1. **Problem Statement:** A particle moves along a straight line passing through a fixed point O. Its displacement after $t$ seconds is given by $$s = 2t^2 - t^3 + 7t.$$ We are to find: a) The initial acceleration. b) The maximum velocity. c) The time when the particle is instantaneously at rest. --- 2. **Formulas and Rules:** - Velocity $v$ is the first derivative of displacement $s$ with respect to time $t$: $$v = \frac{ds}{dt}.$$ - Acceleration $a$ is the derivative of velocity with respect to time: $$a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}.$$ - To find initial acceleration, evaluate $a$ at $t=0$. - To find maximum velocity, find critical points of $v$ by setting $a=0$ and test for maxima. - To find when particle is at rest, solve $v=0$. --- 3. **Step-by-step solution:** **a) Initial acceleration:** - Differentiate $s$ to get velocity: $$v = \frac{d}{dt}(2t^2 - t^3 + 7t) = 4t - 3t^2 + 7.$$ - Differentiate $v$ to get acceleration: $$a = \frac{dv}{dt} = 4 - 6t.$$ - Initial acceleration is $a$ at $t=0$: $$a(0) = 4 - 6 \times 0 = 4.$$ **Answer:** Initial acceleration is $4$ ms$^{-2}$. **b) Maximum velocity:** - Maximum velocity occurs where acceleration $a=0$: $$4 - 6t = 0 \implies 6t = 4 \implies t = \frac{4}{6} = \frac{2}{3}.$$ - Evaluate velocity at $t=\frac{2}{3}$: $$v\left(\frac{2}{3}\right) = 4\times \frac{2}{3} - 3\left(\frac{2}{3}\right)^2 + 7 = \frac{8}{3} - 3 \times \frac{4}{9} + 7 = \frac{8}{3} - \frac{12}{9} + 7 = \frac{8}{3} - \frac{4}{3} + 7 = \frac{4}{3} + 7 = \frac{4}{3} + \frac{21}{3} = \frac{25}{3} \approx 8.33.$$ - Check if this is maximum by second derivative test: $$\frac{da}{dt} = -6 < 0,$$ so $t=\frac{2}{3}$ is a maximum point. **Answer:** Maximum velocity is $\frac{25}{3} \approx 8.33$ ms$^{-1}$. **c) Time when particle is instantaneously at rest:** - Set velocity to zero: $$4t - 3t^2 + 7 = 0 \implies -3t^2 + 4t + 7 = 0.$$ - Multiply both sides by $-1$ for clarity: $$3t^2 - 4t - 7 = 0.$$ - Use quadratic formula: $$t = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 3 \times (-7)}}{2 \times 3} = \frac{4 \pm \sqrt{16 + 84}}{6} = \frac{4 \pm \sqrt{100}}{6} = \frac{4 \pm 10}{6}.$$ - Two solutions: $$t_1 = \frac{4 + 10}{6} = \frac{14}{6} = \frac{7}{3} \approx 2.33,$$ $$t_2 = \frac{4 - 10}{6} = \frac{-6}{6} = -1.$$ - Negative time is not physically meaningful here, so discard $t=-1$. **Answer:** Particle is instantaneously at rest at $t = \frac{7}{3} \approx 2.33$ seconds. --- **Final answers:** - Initial acceleration: $4$ ms$^{-2}$ - Maximum velocity: $\frac{25}{3} \approx 8.33$ ms$^{-1}$ - Time at rest: $\frac{7}{3} \approx 2.33$ seconds