Answer:
The velocity function (v(t)) is ((\frac{1}{2}t^2 + 4, 8t - 2)).The position function (r(t)) is ((\frac{1}{6}t^3 + 4t + 7, 4t^2 - 2t + 6)).
Explanation:
First, let's integrate the acceleration to find the velocity:[\int a(t) dt = \int (t, 8) dt = \left(\frac{1}{2}t^2 + C_1, 8t + C_2\right)]Using the initial velocity condition (v(0) = (4, -2)), we can find the constants (C_1) and (C_2):[\frac{1}{2}(0)^2 + C_1 = 4 \implies C_1 = 4] [8(0) + C_2 = -2 \implies C_2 = -2]So, the velocity function (v(t)) is ((\frac{1}{2}t^2 + 4, 8t - 2)).Now, let's integrate the velocity to find the position:[\int v(t) dt = \int (\frac{1}{2}t^2 + 4, 8t - 2) dt = \left(\frac{1}{6}t^3 + 4t + C_3, 4t^2 - 2t + C_4\right)]Using the initial position condition (r(0) = (7, 6)), we can find the constants (C_3) and (C_4):[\frac{1}{6}(0)^3 + 4(0) + C_3 = 7 \implies C_3 = 7] [4(0)^2 - 2(0) + C_4 = 6 \implies C_4 = 6]So, the position function (r(t)) is ((\frac{1}{6}t^3 + 4t + 7, 4t^2 - 2t + 6)).To summarize:The velocity function (v(t)) is ((\frac{1}{2}t^2 + 4, 8t - 2)).The position function (r(t)) is ((\frac{1}{6}t^3 + 4t + 7, 4t^2 - 2t + 6)).