# Homework Help: Derive an expression for the x- component of velocity.

1. Aug 21, 2016

### JohnTravolski

[Mentor note: Thread moved from technical forums so no formatting template shown]

Unfortunately, I'm completely lost with this question:

1. A particle is moving along the x-axis. Its position as a function of time is given as
x=bt-ct^2
Derive an expression for the x- component of velocity.
and
Derive an expression for the x-component of the particle’s acceleration, ax.

The equation that I have to begin with (I have to start with it and then change it into the answer) MUST be on this sheet:
http://web.mst.edu/~vojtaa/engphys1/handouts/OSEs.pdf
Unfortunately, I have no idea which one to use. Can somebody point me in the right direction?

Last edited by a moderator: Aug 21, 2016
2. Aug 21, 2016

### Merlin3189

Kinematics means "the branch of mechanics that deals with pure motion, without reference to the masses or forces involved in it."

As for the rest,
What are you told about? What are you asked to find? (I was thinking maybe you could say this in words, rather than symbols which you may not understand.)
Then, what equation on your sheet tells you about the relationship between these things? (Edit: This is where we find out whether you do understand the relevant symbols.)

3. Aug 21, 2016

### JohnTravolski

As I understand it, if I were to turn the velocity into the hypotenuse of a right triangle, the x component would be the triangle's height since I think T would represent the x-axis and the y-axis would represent motion along the x-axis. However, I don't know how to relate a right triangle to any of the equations on the sheet.

4. Aug 21, 2016

### PeroK

What do you know about the relationship between position, velocity and acceleration? Forget about the handout for a moment.

5. Aug 21, 2016

### JohnTravolski

Velocity is the derivative of Position while Acceleration is the derivative of Velocity.

6. Aug 21, 2016

### PeroK

If you know that, then what's the problem?

7. Aug 21, 2016

### JohnTravolski

I don't understand what "the x-component of velocity" in the question is referring to given that none of the formulas given on the sheet involve right triangles. I don't know how to derive the formula for the x component given the equations on the sheet since I'm required to start there.

8. Aug 21, 2016

### PeroK

Can you solve the problem without the sheet?

The question is:

1. A particle is moving along the x-axis. Its position as a function of time is given as
x=bt-ct^2
Derive an expression for the x- component of velocity.
and
Derive an expression for the x-component of the particle’s acceleration, ax.

If you can solve the problem, you can work out which formula you used later.

9. Aug 21, 2016

### JohnTravolski

Without the sheet, I would derive the equation given to me to get:
Velocity = -2ct + b
Meaning that, at T=0, the x component of velocity would = b
since the velocity is just a line with a negative slope with the x-axis representing time and the y-axis representing the x-axis that the particle moves along.

I have no idea if this is the answer that's being looked for, though, because I don't know if the line is supposed to be confined to just the first quadrant or not or if if I'm even supposed to use T=0.

10. Aug 21, 2016

### PeroK

In other words, to get the velocity you differentiated (with respect to time) the expression for position. Which is correct.

But, someone or something has seriously confused you, as the rest of your post makes no sense.

In this case you have one-dimensional motion (in the x-direction). Velocity is the slope of the graph of position against time. That is just the same as saying velocity is the derivative of position with respect to time.

In general, position can be a 3-dimensional vector, so velocity will be the derivative of the position vector with respect to time (and this is no longer easy to draw on a graph). You'd need 3D for your x-y-z coordinates and then another axis for time.

If you look at your sheet, you may spot some vector derivatives, which are the generalisation of 1D differentiation.

11. Aug 21, 2016

### JohnTravolski

So I assume that this is the one that I have to use, then?

One of the reasons I'm confused is that these equations have appeared nowhere in my assigned reading and I've had absolutely no experience with them yet.

12. Aug 21, 2016

### PeroK

Yes, they are the vector equations which represent a generalisation to 2D of what you do in 1D. Although, they ought to be 3D, really.

I would have said:

"Velocity is the slope of the graph with time along the horizontal axis and position (in this case in the x-direction) along the vertical axis."

The horizontal and vertical axes are often the x- and y-axes, but they can be anything you choose. In this case, the horizontal axis is time.

13. Aug 21, 2016

### JohnTravolski

Alright, I appreciate the rewording there, your version is less confusing.

I suppose I'll have to look up how to use that formula, though, since I've never used it so far.

14. Aug 21, 2016

### Merlin3189

IMO what you need is to understand what these formulae mean.
$\vec{r} = x\hat{i} + y\hat{j}$ means that position vector $\vec{r}$ has an x component and a y component in the directions of the unit vectors $\hat{i} and \hat{j}$ (which are implicitly orthogonal, I think)
I'm not sure why that is there other than to affirm the notation they want to use. But then the real relations,
$\Large \vec{v} = \frac{d\vec{r}}{dt}$ tells you that the velocity (vector) is the time derivative of the position vector and
$\Large \vec{a} =\frac{d\vec{v}}{dt}$ tells you that the acceleration (vector) is the time derivative of the velocity.
That's what you need to know to answer this question. You are given an equation for the position vector (albeit a one dimensional vector whose y (and z) component is always zero) and asked to find the velocity and acceleration.
You have done the first, finding $\vec{v}$ by finding the time derivative of the position vector.
And now you can find the acceleration.

I'm not sure how you could "use" these equations, other than as an aide memoire to the relation between position, velocity and acceleration, and to the notation that is used.

Edited: to change font size in Latex, for readability

15. Aug 21, 2016

### JohnTravolski

Okay, I almost understand, but there's still one detail I'm missing:
Yes, I did find the derivative of the position equation that was given to me, but I still don't understand how that derivative I obtained correlates to an "x-component."

In the derivative I obtained:
Velocity = -2ct + b
What corresponds to the x-component and to the time-component?

16. Aug 21, 2016

### PeroK

In this case, the x-component is all you've got. The motion is all in the x-direction, so:

$v = v_x = -2ct + b$

Or, more precisely:

$\vec{v} = (v_x, 0, 0) = (-2ct + b, 0, 0)$

17. Aug 21, 2016

### JohnTravolski

OH, now I understand! I didn't realize it was that simple.

To tell you the truth, I actually haven't even had Physics class at all this year yet and I'm just trying to get a head-start on my homework, but as I said, none of the assigned reading has covered this.

Hopefully my instructor will give some examples of these during lecture on Tuesday.

18. Aug 21, 2016

### Merlin3189

I think the question has tried to hard to be simple and made itself confusing.
The particle is moving in only one dimension, "along the x axis", so the x component of the position vector IS the position vector, the x component of the velocity IS the velocity and the x component of the acceleration IS the acceleration. You have found the x component of the velocity, because it has no y or z component, because it moves only along the x axis.
The velocity = b -2ct and the x component of velocity = b -2ct The y component and z component of velocity each equals 0.

Perhaps you could add, from the first formula, that $\vec{v} = (b -2ct)\vec{i}$ to show that this is a one dimensional vector?

19. Aug 22, 2016

### Kajal Sengupta

I agree with Merlin. Since the equation for position vector is given in x-direction so the velocity and acceleration in x- direction will be obtained by simply differentiating the given expression which you have done. Had the question not mentioned about finding x-component of velocity and acceleration it could raise doubt in mind as to whether other directions also need to be considered.

20. Aug 23, 2016

### Dodgerson

Im taking the same course. thanks for the help..