a particle moves along the x-axis so that at time t >(or equal to) 0 its position is given by x(t)=2t^3 - 21t^2 + 72t-53. At what time t is the particle at rest? ...?

Particle is at rest when Velocity = 0
V = X ' ( t ) = 6t^2 - 42t + 72 = 0
V = t^2 - 7t + 12 = 0
Solve for t:t1 = 3t2 = 4Particle is at rest at t = 3 and t = 4.

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CarliReifsteck CarliReifsteck · Answer 2 · 2019-10-22T08:07:38+02:00

Answer:

A particle will be rest firstly at t = 3 sec and again t = 4 sec.

Explanation:

Given that,

The equation of position,

[tex]x(t)=2t^3-21t^2+72t-53[/tex]

We need to calculate the velocity

On differentiating equation of position

[tex]v=\dfrac{dx(t)}{dt}=6t^2-42t+72[/tex]

A particle is at rest when the velocity equal to zero.

So, [tex]6t^2-42t+72=0[/tex]

[tex]t^2-7t+12=0[/tex]

After solution,

[tex]t = 3\ sec[/tex] , [tex] t=4\ sec[/tex]

Hence, A particle will be rest firstly at t = 3 sec and again t = 4 sec.