Let's break down the problem step-by-step: --- ## Step 1: Find the value of \( t \) at point B **Given:** - The particle starts from rest at \( A \) (\( t = \)) - Comes to rest at \( B \) - Acceleration: \( a = 6t - 2t^2 \) At point \( B \), the particle's **velocity is zero** since it comes to rest. ### Let's find the velocity function: Velocity is the integral of acceleration: \[ v = \int a \, dt = \int (6t - 2t^2) \, dt = 3t^2 - \frac{2}{3}t^3 + C \] Since the particle starts from rest, \( v = \) when \( t = \), so \( C = \). Thus, \[ v = 3t^2 - \frac{2}{3}t^3 \] At point B, \( v = \) (again, since it comes to rest): \[ 3t^2 - \frac{2}{3}t^3 = \] Factor \( t^2 \): \[ t^2 \left(3 - \frac{2}{3}t \right) = \] So \( t = \) or \( 3 - \frac{2}{3}t = \). Ignoring \( t = \) (since it's the starting point): \[ 3 = \frac{2}{3}t \implies t = \frac{9}{2} \] **Explanation:** We found \( t \) at point B by setting velocity to zero and solving for \( t \). --- ## Step 2: Find the distance traveled from A to B Distance is the **integral of velocity** from \( t = \) to \( t = \frac{9}{2} \): \[ s = \int_^{9/2} v \, dt = \int_^{9/2} \left(3t^2 - \frac{2}{3}t^3 \right) dt \] Evaluate: \[ = \left[ t^3 - \frac{1}{6} t^4 \right]_^{9/2} \] Plug in \( t = \frac{9}{2} \): First, \( \left(\frac{9}{2}\right)^3 = \frac{729}{8} \) Next, \( \left(\frac{9}{2}\right)^4 = \frac{6561}{16} \) So, \[ s = \frac{729}{8} - \frac{1}{6} \times \frac{6561}{16} = \frac{729}{8} - \frac{6561}{96} \] Get common denominator (96): \[ \frac{729}{8} = \frac{729 \times 12}{96} = \frac{8748}{96} \] So, \[ s = \frac{8748 - 6561}{96} = \frac{2187}{96} \] **Explanation:** We integrated the velocity function over the time interval to find the total distance traveled. --- ## **Final Answers** 1. **Time at point B:** \( t = \frac{9}{2} \) seconds 2. **Distance from A to B:** \( \boxed{\frac{2187}{96}} \) units