Particle Velocity at Instaneous Rest: 3${t}^{2}-{t}^{2}+C$

In summary, the conversation discusses the concept of instantaneous rest and the calculation of acceleration using the derivative of velocity. The person also mentions issues with uploading an image on a Samsung tablet. The final answer for the acceleration is determined to be -6 m/s^2.
  • #1
karush
Gold Member
MHB
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View attachment 5147
$\int v(t) dt = 3{t}^{2}-{t}^{2}+C$
What does is next at instaneous rest mean?
Wouldn't that be zero?
 
Last edited:
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  • #2
What have you tried so far?
 
  • #3
See OP
I have to use a cell phone to upload an image. The upload button does not appear on a Samsung tablet. Not to mention all page clipping on the image manager. 😥

The answers were
(a) $-6 \ m \ {s}^{-2}$
(b) 4 m
 
Last edited:
  • #4
Instantaneous rest means that there is a point in time where the particle is not moving - so has a zero velocity.
 
  • #5
$v\left(t\right)=0$ at t=0 and t=2
$a\left(2\right)=4$
It should be -6?
 
  • #6
Acceleration is the derivative of velocity, not the antiderivative.
 
  • #7
Got it, yeah that worked
 

FAQ: Particle Velocity at Instaneous Rest: 3${t}^{2}-{t}^{2}+C$

What is the significance of "Particle Velocity at Instaneous Rest" in the equation 3${t}^{2}-{t}^{2}+C$?

The phrase "Particle Velocity at Instaneous Rest" refers to the velocity of a particle at a specific moment in time when it is at rest. In this equation, the velocity of the particle is represented by the term 3${t}^{2}-{t}^{2}+C$, where t is the time and C is a constant. This term tells us how the velocity of the particle changes over time.

How is the velocity of the particle calculated in this equation?

The velocity of the particle is calculated by taking the derivative of the equation with respect to time. This gives us the rate of change of the particle's position over time, which is the particle's velocity.

What does the term C represent in the equation 3${t}^{2}-{t}^{2}+C$?

The term C represents the constant of integration. This constant is added to the equation when taking the derivative to account for any initial conditions, such as the particle's starting position or velocity.

Can this equation be used to find the position of the particle?

No, this equation only represents the velocity of the particle. To find the position, we would need to integrate the equation with respect to time. The position of the particle can be found by taking the integral of the velocity equation and adding any initial conditions represented by the constant C.

What is the difference between instantaneous rest and constant rest in this equation?

Instantaneous rest refers to the particle being at rest at a specific moment in time, while constant rest refers to the particle remaining at rest for a continuous period of time. In this equation, the term 3${t}^{2}-{t}^{2}+C$ represents the velocity of the particle at instantaneous rest, while a constant value of 0 for this term would represent the particle at constant rest.

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