Let's solve the problem step by step. **Given:** Velocity: \( v = 2t - t^2 \) (cm/s) \( t \) = time in seconds after leaving O. --- ### **1.1 The value of \( t \) at A (when the particle comes to rest)** The particle comes to rest when \( v = 0 \): \[ 2t - t^2 = 0 \] \[ t(2 - t) = 0 \] So, \( t = 0 \) or \( t = 2 \). - \( t = 0 \): the starting time at O, - \( t = 2 \): when the particle comes to rest at A. **Final Answer:** \[ \boxed{t = 2\ \text{seconds}} \] --- ### **1.2 The acceleration, \( a \) at A** Acceleration is the derivative of velocity with respect to time: \[ a = \frac{dv}{dt} \] \[ v = 2t - t^2 \implies \frac{dv}{dt} = 2 - 2t \] At \( t = 2 \): \[ a = 2 - 2 \times 2 = 2 - 4 = -2\ \text{cm/s}^2 \] **Final Answer:** \[ \boxed{a = -2\ \text{cm/s}^2} \] --- ### **1.3 The distance, \( s \), from O to A** Distance is obtained by integrating velocity with respect to time from \( t = 0 \) to \( t = 2 \): \[ s = \int_{0}^{2} v\,dt = \int_{0}^{2} (2t - t^2)\,dt \] \[ = \int_{0}^{2} 2t\,dt - \int_{0}^{2} t^2\,dt \] Calculate each part: - \( \int 2t\,dt = t^2 \) - \( \int t^2\,dt = \frac{t^3}{3} \) So, \[ s = \left[ t^2 - \frac{t^3}{3} \right]_{0}^{2} \] Plug in \( t = 2 \): \[ s = \left(2^2 - \frac{2^3}{3}\right) - (0^2 - \frac{0^3}{3}) \] \[ = (4 - \frac{8}{3}) - 0 \] \[ = \frac{12}{3} - \frac{8}{3} \] \[ = \frac{4}{3}\ \text{cm} \] **Final Answer:** \[ \boxed{s = \frac{4}{3}\ \text{cm}} \] --- ### **1.4 The total distance covered from \( t = 0 \) to \( t = 4 \)** First, let's check if the velocity changes sign between \( t = 0 \) and \( t = 4 \): \[ v = 2t - t^2 \] - At \( t = 0 \): \( v = 0 \) - At \( t = 2 \): \( v = 0 \) - At \( t = 4 \): \( v = 8 - 16 = -8 \) (negative, so particle reverses direction after \( t = 2 \)) To find the total distance, sum the absolute values of displacement from \( t = 0 \) to \( t = 2 \) and from \( t = 2 \) to \( t = 4 \): #### **Step 1:** Distance from \( t = 0 \) to \( t = 2 \): Already calculated above: \[ s_1 = \frac{4}{3}\ \text{cm} \] #### **Step 2:** Distance from \( t = 2 \) to \( t = 4 \): \[ s_2 = \left| \int_{2}^{4} (2t - t^2)\,dt \right| \] Calculate: \[ = \left| \left[ t^2 - \frac{t^3}{3} \right]_{2}^{4} \right| \] \[ = \left| \left(4^2 - \frac{4^3}{3}\right) - \left(2^2 - \frac{2^3}{3}\right) \right| \] \[ = \left| \left(16 - \frac{64}{3}\right) - (4 - \frac{8}{3}) \right| \] \[ = \left| \left(\frac{48}{3} - \frac{64}{3}\right) - \left(\frac{12}{3} - \frac{8}{3}\right) \right| \] \[ = \left| \left(-\frac{16}{3}\right) - \left(\frac{4}{3}\right) \right| \] \[ = \left| -\frac{16}{3} - \frac{4}{3} \right| \] \[ = \left| -\frac{20}{3} \right| = \frac{20}{3}\ \text{cm} \] #### **Total distance:** \[ \text{Total distance} = s_1 + s_2 = \frac{4}{3} + \frac{20}{3} = \frac{24}{3} = 8\ \text{cm} \] **Final Answer:** \[ \boxed{8\ \text{cm}} \] --- ## **Summary of Answers** 1. **The value of \( t \) at A:** \(\boxed{2\ \text{seconds}}\) 2. **The acceleration at A:** \(\boxed{-2\ \text{cm/s}^2}\) 3. **The distance from O to A:** \(\boxed{\frac{4}{3}\ \text{cm}}\) 4. **The total distance from \( t = 0 \) to \( t = 4 \):** \(\boxed{8\ \text{cm}}\)