What Is the Magnitude of the Particle's Acceleration When Velocity Is Zero?

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SUMMARY

The particle's acceleration when its velocity is zero is 24 m/s². The position function is defined as x = 24t - 2.0t³. The velocity is determined by differentiating the position function, resulting in v = 24 - 6t². Setting this equal to zero yields t = ±2 seconds. The acceleration is calculated using a = -12t, leading to a magnitude of 24 m/s², regardless of the time value being positive or negative.

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Homework Statement


A particle moving along the x-axis has a position given by
x = 24t - 2.0t3 m, where t is
measured in s. What is the magnitude of the acceleration of the particle at the instant
when its velocity is zero?

Homework Equations



v=dx/dt

a=dv/dt=d2x/dt2

The Attempt at a Solution



v = dx/dt = 24 - 6t2 = 0 =>t2 = 4 => t= ± 2

a = dv/dt = d2x/dt2 = -12t

=> a = -12 (2) = -24 m/s2

or a = -12(-2) = 24 m/s2

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the right answer is a = 24 m/s2
but why??
 
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Time cannot be negative so your second solution, although meaningful in a mathematical sense, can be discarded. But this observation not withstanding, the problem asks for the magnitude of the acceleration.
 
PhanthomJay said:
Time cannot be negative so your second solution, although meaningful in a mathematical sense, can be discarded. But this observation not withstanding, the problem asks for the magnitude of the acceleration.


Yes, that's it.
Thank you.
the magnitude is 24 whether time = 2 or -2.
 

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