How to Solve an Object's Velocity and Acceleration Using Vectors?

  • Thread starter tarletontexan
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In summary: Makes total sense now!In summary, the homework statement is asking for the displacement, velocity, and acceleration of an object that moves according to the equations x=3m exp(-2/s t), y=4msin(3/s t), z=-5cos(3/s t).
  • #1
tarletontexan
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Homework Statement



an object moves according to x=3m exp(-2/s t), y=4msin(3/s t), z=-5cos(3/s t) find the velocity and the acceleration.

Homework Equations





The Attempt at a Solution


I am unsure what this means, i am doing this for extra credit for my physics class in college(civil engineering major, fresh) and would like an idea as to what i need to do to solve this problem
 
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  • #2
You're given the position, or displacement, of the object as [tex]\vec{s}(t) = (x(t),y(t),z(t)).[/tex] How are the velocity and acceleration related to the position?
 
  • #3
well they're the derivatives with respect to x...can it really be that simple?
 
  • #4
You mean derivatives with respect to time I think. Yes, it's kind of simple but it does illustrate a few concepts.
 
  • #5
well what concepts are those??, I am just a newbie afterall...
 
  • #6
tarletontexan said:
well what concepts are those??, I am just a newbie afterall...

Vectors and their components, derivatives of common functions, velocity and acceleration.
 
  • #7
so your tellin me that these functions are vectors? Ok i can dig it, so i take the derivative with respect to time and i'll have the velocity, take it again and i'll have the acceleration. That seems to be too simple, but i'll try it
 
  • #8
tarletontexan said:
so your tellin me that these functions are vectors? Ok i can dig it, so i take the derivative with respect to time and i'll have the velocity, take it again and i'll have the acceleration. That seems to be too simple, but i'll try it

What I mean is that the position of the object is a vector whose components are the specified functions. While the concepts are simple, you still have to know that the time derivative of the vector is the vector composed of the derivatives of the components, etc.
 
  • #9
ok i see where your comin from the derivative of the vector with respect to time is the vector with the derivatives of the components with respect to time
 

What is the "Forces extra credit problem"?

The "Forces extra credit problem" is a physics problem that involves using the principles of forces and motion to solve a given scenario. It is commonly assigned as an extra credit assignment in physics classes.

Why is the "Forces extra credit problem" important?

The "Forces extra credit problem" helps students apply their knowledge of forces and motion in a practical and challenging way. It also encourages critical thinking and problem-solving skills.

How do I approach solving the "Forces extra credit problem"?

The first step in solving the "Forces extra credit problem" is to carefully read and understand the given scenario. Then, identify all the forces acting on the objects in the scenario and draw a free body diagram to visualize them. Next, use Newton's laws of motion to set up and solve equations to find the desired quantity.

What are some common mistakes to avoid when solving the "Forces extra credit problem"?

Some common mistakes to avoid when solving the "Forces extra credit problem" include forgetting to include all the relevant forces, using incorrect equations, and not carefully checking the units of the final answer. It is also important to clearly label all variables and show all work.

How can I improve my skills in solving the "Forces extra credit problem"?

To improve your skills in solving the "Forces extra credit problem", it is important to practice regularly and seek help from your teacher or peers if you get stuck. You can also review and understand the solutions to previous problems to identify any areas for improvement.

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