# Homework Help: Acceleration of Particle

1. May 8, 2016

### Quantum Singularity

1. The problem statement, all variables and given/known data
So I am studying for my finals at the moment, and I came across a problem that I am not really sure how to assess. I am given that the velocity of a particle is determined by vx=12t2-5t, vy=15t3-6t. It wants me to find when the acceleration of the particle will be zero at time t. Because the equation is in parametric form, I am kind of confused by it, and am unsure of what I am supposed to use to determine the acceleration at a given time. The answer will also be never, but I am unsure how that is determined.

2. Relevant equations
I thought maybe because acceleration is determined from the derivative of velocity, use (dy/dt)/(dx/dt) but after researching a little bit, I found the equation:
||a||=√((d2x/dt2)2+(d2y/dt2)2)

3. The attempt at a solution
So using the second equation:
||a||=√(242+(90t)2)
0=√(576+8100t2)
0=576+8100t2
-576/8100=t2
√(-576/8100)=t
So this would seem like it is the correct way to assess the problem, considering t doesn't exist for a=0. Did I do this right? Or is there another way to do it? If so, what is the correct equation(s) to use?

2. May 8, 2016

### Staff: Mentor

Welcome to the PF.

That doesn't look right to me. You are given the equations for the x and y components of velocity, so to find the respective accelerations, you just take the time derivative of each component. That gives you ax and ay, and presumably both of those have to be zero at the same time to give you an overall zero acceleration at a time t.

3. May 8, 2016

### nrqed

Watch out, you have to take a first derivative of the expressions they give since what they give are the components of the velocity. Another way of doing it is to simply find $a_x$ and $a_y$ separately and find the time(s) at which each is zero. If there is a common value of t at which they are both are zero, that's answer.

4. May 8, 2016

### Quantum Singularity

So then:
ax=24t-5
ay=45t-6
set them equal to 0:
ax=0
0=24t-5
5=24t
5/24=t
ay=0
0=45t-6
6=45t
6/45=t
They aren't equal, so the acceleration is never 0.
That makes a lot more sense then what I was doing. Thank you!

5. May 8, 2016

### SteamKing

Staff Emeritus
This is why searching for random equations often gets you into trouble with physics. You must understand what the equation is describing in order to use it correctly.

Your "relevant equation" is simply the vector norm of two acceleration components, which are each calculated knowing the position of a particle as a function of time, not the velocity of the particle as a function of time, which is the case you are analyzing here.

6. May 8, 2016

### Quantum Singularity

Yeah, I have found that with physics. The book I have for my class doesn't really have anything on it though, so I figured use a calculus equation for it initially, but I realized that wouldn't work because I believe that just gets you the slope.

7. May 8, 2016

### nrqed

Good job.

But your initial approach is also good, I suggested that other way just because I thought it might be easier. What you found was the magnitude of the acceleration (well, if you use first order derivative instead of second order) and the magnitude is zero only if the acceleration vector itself is zero so you were right that setting that to zero allows to find if there were values of "t" that solved it was one way to find the answer. You may try it again and you should find that there are no solutions with a real value of time, which shows again that the acceleration is never zero.