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|| CALC The vertical position of a particle is given by the function y=(t^2-4t+2) m, where t is in s. At what time does the particle have a turning point in its motion? What is the particle’s position at that time?

Sagot :

KitdaAneekidn

Answer:

-2 meters

Explanation:

Start by taking the derivative of the vertical position function. From there, find where the derivative equals 0. To ensure that this point is a turning point, take the second derivative and ensure that it is not equal to 0.

Solving:

[tex]\( y = t^2 - 4t + 2 \)[/tex]

[tex]\[\frac{dy}{dt} = \frac{d}{dt}(t^2 - 4t + 2) = \boxed{2t - 4}\][/tex]

[tex]2t - 4 = 0\\2t = 4\\\\\boxed{t = 2}[/tex]

[tex]\[\frac{d^2y}{dt^2} = \frac{d}{dt}(2t - 4) = \boxed{2}\][/tex]

Since the second derivative is greater than 0, it is a local minimum which means that the sign changes at t = 2 in the first derivative. From here we plug in 2 for t in the original function to find the particle's position.

[tex]y(2) = (2)^2 - 4(2) + 2\\\\y(2) = 4 - 8 + 2\\\\\boxed{y(2) = -2\\}[/tex]

Therefore, the particle has a turning point at t = 2 with a position of -2 meters.

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