1. Problem 1: Find the velocity at $t=2$ if $s(t)=2t^2-5t+3$. 2. Since $s(t)$ is a position function, the velocity is $v(t)=s'(t)$. 3. Compute the derivative: $v(t)=\frac{d}{dt}(2t^2-5t+3)=4t-5$. 4. Evaluate at $t=2$: $v(2)=4(2)-5=8-5=3$. 5. Conclusion: the velocity is $3$ m/s, so the correct choices are a) and c) (both are 3 m/s). 6. Problem 2: Find the acceleration at $t=1$ if $x(t)=4t^3-2t+10$. 7. Identify that $x(t)$ is a position function, so velocity is $v(t)=x'(t)$ and acceleration is $a(t)=x''(t)$. 8. Differentiate: $v(t)=x'(t)=\frac{d}{dt}(4t^3-2t+10)=12t^2-2$. 9. Differentiate again to get acceleration: $a(t)=x''(t)=\frac{d}{dt}(12t^2-2)=24t$. 10. Evaluate at $t=1$: $a(1)=24(1)=24$. 11. Conclusion: the acceleration is $24$ m/s^2, so the correct choice is b) 24 m/s^2. 12. Problem 3: For $s(t)=t^4-3t^2+6t$, find the magnitude of acceleration when the velocity is zero. 13. Compute velocity and acceleration formulas: $v(t)=s'(t)=\frac{d}{dt}(t^4-3t^2+6t)=4t^3-6t+6$ and $a(t)=s''(t)=\frac{d}{dt}(4t^3-6t+6)=12t^2-6$. 14. Set velocity to zero to find times: solve $4t^3-6t+6=0$. 15. This is a cubic equation without simple rational roots, so we find real root(s) numerically by bracketing and refinement. 16. Evaluate the cubic at sample points: $f(-1.6)=4(-1.6)^3-6(-1.6)+6\approx-0.784$ and $f(-1.56)\approx0.176$, so a root lies near $t\approx-1.5707$ by interpolation. 17. Evaluate acceleration at that root: $a(t)=12t^2-6\approx12(1.5707^2)-6\approx23.604$. 18. The magnitude of the acceleration at the time(s) when velocity is zero is approximately $23.60$ m/s^2, so none of the provided choices (0, 6, 12, 18) match the correct value. 19. Problem 4: For $s(t)=t^3-6t^2+9t$, find the time(s) when velocity is zero. 20. Differentiate to get velocity: $v(t)=s'(t)=3t^2-12t+9$. 21. Factor the quadratic: $v(t)=3(t^2-4t+3)=3(t-1)(t-3)$. 22. Set $v(t)=0$ and solve: $(t-1)(t-3)=0$, so $t=1$ or $t=3$. 23. Conclusion: the velocity is zero at $t=1$ s and $t=3$ s, so the correct choice is c) $t=1$ s and $t=3$ s. 24. Key ideas and how to decide what to do when you see a problem: 25. If you are given a position function (often labeled $s(t)$ or $x(t)$), find velocity by differentiating once: $v(t)=s'(t)$. 26. Find acceleration by differentiating velocity once more or the position twice: $a(t)=v'(t)=s''(t)$. 27. If you are given a velocity function and asked for acceleration, differentiate the velocity. 28. If you are given acceleration and asked for velocity, integrate and use initial conditions if provided. 29. Keywords that hint what to do: the words "position" or a function labeled $s(t)$ or $x(t)$ imply you should differentiate to get velocity; the words "velocity" or $v(t)$ imply differentiate to get acceleration; "acceleration" implies you have the second derivative or must integrate if asked for velocity. 30. When asked "when is velocity zero?" set $v(t)=0$ and solve; when asked "what is the acceleration at that time?" evaluate $a(t)$ at the solutions. 31. Numerical roots of polynomials that do not factor nicely can be found using graphing, a calculator, or numerical methods (bisection, Newton's method), and then you evaluate $a(t)$ at those numerical roots.