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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 $28 s$ 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?

33$s$

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in the first situation. We have two forces acting on the same direction with the same intensity. Let me call this force F. Then in the first situation, you don't seconds. Law tells us the following the net force is equal. The mass times acceleration. The net force is because true times f and it goes to the mass times acceleration. Then the acceleration is he close to two times have divided by the mass. Now we can relate the acceleration with the displacement, using the following relation. Remember that displacement is it close to the initial velocity times that I'm interval plus acceleration times the time interval squared, divided by truth. The initial velocity easy close to zero As the probe departs for rest, then its placement is equals. Two declaration times that time squared, divided by truth. Then we can now. So for the time together following two times the displacement is the coast with excavation times that I'm interrupt squared. Then the time interrupt squared. Is it close to two times displacement, divided by the acceleration, Then the time interval is a question. The square it off two times displacement divided by acceleration, and it's equal to 20 seconds in the situation. Now we can repeat this calculation for the second situation. But on the second situation we have two forces that acts pepper perpendicular to one another. So half here and half. Here we begin by couple eating. What is the magnitude after Net force? We have the following. There is one force that is acting on the axe access another force that is acting on the Y axis. And the result is a net force that is some kind of gag a note. So these will be the net force. What is the management of the Net force? It's very easy to solve this question because we have ah rectangle triangle here so we can use the Pythagorean theorem to discover what is the value off the net force. So the net force squared is eco's to f squared plus f squared. Then the net force squared is equal to two times the force squared. So then that force is it close to the square it off True times The force f now using youto second law, we have the following the net force is it goes to the mass finds that solution off the probe. Let me call it a prime because this is a different acceleration. Because this is a different situation. So square root of truth times f is it goes to the mass times the new acceleration. So the new acceleration is it goes to the square root of truth times f divided by am we can relate these acceleration with this order acceleration by noting the following a easy goes to the square root of two times a prime. Then we can use these relations only changing a by a prime to get the following Dr. T is equal to the square root of two times not as divided by a prime but a prime is because to a divided by the square it off too. So we have true Delta s divided by a divided by spirit of truth. This is it cost you square it off too. Delta s divided by a times This quite a tough truth Then we have delta t is equal to the square root off two times Delta s divided by a the previous acceleration times the square root off the square root. So we have the fourth square off True. The first term. We know that it's a close to 28 seconds. We notice from the pretty real situation then in this situation, the time interval is that close to 28 times the fourth square off, too, and this is approximately 33 seconds.

Brazilian Center for Research in Physics