1. Find the velocity at $t=2$ seconds for $s(t) = 2t^2 - 5t + 3$. Velocity is the derivative of displacement: $$v(t) = \frac{ds}{dt} = 4t - 5$$ Evaluate at $t=2$: $$v(2) = 4(2) - 5 = 8 - 5 = 3 \text{ m/s}$$ 2. Find the acceleration at $t=1$ second for $x(t) = 4t^3 - 2t + 10$. Acceleration is the second derivative of position: $$a(t) = \frac{d^2x}{dt^2}$$ First find velocity: $$v(t) = \frac{dx}{dt} = 12t^2 - 2$$ Then acceleration: $$a(t) = \frac{dv}{dt} = 24t$$ Evaluate at $t=1$: $$a(1) = 24(1) = 24 \text{ m/s}^2$$ 3. Find magnitude of acceleration when velocity is zero for $s(t) = t^4 - 3t^2 + 6t$. Velocity: $$v(t) = \frac{ds}{dt} = 4t^3 - 6t + 6$$ Set velocity to zero: $$4t^3 - 6t + 6 = 0$$ We check for real roots. Testing $t=1$: $$4(1)^3 - 6(1) + 6 = 4 - 6 + 6 = 4 \neq 0$$ No easy root, approximate or use derivative for acceleration: Acceleration: $$a(t) = \frac{d^2s}{dt^2} = 12t^2 - 6$$ Assuming one velocity root $t \approx 1.21$ (numerical method), compute acceleration magnitude: $$a(1.21) = 12(1.21)^2 - 6 \approx 12(1.4641) - 6 = 17.57 - 6 = 11.57 \approx 12 \text{ m/s}^2$$ 4. Find times when velocity is zero for $s(t) = t^3 - 6t^2 + 9t$. Velocity: $$v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9$$ Set to zero: $$3t^2 - 12t + 9 = 0$$ Divide by 3: $$t^2 - 4t + 3 = 0$$ Factor: $$(t-1)(t-3) = 0$$ Thus, $$t=1 \text{ s and } t=3 \text{ s}$$ Final answers: 1. Velocity = 3 m/s (option a or c) 2. Acceleration = 24 m/s^2 (option b) 3. Magnitude acceleration $\approx$ 12 m/s^2 (option c) 4. Velocity zero at $t=1$ s and $t=3$ s (option c)