Answer:29.01 s
Step-by-step explanation:
Given
First two engine applies Force in same direction
Considering F be the magnitude of each force then
net Force is 2F
Let the distance travel be s

here

where m is the mass of Space Probe

----1
If the force actin in perpendicular direction
then




------2
From 1 & 2 we get

t=29.01 s