Ответ:
Решение:
- Скорость \( v(t) = s'(t) = \frac{d}{dt}(5t^2 + 2t - 3) = 10t + 2 \). В момент \( t = 2 \): \( v(2) = 10(2) + 2 = 20 + 2 = 22 \).
- Ускорение \( a(t) = v'(t) = s''(t) \). \( v(t) = s'(t) = \frac{d}{dt}(t^3 - 4t^2 + 6t) = 3t^2 - 8t + 6 \). \( a(t) = v'(t) = \frac{d}{dt}(3t^2 - 8t + 6) = 6t - 8 \). В момент \( t = 1 \): \( a(1) = 6(1) - 8 = 6 - 8 = -2 \).
- Путь \( S = \int_{0}^{2} v(t) dt = \int_{0}^{2} (6t^2 - 2t + 1) dt \). \( S = \left[ 6\frac{t^3}{3} - 2\frac{t^2}{2} + t \right]_{0}^{2} = \left[ 2t^3 - t^2 + t \right]_{0}^{2} = (2(2^3) - 2^2 + 2) - (0) = (2(8) - 4 + 2) = 16 - 4 + 2 = 14 \).
Ответ: 1. 22; 2. -2; 3. 14
