X and Y components of velocity and acceleration vectors

[marksyncm]

Homework Statement

(Translating from a Polish high school textbook, so if anything is unclear please let me know).

An object moves on a trajectory described by the parabola $y=\frac{1}{2\lambda}x^2$ such that the $x$ component of its velocity is constant and equal to $v_0$. The $\lambda$ parameter, which is in units of length, is constant. At time $t=0$ the object was located at $(x_0, y_0) = (0,0)$.

Show how the following depend on time:

a) The $x$ and $y$ coordinates of the velocity vector.
b) The $x$ and $y$ coordinates of the tangential acceleration vector.

Homework Equations

None

The Attempt at a Solution

For part (a), I took advantage of the fact that $v_0$ in the $x$ direction is constant and rewrote the parabola equation as:

$$y=\frac{1}{2\lambda}(tv_0)^2$$

Differentiating, I get that the change in $y$ with respect to time = $\frac{v_0^2t}{\lambda}$

So in answer to (a), the $x$ and $y$ coordinates of the velocity vector are

$$(v_0, \frac{v_0^2t}{\lambda})$$

Unfortunately, I'm at a loss as to how to tackle (b). I started by differentiating velocity with respect to time, which yields $\frac{v_0^2}{\lambda}$, but don't know how to proceed from here.

 

[Celso]

The velocity vector is the vector whose components are the derivates of the respective position components. The position vector can be parameterized as $(x,\frac {1}{2\lambda}x^2)$

 

[marksyncm]

Thanks for the reply, but I am not sure how I can use that information to find the $x$ and $y$ coordinates of the tangential acceleration vector. Can you elaborate?

 

[Celso]

Just as the velocity vector is defined as the rate of change of the position vector in an infinitesimal amount of time, and it is tangencial to it (geometrical motivation of the derivate), the acceleration vector is the change of the velocity vector in an infinitesimal amount of time, hence it's also tangential.
In the position vector $S = (x, y)$, we can write it as $S = (x, \frac {1}{2\lambda}x^2)$ the first component in the $x$ direction, the second in the $y$ direction (we know that the second component is this because it was given that $y = \frac {1}{2\lambda}x^2$, if this is confusing to you, you might should take a look into parametrization)

 

[Celso]

all I'm saying is that to find the acceleration vector thought the position vector, you must derivate it twice
$a = (\frac {d^2}{dx^2} x, \frac {d^2}{dx^2} \frac {1}{2\lambda}x^2)$

 

[marksyncm]

Thanks for the explanation. I'm aware that the acceleration vector is the derivative of the velocity with respect to time, but I feel like the question has a bit of subtlety to it that's confusing me. Basically, simply taking the derivative of the position vector doesn't work here because that would get me the change of Y with respect to X, but that's not what the question is asking for - it's asking for the change of Y (and X) coordinates of velocity with respect to time. The same applies to the acceleration vector.

 

[Celso]

Thanks for the explanation. I'm aware that the acceleration vector is the derivative of the velocity with respect to time, but I feel like the question has a bit of subtlety to it that's confusing me. Basically, simply taking the derivative of the position vector doesn't work here because that would get me the change of Y with respect to X, but that's not what the question is asking for - it's asking for the change of Y (and X) coordinates of velocity with respect to time. The same applies to the acceleration vector.

lol I'm sorry, I completely misunderstood the problem. It's a little bit more complex, the first thought that came to my mind though was writing $v_{0}$ as $\frac {dx}{dt}$
Then find $y$ applying the chain rule (we don't know how $y$ changes with respect to time, but we know how $y$ changes with respect to $x$ and how $x$changes with respect to time. $\frac {dy}{dt} = \frac {dy}{dx} \frac {dx}{dt}$

 

[marksyncm]

Thank you, this gets me the velocity coordinates with respect to time (see the original posting - I already managed to find that). Do you have suggestions for how to move forward from the velocity vector to get the change of the x and y coordinates of the acceleration vector with time?

 

[Celso]

Thank you, this gets me the velocity coordinates with respect to time (see the original posting - I already managed to find that). Do you have suggestions for how to move forward from the velocity vector to get the change of the x and y coordinates of the acceleration vector with time?

try doing the same thing, after finding the vector $V = (\frac{dx}{dt}, \frac{dy}{dx} \frac{dx}{dt}) = (v_{0}, \frac{x}{\lambda} v_{0})$, $a = \frac{d}{dt}\vec{V}$
$a =(\frac{dv_{0}}{dt}, \frac{d}{dt}\frac{v_{0}x}{\lambda}\frac{dx}{dy})$
I hope this is correct now

 

[ehild]

The Attempt at a Solution

For part (a), I took advantage of the fact that $v_0$ in the $x$ direction is constant and rewrote the parabola equation as:

$$y=\frac{1}{2\lambda}(tv_0)^2$$

Differentiating, I get that the change in $y$ with respect to time = $\frac{v_0^2t}{\lambda}$

So in answer to (a), the $x$ and $y$ coordinates of the velocity vector are

$$(v_0, \frac{v_0^2t}{\lambda})$$

Unfortunately, I'm at a loss as to how to tackle (b). I started by differentiating velocity with respect to time, which yields $\frac{v_0^2}{\lambda}$, but don't know how to proceed from here.

$\frac{v_0^2}{\lambda}$ is the y component of the acceleration vector. What is the x component?
You need the tangential acceleration. The tangent is parallel to the velocity vector. So you need the component of the acceleration parallel to the velocity vector. How do you get the projection of a vector to a given direction?

 

[marksyncm]

What is the x component?

Isn't the $x$ component equal to zero because the object is not accelerating in the $x$ direction?

 

[haruspex]

Isn't the $x$ component equal to zero because the object is not accelerating in the $x$ direction?

Yes.

 

[marksyncm]

If the $x$ component is equal to zero, doesn't this mean that the acceleration vector is pointing straight up (parallel to the $y$ axis), so that it can never be parallel to the velocity vector?

 

[haruspex]

If the $x$ component is equal to zero, doesn't this mean that the acceleration vector is pointing straight up (parallel to the $y$ axis), so that it can never be parallel to the velocity vector?

It will not as a whole be parallel to the velocity vector, but unless it is orthogonal to it there will be a component parallel to the velocity vector.

 

[ehild]

If the $x$ component is equal to zero, doesn't this mean that the acceleration vector is pointing straight up (parallel to the $y$ axis), so that it can never be parallel to the velocity vector?

The red vector is the acceleration, the blue is the tangent vector to the curve, parallel with the velocity. The tangential acceleration was the question, that is the component of acceleration in the tangential direction. Is it really zero? How do you determine it?

 

[marksyncm]

Thank you for the image, it's very helpful.

Is the tangential acceleration equal to the acceleration vector times the cosine of the angle between the acceleration vector and the velocity vector? Is that angle $\alpha$ equal to $\frac{\pi}{4} - \arctan(y/x)$, where $y$ and $x$ are the coordinates of the velocity vector? (EDIT: on second though, I don't think this approach would work in all cases).

Alternatively, could we maybe obtain the angle using the vector product? So $$\frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|} = \cos(\alpha)$$

Or is there something much more obvious that I'm missing?

 

[haruspex]

using the vector product?

You mean the scalar product, but yes, that is the idea.
See also http://www.maths.usyd.edu.au/u/MOW/vectors/vectors-10/v-10-1.html.

 

[marksyncm]

Yes, the scalar product.

Thank you all, I have learned a lot from this thread and from the questions that were asked.

 

[ehild]

Instead of the angle, you need the projection of the acceleration vector onto the tangent unit vector which is $\frac{\vec{v} \cdot \vec{a}}{|\vec{v}| }$ That is the magnitude of the tangential acceleration.
The other approach is: the tangential acceleration along a curve is the time derivative of the speed. Make it a vector by multiplying with the tangential unit vector .

 

[marksyncm]

Thanks, I will take a look at this answer soon (as I'm in the middle of creating another thread).

 

[marksyncm]

Instead of the angle, you need the projection of the acceleration vector onto the tangent unit vector which is

Yes, thank you.

The other approach is: the tangential acceleration along a curve is the time derivative of the speed. Make it a vector by multiplying with the tangential unit vector.

I'm not sure I follow. I understand that taking the derivative of the velocity gives us the equation for acceleration, but I do not understand what is meant by "multiplying with the tangential unit vector| to turn it into a vector. For example, differentiating velocity I get $\frac{dv}{dt} = \frac{v_0^2}{\lambda}$. What happens now?

 

[marksyncm]

I still have a follow-up question.

We were able to determine the magnitude of the tangential acceleration, but I just realized that the question asks for the $x$ and $y$ coordinates of that vector, not its magnitude. I'm not sure how to get the coordinates from here given that what we have is a scalar value?

 

[ehild]

I still have a follow-up question.

We were able to determine the magnitude of the tangential acceleration, but I just realized that the question asks for the $x$ and $y$ coordinates of that vector, not its magnitude. I'm not sure how to get the coordinates from here given that what we have is a scalar value?

The tangential acceleration is a vector. It has magnitude and direction parallel with the tangent of the parabola.

 

[ehild]

I still have a follow-up question.

We were able to determine the magnitude of the tangential acceleration, but I just realized that the question asks for the $x$ and $y$ coordinates of that vector, not its magnitude. I'm not sure how to get the coordinates from here given that what we have is a scalar value?

What did you get for the magnitude of the tangential acceleration? It is not $\frac{v_0^2}{\lambda}$

 

[marksyncm]

What did you get for the magnitude of the tangential acceleration? It is not $\frac{v_0^2}{\lambda}$

I got this for the magnitude.

$$\frac{v_{0}^3t}{\lambda^2 \sqrt{1+\frac{t^2}{\lambda^2}}}$$

Does it make sense?

 

[ehild]

I got this for the magnitude.

$$\frac{v_{0}^3t}{\lambda^2 \sqrt{1+\frac{t^2}{\lambda^2}}}$$

Does it make sense?

Check the expression under the square root.