Answer :
Final answer:
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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