1. **Problem 1: Position and velocities of a particle** Given: $$s = t^3 - 6t^2 - 15t + 7$$ ft, $$t$$ in seconds. Find total distance traveled at $$t=10s$$, average velocity, average speed, instantaneous velocity, acceleration. 2. First, calculate position at $$t=10$$: $$s(10) = 10^3 - 6 \cdot 10^2 - 15 \cdot 10 + 7 = 1000 - 600 - 150 + 7 = 257 \text{ ft}$$ 3. Velocity is derivative of position: $$v = \frac{ds}{dt} = 3t^2 - 12t - 15$$ At $$t=10$$: $$v(10) = 3\cdot 10^2 - 12\cdot 10 - 15 = 300 - 120 - 15 = 165 \text{ ft/s}$$ 4. Acceleration is derivative of velocity: $$a = \frac{dv}{dt} = 6t - 12$$ At $$t=10$$: $$a(10) = 6 \cdot 10 - 12 = 48 \text{ ft/s}^2$$ 5. To find total distance traveled, find when velocity changes sign (turning points). Solve $$3t^2 - 12t - 15 = 0$$ Divide by 3: $$t^2 - 4t - 5 = 0$$ $$t = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}$$ So, roots at $$t = -1$$ (discard negative time) and $$t=5$$. 6. Calculate positions at $$t=0, 5, 10$$ to get distance segments: $$s(0) = 0 -0 -0 +7 = 7$$ $$s(5) = 125 - 150 - 75 + 7 = -93$$ $$s(10) = 257$$ 7. Distance segments: From 0 to 5: $$| -93 - 7 | = 100$$ ft From 5 to 10: $$| 257 - (-93) | = 350$$ ft Total distance = $$100 + 350 = 450$$ ft 8. Average velocity: $$\frac{s(10) - s(0)}{10 - 0} = \frac{257 - 7}{10} = 25 \text{ ft/s}$$ 9. Average speed is total distance / total time: $$\frac{450}{10} = 45 \text{ ft/s}$$ --- 10. **Problem 2: Position and distance traveled** Given: $$s = 1.5t^3 - 13.5t^2 + 22.5t$$ ft, $$t$$ in seconds. Find position at $$t=6s$$ and total distance traveled in 6 s. 11. Calculate position at $$t=6$$: $$s(6) = 1.5 \cdot 6^3 - 13.5 \cdot 6^2 + 22.5 \cdot 6 = 1.5 \cdot 216 - 13.5 \cdot 36 + 135 = 324 - 486 + 135 = -27 \text{ ft}$$ 12. Find velocity derivative: $$v = \frac{ds}{dt} = 4.5 t^2 - 27 t + 22.5$$ 13. Find roots of $$v=0$$ to locate turning points: $$4.5 t^2 - 27 t + 22.5 = 0$$ Divide all by 4.5: $$t^2 - 6 t + 5 = 0$$ Solve quadratic: $$t = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}$$ Roots: $$t=1$$ and $$t=5$$ 14. Calculate positions at $$t=0, 1, 5, 6$$: $$s(0)=0$$ $$s(1) = 1.5 - 13.5 + 22.5 = 10.5$$ $$s(5) = 1.5 \cdot 125 - 13.5 \cdot 25 + 22.5 \cdot 5 = 187.5 - 337.5 + 112.5 = -37.5$$ $$s(6) = -27$$ (from above) 15. Compute distances: 0 to 1: $$|10.5 - 0|=10.5$$ ft 1 to 5: $$|-37.5 - 10.5| = 48$$ ft 5 to 6: $$|-27 - (-37.5)|=10.5$$ ft 16. Total distance = $$10.5 + 48 + 10.5 = 69$$ ft --- 17. **Problem 3: Airplane acceleration** Given velocity: $$v = \frac{80}{1+0.32t}$$ m/s, find acceleration at $$t=3s$$. 18. Acceleration is derivative of velocity: Write velocity as $$v = 80 (1 + 0.32 t)^{-1}$$ 19. Derivative: $$a = \frac{dv}{dt} = 80 \cdot (-1) \cdot (1 + 0.32 t)^{-2} \cdot 0.32 = -25.6 (1 + 0.32 t)^{-2}$$ 20. At $$t=3$$: $$a = -25.6 \left(1 + 0.32 \cdot 3 \right)^{-2} = -25.6 (1 + 0.96)^{-2} = -25.6 (1.96)^{-2}$$ 21. Calculate $$1.96^{-2} = \frac{1}{1.96^2} = \frac{1}{3.8416} \approx 0.2604$$ 22. Acceleration: $$a \approx -25.6 \times 0.2604 = -6.67 \text{ m/s}^2$$ (negative sign indicates deceleration) **Final answers:** 1) Total distance = 450 ft, average velocity = 25 ft/s, average speed = 45 ft/s, instantaneous velocity at 10s = 165 ft/s, acceleration at 10s = 48 ft/s². 2) Position at 6s = -27 ft, total distance traveled = 69 ft. 3) Acceleration at 3s = -6.67 m/s².