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 21.2 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

25.2 s

Step-by-step explanation:

In an uniformly accelerated motion starting from rest, the time taken for the motion is given by the equation

`d=(1)/(2)at^2 \rightarrow t=\sqrt{(2d)/(a)}`

where t is the time, d is the distance covered, a is the acceleration.

We also know that the acceleration can be found by using Newton's second law:

`\sum F = ma \rightarrow a=(\sum F)/(m)`

where
`\sum F` is the net force on the object and m its mass. Substituting into the previous equation,

`t=\sqrt{(2md)/(\sum F)}` (1)

So we see that the time taken for the motion is inversely proportional to the square root of the net force.

Let's consider now the space probe in the problem. Let's call F the magnitude of the force generated by each engine.

When the two forces are applied in the same direction, the net force on the space probe is

`\sum F = F+F = 2F`

But when the two forces are applied perpendicularly, the net force is

`\sum F' = √(F^2+F^2)=√(2) F`

Using eq.(1) we can write:

`(t)/(t')=\sqrt{(\sum F')/(\sum F)}`

where t' is the new duration of the motion. Solving for t',

`t'=\sqrt{(\sum F)/(\sum F')}t=\sqrt{(2F)/(√(2)F)}(21.2 s)=25.2 s`