When will the acceleration of a particle be zero?

In summary: Homework Statement 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.Homework EquationsI thought maybe because acceleration is determined from the derivative of velocity, use
  • #1
Quantum Singularity
8
1

Homework Statement


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.

Homework 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)

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?
 
Physics news on Phys.org
  • #2
Quantum Singularity said:

Homework Statement


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.

Homework 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)

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?
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.
 
  • Like
Likes Quantum Singularity
  • #3
Quantum Singularity said:

Homework Statement


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.

Homework 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)

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?
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.
 
  • Like
Likes Quantum Singularity and berkeman
  • #4
nrqed said:
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.
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!
 
  • Like
Likes berkeman
  • #5
Quantum Singularity said:

Homework Statement


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.

Homework 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)

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?
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.
 
  • Like
Likes Quantum Singularity
  • #6
SteamKing said:
You're already starting with the velo

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.
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
Quantum Singularity said:
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.
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.
 
  • Like
Likes Quantum Singularity

1. What is acceleration of a particle?

The acceleration of a particle is the rate at which its velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. It is usually measured in meters per second squared (m/s^2).

2. How is acceleration of a particle calculated?

The acceleration of a particle can be calculated by dividing the change in velocity by the change in time. It can also be calculated by taking the derivative of the particle's velocity with respect to time, or by using the equation a = F/m, where a is acceleration, F is net force, and m is mass.

3. What causes acceleration of a particle?

Acceleration of a particle can be caused by several factors, including the application of a force, a change in direction of motion, or a change in speed. In order for a particle to accelerate, there must be a net force acting on it.

4. How does acceleration affect the motion of a particle?

Acceleration affects the motion of a particle by changing its velocity. If the acceleration is in the same direction as the velocity, the particle will speed up. If the acceleration is in the opposite direction, the particle will slow down. If the acceleration is perpendicular to the velocity, it will cause the particle to change direction.

5. What is the difference between acceleration and velocity?

Acceleration and velocity are both related to the motion of a particle, but they are not the same thing. Velocity is the rate of change of displacement, while acceleration is the rate of change of velocity. In other words, velocity tells you how fast and in what direction the particle is moving, while acceleration tells you how the velocity is changing over time.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
258
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
856
Replies
8
Views
783
  • Introductory Physics Homework Help
Replies
11
Views
965
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
163
  • Introductory Physics Homework Help
Replies
7
Views
798
Back
Top