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

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The discussion centers on the equation for particle velocity, specifically addressing the concept of instantaneous rest, where the particle's velocity is zero. Participants clarify that at t=0 and t=2, the velocity v(t) equals zero, indicating moments of rest. The acceleration, derived from the velocity function, is confirmed to be -6 m/s² at t=2. There is also a technical issue mentioned regarding image uploads on a Samsung tablet. Overall, the conversation highlights the relationship between velocity and acceleration in the context of particle motion.
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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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