1. The problem states: A particle's position is $s = t^3 - 6t^2 - 15t + 7$ ft and $t$ is in seconds. Find total distance traveled at $t=10$ s, average velocity, average speed, and instantaneous velocity and acceleration at $t=10$ s. 2. First, find position at $t=10$: $s(10) = 10^3 - 6(10)^2 - 15(10) + 7 = 1000 - 600 - 150 +7 = 257$ ft. 3. Total distance requires checking positions where velocity changes sign. Velocity $v = \frac{ds}{dt} = 3t^2 - 12t - 15$. 4. Solve $v=0$: $3t^2 - 12t - 15 = 0 \implies t^2 - 4t - 5 = 0 \implies (t-5)(t+1)=0$. Relevant root $t=5$. 5. Calculate position at $t=0$, $t=5$, and $t=10$: $s(0)=7$ ft, $s(5)=125 - 150 - 75 + 7 = -93$ ft, $s(10)=257$ ft. 6. Distances: from $0$ to $5$ is $|7 - (-93)| = 100$ ft, from $5$ to $10$ is $| -93 - 257| = 350$ ft. Total distance = $100 + 350 = 450$ ft. 7. Average velocity = $\frac{s(10) - s(0)}{10 - 0} = \frac{257 - 7}{10} = 25$ ft/s. 8. Average speed = total distance / total time = $450 / 10 = 45$ ft/s. 9. Instantaneous velocity at $t=10$: $v(10) = 3(10)^2 - 12(10) - 15 = 300 - 120 - 15 = 165$ ft/s. 10. Acceleration $a = \frac{dv}{dt} = 6t - 12$, so $a(10) = 6(10) - 12 = 48$ ft/s². --- 1. Problem 2: Position $s = 1.5t^3 - 13.5t^2 + 22.5t$, find position at $t=6$ and total distance traveled. 2. Calculate position at key points: s(0) = 0, s(3) = 1.5(27) - 13.5(9) + 22.5(3) = 40.5 - 121.5 + 67.5 = -13.5 ft, s(6) = 1.5(216) - 13.5(36) + 22.5(6) = 324 - 486 + 135 = -27 ft. 3. Distance from 0 to 3 s: $|0 - (-13.5)|=13.5$ ft, from 3 to 6 s: $|-13.5 - (-27)|=13.5$ ft. 4. Total distance is $13.5 + 13.5 = 27$ ft. The position at $t=6$ is $-27$ ft. --- 1. Problem 3: Velocity $v = \frac{80}{1 + 0.32t}$ m/s. Find acceleration at $t=3$ s. 2. Acceleration $a = \frac{dv}{dt} = -80 \cdot \frac{0.32}{(1 + 0.32t)^2} = -\frac{25.6}{(1 + 0.32t)^2}$. 3. At $t=3$: $a = -\frac{25.6}{(1 + 0.96)^2} = -\frac{25.6}{(1.96)^2} = -\frac{25.6}{3.8416} \approx -6.67$ m/s². --- 1. Problem 4: Motorcycle speed-time graph given; find total distance traveled until stop at $t=15$ s. 2. From $0$ to $4$ s: $v=1.25t$, distance $= \int_0^4 1.25t \, dt = 1.25 \times \frac{4^2}{2} = 1.25 \times 8 = 10$ m. 3. From $4$ to $10$ s: $v=5$ m/s constant, distance = $5 \times (10 - 4) = 30$ m. 4. From $10$ to $15$ s: $v = -t + 15$, distance $= \int_{10}^{15} (-t + 15) dt = [ -\frac{t^2}{2} + 15t ]_{10}^{15} = ( -\frac{225}{2} + 225 ) - ( -\frac{100}{2} + 150 ) = ( -112.5 + 225 ) - ( -50 + 150 ) = 112.5 - 100 = 12.5$ m. 5. Total distance = $10 + 30 + 12.5 = 52.5$ m. --- 1. Problem 5: Initial speed 70 km/h, acceleration 6000 km/h², find time to reach 120 km/h and distance traveled. 2. Time: $t = \frac{v - v_0}{a} = \frac{120 - 70}{6000} = \frac{50}{6000} = 0.00833$ h = 30 s. 3. Distance: $s = v_0 t + \frac{1}{2} a t^2 = 70 \times 0.00833 + 0.5 \times 6000 \times (0.00833)^2 = 0.583 + 0.208 = 0.791$ km = 791 m. --- 1. Problem set 4, problem 1: $a=2t -6$, initial velocity 0. Find velocity at $t=6$ and position at $t=11$. 2. $v = \int a \, dt = \int (2t - 6) dt = t^2 - 6t + C$, at $t=0$, $v=0$, so $C=0$. 3. $v(6) = 36 - 36 = 0$ m/s. 4. Position $s = \int v \, dt = \int (t^2 - 6t) dt = \frac{t^3}{3} - 3t^2 + D$, at $t=0$, take $s=0$ so $D=0$. 5. $s(11) = \frac{11^3}{3} - 3(11)^2 = \frac{1331}{3} - 363 = 443.67 - 363 = 80.67$ m. --- 1. Problem 2: $a = \frac{1}{4} \sqrt{s}$, with $v=0$ and $s=1$ at $t=0$. Find $v$ at $s=2$. 2. Using $a = v \frac{dv}{ds}$, so $v \frac{dv}{ds} = \frac{1}{4} s^{1/2}$. 3. Separate variables: $v dv = \frac{1}{4} s^{1/2} ds$. 4. Integrate from $v=0$ to $v$, and $s=1$ to $2$: $\frac{v^2}{2} = \frac{1}{4} \times \frac{2}{3} (2^{3/2} - 1^{3/2})$. 5. Simplify: $v^2 = \frac{1}{3} (2^{3/2} - 1) = \frac{1}{3} (2 \sqrt{2} - 1) \approx \frac{1}{3} (2.828 -1) = 0.609$. 6. So $v \approx \sqrt{0.609} = 0.78$ m/s. --- 1. Problem 3: Race car velocity and acceleration graph, find max acceleration. 2. From $0$ to $4$ s, $v = 3.5t$, so $a = 3.5$ m/s² max. --- 1. Problem 4: Sandbag dropped from balloon ascending at 6 m/s upward, hits ground at $t=8$ s. Find impact speed and altitude at impact. 2. Initial velocity $v_0 = 6$ m/s upward, acceleration $g = -9.8$ m/s². 3. Velocity at impact: $v = v_0 - g t = 6 - 9.8 imes 8 = 6 - 78.4 = -72.4$ m/s (downward). 4. Displacement: s = $v_0 t - \frac{1}{2} g t^2 = 6 imes 8 - 0.5 imes 9.8 imes 64 = 48 - 313.6 = -265.6$ m (downward from release point). 5. Balloon altitude at impact is 265.6 m above ground. --- 1. Problem 5: Steel ball velocity with acceleration $a = 2.4 - 0.6 v$, initial $v=0$. Find $v$ at $t=2$ s. 2. Solve $\frac{dv}{dt} = 2.4 - 0.6 v$. Rearranged: $\frac{dv}{2.4 - 0.6 v} = dt$. 3. Integrate: $\int \frac{dv}{2.4 - 0.6 v} = \int dt$. 4. Let $u=2.4 - 0.6 v$, then $du = -0.6 dv \implies dv = -\frac{du}{0.6}$. 5. Substituting: $\int \frac{-du/0.6}{u} = t + C$. 6. Integral is: $-\frac{1}{0.6} \int \frac{du}{u} = t + C \Rightarrow -\frac{1}{0.6} \ln |u| = t + C$. 7. Solve for $v$: $\ln |2.4 - 0.6 v| = -0.6 t + C'$. 8. At $t=0$, $v=0$: $\ln 2.4 = C'$. 9. Thus $\ln (2.4 - 0.6 v) = -0.6 t + \ln 2.4$. 10. Exponentiate: $2.4 - 0.6 v = 2.4 e^{-0.6 t}$. 11. Solve for $v$: $v = \frac{2.4}{0.6} (1 - e^{-0.6 t}) = 4 (1 - e^{-0.6 t})$ cm/s. 12. At $t=2$ s, $v = 4 (1 - e^{-1.2}) \approx 4 (1 - 0.301) = 4 (0.699) = 2.796$ cm/s. Final answers are boxed in the explanations above.