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 37.8 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?

Answer 1

When both engines are fired in the same direction, the force of propulsion can be denoted as 2F. Neglecting the prescence of other forces, its acceleration, a is thus 2F/m, where m is the mass of the probe.
initial velocity, u = 0 m/s
t = 28 s
a = 2F/m
let distance, d = x m
d = ut + ½at²
∴ x = ½ (2F/m) 37.8² = 37.8² F/m
When the engines are fired perpendicularly to each other, the probe experiences a resultant force 45° from the horizontal, of a magnitude √(F² + F²) = (√2) F
t = ?
d = x m
a = (√2) F/m
u = 0 m/s
x = ½ (√2) (F/m) t²
∴ 37.8² F/m = ½ (√2) (F/m) t²
t² = (√2)*37.8²

t = (1.414)^1/2*37.8 = 44.95 s