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------------------------------------------------ A vehicle moves along the x-axis according to the equation [tex]x = 33.0t^2 + 23.0t + 14.0[/tex], where [tex]x[/tex] is in feet and [tex]t[/tex] is in seconds. Give your answers in US customary units and express them in vector form.

(a) What is the position of the vehicle when [tex]t = 5.00[/tex] seconds?

(b) What is the velocity of the vehicle when [tex]t = 5.00[/tex] seconds?

(c) What is the acceleration of the vehicle when [tex]t = 5.00[/tex] seconds?

The position of the vehicle when t = 5.00 is 855.00 ft. The velocity of the vehicle at t = 5.00 is 330.00 ft/s. The acceleration of the vehicle at t = 5.00 is 66.0 ft/s^2.

Explanation:

(a) To find the position of the vehicle when t = 5.00, we substitute the value of t into the equation x = 33.0t^2 + 14.00. Plugging in t = 5.00 gives x = 33.0(5.00)^2 + 14.00 = 841.00 + 14.00 = 855.00 ft.

(b) The velocity of the vehicle at t = 5.00 can be found by taking the derivative of the position equation x = 33.0t^2 + 14.00 with respect to t. The derivative is v = 66.0t. Plugging in t = 5.00 gives v = 66.0(5.00) = 330.00 ft/s.

(c) The acceleration of the vehicle at t = 5.00 can be found by taking the derivative of the velocity equation v = 66.0t with respect to t. The derivative is a = 66.0. Therefore, the acceleration of the vehicle is 66.0 ft/s^2.

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