law
2T = ma ( where a is acceleration of the probe)
now let distacne to be travelled be d
starting from rest
d = 0.5at^2 = 0.5*(2T/m)t^2
given t = 30.3 s
d = 0.5*(2T/m)*918.09

now when the forces are perpendivular to each other
let one force make angle theta with the direction of the motion, then the other one makes 90 - theta
then from force balance
Tcos(theta) = Tsin(theta)
theta = 45 deg

also., T(cos(45) + sin(45)) = m*a
and d = 0.5(at^2)
d = 0.5(T(cos(45) + sin(45))/m)t^2

equating with previous equaiton
0.5(T(cos(45) + sin(45))/m)t^2 = 0.5*(2T/m)*918.09
((cos(45) + sin(45)))t^2 = (2)*918.09
cos(45) = sin(45) = 1/sqroot(2)
cos(45) + sin(45) = 2/sqroot(2) = sqroot(2)
t^2 = sqroot(918.09)
t = 36.032 s
so this time the probe takes 36.032 s to travel the same distance

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