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

Given data:

X = t³ - 3t²

Y = 2t³ - 3t² - 12t

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

First step : Determine the first derivative of eac equations

dx = 3t² - 6t

dy = 6t² - 6t - 12

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

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

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Attached below is the complete question

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?

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