Position Vectors, Velocity Vectors, and Acceleration Vectors

[niyati]

The coordinates of an object moving in the xy plane vary with time according to the equations x = -(5.00 m)sin(wt) and y = (4.00 m) - (5.00 m)cos(wt), where w is a constant and t is in seconds. (a) Determine the components of velocity and components of acceleration at t = 0. (b) Write expressions for the position vector, the velocity vector, and the acceleration vector at any time t > 0. (c) Describe the path of the object in an xy plot.

(a) When time is zero, the positions of x and y are 0 and 4, respectively. I am wondering how I can determine a velocity of a vertical line. I don't think it is zero, as that would be a horizontal line, and to have no slope would mean that there isn't a velocity. Where there is no velocity, there is no acceleration, so to find the components would be impossible.

(b) My problem here is that there are two different equations dealing with the components. It is not an independent variable as x and the dependent variable as y. However, if I squared both equations, added them, and then took the square root (like find the length of the hypotenuse), would that be an equation for the position? And since it says at any time greater than zero, am I to take the derivative of such equation to get the velocity, and again for the acceleration?

(c) I...really don't know what to do with this part, but I'm positive that this has something to do with part (b) (...duh), which, well, I'm not getting either.

I think I'm over-complicating things, especially in part (b), because nothing in this portion of my chapter (...the beginning) did anything this weird.

Help?

 

[rootX]

The coordinates of an object moving in the xy plane vary with time according to the equations x = -(5.00 m)sin(wt) and y = (4.00 m) - (5.00 m)cos(wt), where w is a constant and t is in seconds. (a) Determine the components of velocity and components of acceleration at t = 0. (b) Write expressions for the position vector, the velocity vector, and the acceleration vector at any time t > 0. (c) Describe the path of the object in an xy plot.

(a) When time is zero, the positions of x and y are 0 and 4, respectively. I am wondering how I can determine a velocity of a vertical line. I don't think it is zero, as that would be a horizontal line, and to have no slope would mean that there isn't a velocity. Where there is no velocity, there is no acceleration, so to find the components would be impossible.

(b) My problem here is that there are two different equations dealing with the components. It is not an independent variable as x and the dependent variable as y. However, if I squared both equations, added them, and then took the square root (like find the length of the hypotenuse), would that be an equation for the position? And since it says at any time greater than zero, am I to take the derivative of such equation to get the velocity, and again for the acceleration?

(c) I...really don't know what to do with this part, but I'm positive that this has something to do with part (b) (...duh), which, well, I'm not getting either.

I think I'm over-complicating things, especially in part (b), because nothing in this portion of my chapter (...the beginning) did anything this weird.

Help?

{a}_{x}=\frac{d}{d\,t}\,{v}_{x}
{v}_{x}=\frac{d}{d\,t}\,{s}_{x}
{v}_{y}=\frac{d}{d\,t}\,{s}_{y}
{a}_{y}=\frac{d}{d\,t}\,{v}_{y}

b) it's asking for the position vector...
and so use those parametric equations.


c) I would say it has nothing to do with b.
just eliminate t, and combine those two equations so as to make y>>x

 

[SocratesOscar]

I have a big problem here, i don't understand a word, can someone explain it from the beginning ?

 

[Pyrrhus]

I have a big problem here, i don't understand a word, can someone explain it from the beginning ?

What you don't understand?

This is a textbook problem. Start from your book definitions (or read rootx reply) of velocity, and acceleration. Remember that because this is a 2D movement, your vectors must account for both x and y components.