Question

S(t) =t³-2t+5, t=3 найти скорость и ускорение в указанности монотонности

Answer 1

Ответ:

Пошаговое объяснение:

в конце a=6*3=18

Answer 2

Task: Find the velocity and acceleration of the function S(t) = t³ - 2t + 5 at t = 3, given the monotonicity.

To find the velocity and acceleration of the function S(t) = t³ - 2t + 5 at t = 3, we need to differentiate the function with respect to time.

Velocity:

The velocity of an object is the derivative of its position function with respect to time. In this case, the position function is S(t) = t³ - 2t + 5.

To find the velocity, we differentiate S(t) with respect to t:

S'(t) = d(S(t))/dt = 3t² - 2

Now, we can substitute t = 3 into the velocity equation to find the velocity at t = 3:

S'(3) = 3(3)² - 2 = 3(9) - 2 = 27 - 2 = 25

Therefore, the velocity of the function S(t) = t³ - 2t + 5 at t = 3 is 25.

Acceleration:

The acceleration of an object is the derivative of its velocity function with respect to time. In this case, the velocity function is S'(t) = 3t² - 2.

To find the acceleration, we differentiate S'(t) with respect to t:

S''(t) = d(S'(t))/dt = d²(S(t))/dt² = 6t

Now, we can substitute t = 3 into the acceleration equation to find the acceleration at t = 3:

S''(3) = 6(3) = 18

Therefore, the acceleration of the function S(t) = t³ - 2t + 5 at t = 3 is 18.

Monotonicity:

To determine the monotonicity of the function S(t) = t³ - 2t + 5, we need to analyze the sign of its derivative.

The derivative of S(t) is given by S'(t) = 3t² - 2. To determine the monotonicity, we need to find the intervals where the derivative is positive or negative.

Setting S'(t) > 0, we have:

3t² - 2 > 0

Solving this inequality, we find:

t > sqrt(2/3) or t < -sqrt(2/3)

Therefore, the function S(t) = t³ - 2t + 5 is increasing for t > sqrt(2/3) and decreasing for t < -sqrt(2/3).

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