1. **Problem statement:** We have a cannon ball following the path $$\vec{r}(t) = \langle 10t^3, -10t^2, 35 \rangle$$ meters. We need to find velocity, acceleration, instantaneous velocities at specific times, and average acceleration. 2. **Velocity formula:** Velocity is the first derivative of position with respect to time: $$\vec{v}(t) = \frac{d\vec{r}}{dt} = \left\langle \frac{d}{dt}(10t^3), \frac{d}{dt}(-10t^2), \frac{d}{dt}(35) \right\rangle$$ 3. **Calculate velocity components:** $$\frac{d}{dt}(10t^3) = 30t^2$$ $$\frac{d}{dt}(-10t^2) = -20t$$ $$\frac{d}{dt}(35) = 0$$ So, $$\vec{v}(t) = \langle 30t^2, -20t, 0 \rangle$$ m/s 4. **Acceleration formula:** Acceleration is the derivative of velocity: $$\vec{a}(t) = \frac{d\vec{v}}{dt} = \left\langle \frac{d}{dt}(30t^2), \frac{d}{dt}(-20t), \frac{d}{dt}(0) \right\rangle$$ 5. **Calculate acceleration components:** $$\frac{d}{dt}(30t^2) = 60t$$ $$\frac{d}{dt}(-20t) = -20$$ $$\frac{d}{dt}(0) = 0$$ So, $$\vec{a}(t) = \langle 60t, -20, 0 \rangle$$ m/s² 6. **Instantaneous velocity at $t=1$ s:** Substitute $t=1$ into velocity: $$\vec{v}(1) = \langle 30(1)^2, -20(1), 0 \rangle = \langle 30, -20, 0 \rangle$$ m/s 7. **Instantaneous velocity at $t=2$ s:** Substitute $t=2$: $$\vec{v}(2) = \langle 30(2)^2, -20(2), 0 \rangle = \langle 120, -40, 0 \rangle$$ m/s 8. **Average acceleration between $t_i=1$ s and $t_f=2$ s:** Average acceleration is change in velocity over change in time: $$\vec{a}_{avg} = \frac{\vec{v}(2) - \vec{v}(1)}{2 - 1} = \langle \frac{120 - 30}{1}, \frac{-40 - (-20)}{1}, \frac{0 - 0}{1} \rangle = \langle 90, -20, 0 \rangle$$ m/s² **Final answers:** - Velocity: $$\vec{v} = \langle 30t^2, -20t, 0 \rangle$$ m/s - Acceleration: $$\vec{a} = \langle 60t, -20, 0 \rangle$$ m/s² - Instantaneous velocity at $t=1$ s: $$\vec{v}_i = \langle 30, -20, 0 \rangle$$ m/s - Instantaneous velocity at $t=2$ s: $$\vec{v}_f = \langle 120, -40, 0 \rangle$$ m/s - Average acceleration between $t=1$ s and $t=2$ s: $$\vec{a}_{avg} = \langle 90, -20, 0 \rangle$$ m/s²