1. **State the problem:** We have a particle traveling such that its distance $s$ in metres at time $t$ seconds is given by the equation: $$s = t^2 - 2t + 5$$ We need to show by differentiation that the acceleration is constant, find its value, calculate the velocity at $t=1$, $t=3$, and $t=4$ seconds, and check by drawing the distance-time graph for $t$ from 0 to 5. 2. **Recall formulas and rules:** - Velocity $v$ is the first derivative of distance $s$ with respect to time $t$: $$v = \frac{ds}{dt}$$ - Acceleration $a$ is the derivative of velocity with respect to time, or the second derivative of distance: $$a = \frac{d^2s}{dt^2}$$ - If acceleration is constant, its value will be a constant number (not dependent on $t$). 3. **Differentiate to find velocity:** Given: $$s = t^2 - 2t + 5$$ Differentiate with respect to $t$: $$v = \frac{ds}{dt} = 2t - 2$$ 4. **Differentiate velocity to find acceleration:** $$a = \frac{dv}{dt} = \frac{d}{dt}(2t - 2) = 2$$ Since $a = 2$ is a constant, acceleration is constant. 5. **Calculate velocity at specific times:** - At $t=1$: $$v = 2(1) - 2 = 0$$ - At $t=3$: $$v = 2(3) - 2 = 6 - 2 = 4$$ - At $t=4$: $$v = 2(4) - 2 = 8 - 2 = 6$$ 6. **Interpretation:** The velocity increases linearly with time, starting from $v=0$ at $t=1$. The acceleration is constant at 2 m/s². 7. **Graph description:** The graph of $s = t^2 - 2t + 5$ is a parabola opening upwards, with vertex near $t=1$. It starts at $s=5$ when $t=0$, dips slightly near $t=1$, then rises as $t$ increases to 5. This confirms the distance-time relation and the constant acceleration. **Final answers:** - Acceleration $a = 2$ m/s² (constant) - Velocity at $t=1$ is 0 m/s - Velocity at $t=3$ is 4 m/s - Velocity at $t=4$ is 6 m/s