Given acceleration a = (9 - 3t) ms^-2.

Morning Micah — let's get this done. a) The velocity ($V$) of the particle at any given time $t$. Step 1: Integrate the acceleration function to find the velocity function. Given acceleration $a = (9 - 3t) \text{ ms}^{-2}$. Velocity $V(t)$ is the integral of acceleration with respect to time: $$V(t) = \int a(t)\,dt = \int (9 - 3t)\,dt$$ $$V(t) = 9t - \frac{3t^2}{2} + C$$ Step 2: Use the initial condition to find the constant of integration $C$. The initial velocity was $1 \text{ ms}^{-1}$, meaning $V(0) = 1$. $$1 = 9(0) - \frac{3(0)^2}{2} + C$$ $$1 = C$$ Step 3: Write the complete velocity function. $$V(t) = 9t - \frac{3t^2}{2} + 1$$ The velocity of the particle at any given time $t$ is $\boxed{\mathbf{V(t) = \left(9t - \frac{3t^2}{2} + 1\right) \text{ ms}^{-1}}}$. b) The maximum velocity of the particle. Step 1: Find the time $t$ when the acceleration is zero, as maximum velocity occurs when acceleration is zero. $$a(t) = 9 - 3t = 0$$ $$3t = 9$$ $$t = 3 \text{ s}$$ Step 2: Substitute this time into the velocity function found in part a) to find the maximum velocity. $$V_{max} = V(3) = 9(3) - \frac{3(3)^2}{2} + 1$$ $$V_{max} = 27 - \frac{3(9)}{2} + 1$$ $$V_{max} = 27 - \frac{27}{2} + 1$$ $$V_{max} = 28 - 13.5$$ $$V_{max} = 14.5 \text{ ms}^{-1}$$ The maximum velocity of the particle is $\boxed{\mathbf{14.5 \text{ ms}^{-1}}}$. c) The distance covered by the particle by the time it attained maximum velocity. Step 1: Integrate the velocity function to find the displacement function. Displacement $s(t)$ is the integral of velocity with respect to time: $$s(t) = \int V(t)\,dt = \int \left(9t - \frac{3t^2}{2} + 1\right)\,dt$$ $$s(t) = \frac{9t^2}{2} - \frac{3t^3}{2 \times 3} + t + D$$ $$s(t) = \frac{9t^2}{2} - \frac{t^3}{2} + t + D$$ Step 2: Use the initial condition for displacement to find the constant of integration $D$. Assume the particle starts at $s=0$ when $t=0$. $$0 = \frac{9(0)^2}{2} - \frac{(0)^3}{2} + 0 + D$$ $$0 = D$$ Step 3: Write the complete displacement function. $$s(t) = \frac{9t^2}{2} - \frac{t^3}{2} + t$$ Step 4: Substitute the time when maximum velocity is attained ($t=3 \text{ s}$ from part b) into the displacement function. $$s(3) = \frac{9(3)^2}{2} - \frac{(3)^3}{2} + 3$$ $$s(3) = \frac{9(9)}{2} - \frac{27}{2} + 3$$ $$s(3) = \frac{81}{2} - \frac{27}{2} + 3$$ $$s(3) = \frac{81 - 27}{2} + 3$$ $$s(3) = \frac{54}{2} + 3$$ $$s(3) = 27 + 3$$ $$s(3) = 30 \text{ m}$$ The distance covered by the particle by the time it attained maximum velocity is $\boxed{\mathbf{30 \text{ m}}}$. Send me the next one 📸