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laila [671]
2 years ago
13

The position of a particle moving in the xy plane is given by the parametric equations x(t)=t^3-3t^2

Physics
1 answer:
Basile [38]2 years ago
4 0

The value of t  for which the particle is at rest is :  t = 2

<u>Given data:</u>

X = t³ - 3t²

Y = 2t³ - 3t² - 12t

<h3>Determine the values of t for which the particle is at rest </h3>

<u>First step : Determine the first derivative of eac equations</u>

dx = 3t² - 6t

dy = 6t² - 6t - 12

<u>Next step : </u><u>determine</u><u> the value of slope ( dy/dx ) = o </u>

dy / dx = ( 6 (t² - t - 2) )  /  ( 3t ( t - 2) )

Therefore

dy / dx = ( 2 ( t + 1 ) (t - 2) ) / ( t ( t -2) )

           = 0

Therefore at ; t = -1 and 2  the particle is at rest

Hence we can conclude that Neglecting the negative value the particle will be at rest when t = 2

Learn more about Object at rest : brainly.com/question/24160573

#SPJ1

<u>Attached below is the complete question</u>

<em>The position of a particle moving in the xy-plane is given by the parametric equations x = t3 - 3t2 and y = 2t3 - 3t2 - 12t. For what values of t is the particle at rest?</em>

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