Let's solve **part (b)** step by step as requested: --- ### Step 1: Find the velocity function and when the particle is at rest Given: - Initial velocity \( v_ = -20 \) m/s - Acceleration: \( a = 12 - 2t \) Recall: \( a = \frac{dv}{dt} \) Integrate to get velocity: \[ \frac{dv}{dt} = 12 - 2t \] \[ dv = (12 - 2t)dt \] Integrate both sides: \[ \int dv = \int (12 - 2t)dt \] \[ v = 12t - t^2 + C \] Use initial condition at \( t= \), \( v = -20 \): \[ -20 = 12() - ()^2 + C \implies C = -20 \] So, \[ v(t) = 12t - t^2 - 20 \] #### When is the particle at rest? Set \( v(t) = \): \[ 12t - t^2 - 20 = \] \[ -t^2 + 12t - 20 = \] \[ t^2 - 12t + 20 = \] \[ t = \frac{12 \pm \sqrt{144 - 80}}{2} = \frac{12 \pm \sqrt{64}}{2} = \frac{12 \pm 8}{2} \] \[ t_1 = 2,\quad t_2 = 10 \] **Explanation:** We found the velocity function and determined the times (\( t=2 \) and \( t=10 \)) when the particle is at rest (velocity = ). --- ### Step 2: Find total distance traveled (\( \leq t \leq 12 \)) Total distance = sum of **absolute values** of displacements in each interval (since velocity changes sign). \[ \text{Displacement} = \int v(t)dt \] Find \( s(t) \) by integrating \( v(t) \): \[ v(t) = 12t - t^2 - 20 \] \[ s(t) = \int (12t - t^2 - 20)dt = 6t^2 - \frac{1}{3}t^3 - 20t + D \] Let \( s() = \Rightarrow D = \), so: \[ s(t) = 6t^2 - \frac{1}{3}t^3 - 20t \] Now, calculate displacement for each interval, using the rest points: - From \( t= \) to \( t=2 \) - From \( t=2 \) to \( t=10 \) - From \( t=10 \) to \( t=12 \) #### Calculate \( s(2), s(10), s(12) \): **At \( t= \):** \[ s() = \] **At \( t=2 \):** \[ s(2) = 6(2^2) - \frac{1}{3}(2^3) - 20(2) = 24 - \frac{8}{3} - 40 = -16 - \frac{8}{3} = -\frac{56}{3} \] **At \( t=10 \):** \[ s(10) = 6(100) - \frac{1}{3}(100) - 200 = 600 - 333.33 - 200 = 66.67 \] **At \( t=12 \):** \[ s(12) = 6(144) - \frac{1}{3}(1728) - 240 = 864 - 576 - 240 = 48 \] **Now find distance for each segment:** - From \( t= \) to \( t=2 \): \( |s(2) - s()| = |-56/3 - | = 56/3 \) - From \( t=2 \) to \( t=10 \): \( |s(10) - s(2)| = |66.67 - (-56/3)| \) \[ 66.67 + 18.67 = 85.34 \] - From \( t=10 \) to \( t=12 \): \( |s(12) - s(10)| = |48 - 66.67| = 18.67 \) **Total distance:** \[ \text{Total distance} = \frac{56}{3} + 85.34 + 18.67 \approx 18.67 + 85.34 + 18.67 = 122.68~\text{m} \] --- ## **Final Answer** \[ \boxed{122.7~\text{m}} \] **Explanation:** We integrated the velocity to get the position function, evaluated the position at the turning points, and summed the absolute values of the displacements to get the total distance traveled.