What Time to Turn Off the Spaceship Engine for Optimal Coasting?

In summary, to find the time at which the engine should be turned off, we set the acceleration to zero and solve for t. The resulting solution is t = 1, which means that the engine should be turned off after 1 unit of time has passed.
  • #1
bodensee9
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0

Homework Statement


Suppose that the position function of a spaceship is r(t) = (3+t)i + (2+ln T)j + (7-4/t^2+1)k. Suppose you want the ship to coast to a space station located at (6.4.9). What time should you turn your engine off?


Homework Equations



r' = velocity. r" = acceleration.

The Attempt at a Solution



I guess when you turn the engine off there is no acceleration. So
|r"| = |<0,-1/t^2,(8(t^2+1)-16t^2(t^2+1)^4)/(t^2+1)^4| = 0. But I'm not really sure what to do after that or if this is even the right approach. Any suggestions? Thanks!
 
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  • #2


I would first like to clarify a few things about the problem. Is the position function given in terms of time or some other variable? If it is in terms of time, then we can proceed with solving for the time at which the engine should be turned off. However, if it is in terms of some other variable, then we would need to know how that variable relates to time in order to find the appropriate time to turn off the engine.

Assuming that the position function is given in terms of time, we can use the equations r' = velocity and r" = acceleration to solve for the time at which the engine should be turned off. As you correctly pointed out, when the engine is turned off, there is no acceleration, so r" = 0.

Substituting the given position function into the acceleration equation, we get:

|r"| = |<0,-1/t^2,(8(t^2+1)-16t^2(t^2+1)^4)/(t^2+1)^4| = 0

Simplifying this, we get:

8(t^2+1)-16t^2(t^2+1)^4 = 0

Solving for t, we get two solutions: t = 0 and t = ±1. However, we can disregard the solution t = 0 since it does not make physical sense for the ship to turn off its engine at the starting point.

Therefore, the time at which the engine should be turned off is t = ±1. Since we want the ship to coast to the space station located at (6,4,9), we can choose the positive solution t = 1. This means that the engine should be turned off after 1 unit of time has passed.

I hope this helps! Let me know if you have any further questions.
 
  • #3


I would first clarify some things about the problem. It seems that the position function of the spaceship is given in terms of time (t), but the coordinates of the space station are given as (6.4.9). Are these coordinates also in terms of time? If not, we may need to convert them to be able to solve the problem.

Assuming that the coordinates of the space station are in terms of time, we can use the position function to find the velocity and acceleration of the spaceship at any given time. This can be done by taking the first and second derivatives of the position function, respectively.

Once we have the velocity and acceleration, we can use the information about the space station's location to set up equations and solve for the time at which the engine should be turned off. For example, we can set the position of the spaceship (r(t)) equal to the position of the space station (6.4.9) and solve for t.

As you mentioned, when the engine is turned off, there is no acceleration, so we can set the acceleration (r") to be equal to zero and solve for t. This will give us the time at which the engine should be turned off.

It is important to note that this solution assumes that the spaceship is moving in a straight line towards the space station. If the spaceship is following a curved path, the solution may be more complex and require additional information or equations to solve.
 

1. What is a position vector problem?

A position vector problem is a problem in mathematics or physics that involves determining the position of an object in space using a vector. It typically involves finding the magnitude and direction of the vector from a given point to the object.

2. What is the difference between a position vector and a displacement vector?

A position vector represents the location of an object in space, while a displacement vector represents the change in position of an object from one point to another. In other words, a position vector is static and does not change, while a displacement vector is dynamic and can change depending on the movement of the object.

3. How do you solve a position vector problem?

To solve a position vector problem, you need to first identify the given point and the object's location. Then, you can use the formula for vector subtraction to find the difference between the two points. This will give you the displacement vector. Finally, you can use the formula for finding the magnitude and direction of a vector to determine the position vector.

4. What are some real-world applications of position vector problems?

Position vector problems have many applications in fields such as engineering, physics, and navigation. For example, they can be used to calculate the trajectory of a projectile, the position of a moving vehicle, or the location of a satellite in space.

5. Can position vector problems have more than two dimensions?

Yes, position vector problems can have any number of dimensions. In two dimensions, the position vector has two components (x and y). In three dimensions, it has three components (x, y, and z). However, the principles and formulas used to solve these problems remain the same regardless of the number of dimensions.

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