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

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The discussion centers on the calculation of particle velocity and acceleration from the equation $\int v(t) dt = 3{t}^{2}-{t}^{2}+C$. Participants clarify that "instantaneous rest" refers to moments when the particle's velocity is zero, specifically at $t=0$ and $t=2$. The correct acceleration at $t=2$ is determined to be $4 \ m/s^2$, while the confusion around the value of $-6 \ m/s^2$ is resolved by emphasizing that acceleration is derived from the velocity function, not its integral.

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$\int v(t) dt = 3{t}^{2}-{t}^{2}+C$
What does is next at instaneous rest mean?
Wouldn't that be zero?
 
Last edited:
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What have you tried so far?
 
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:
Instantaneous rest means that there is a point in time where the particle is not moving - so has a zero velocity.
 
$v\left(t\right)=0$ at t=0 and t=2
$a\left(2\right)=4$
It should be -6?
 
Acceleration is the derivative of velocity, not the antiderivative.
 
Got it, yeah that worked
 

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