수직선 위를 움직이는 점 P의 시각 t(t ≥ 0)에서의
속도 u(t)와 가속도 a(t)가 다음 조건을 만족시킨다.

(가) 0 ≤ t ≤ 2일 때, u(t) = (2t)^3 - 8t이다.
(나) t ≥ 2일 때, a(t) = 6t + 4이다

시각 t = 0에서 t = 3까지 점 P가 움직인 거리를 구하시오. 문제 댓글보고 복붙해서  문제 gpt4한테 풀어보라고 시킴 


We find the constant of integration D by setting s(2) from the first interval equal to s(2) from the second interval:

(1/2)(22)^4 - 42^2 = 2^3 + 22^2 - 4*2 + D ⟹ D = -4.

Therefore, the position function for the interval t ≥ 2 is s(t) = t^3 + 2t^2 - 4t - 4.

To find the distance traveled from t=2 to t=3, we subtract the position at t=2 from the position at t=3:

Distance = |s(3) - s(2)| = |3^3 + 23^2 - 43 - 4 - (2^3 + 22^2 - 42 - 4)| = 9 units.

The total distance traveled from t=0 to t=3 is the sum of the distances traveled in the two intervals:

Total distance = 8 units + 9 units = 17 units.

So, from t=0 to t=3, point P travels a total distance of 17 units on the number line.

This problem is an application of the Fundamental Theorem of Calculus, which connects differentiation and integration, and is a key tool in physics for relating position, velocity, and acceleration.


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