1. **Problem 3.1:** Given position $x = 3t^2 + 2t + 10$ m, find positions at $t=0,1,2,3,4,5$ and average speed between $t=0$ and $t=2$. 2. Use the formula for position and substitute values: $$x(t) = 3t^2 + 2t + 10$$ Calculate each: - $x(0) = 3(0)^2 + 2(0) + 10 = 10$ - $x(1) = 3(1)^2 + 2(1) + 10 = 3 + 2 + 10 = 15$ - $x(2) = 3(2)^2 + 2(2) + 10 = 12 + 4 + 10 = 26$ - $x(3) = 3(3)^2 + 2(3) + 10 = 27 + 6 + 10 = 43$ - $x(4) = 3(4)^2 + 2(4) + 10 = 48 + 8 + 10 = 66$ - $x(5) = 3(5)^2 + 2(5) + 10 = 75 + 10 + 10 = 95$ 3. Average speed between $t=0$ and $t=2$ is: $$\text{Average speed} = \frac{x(2) - x(0)}{2 - 0} = \frac{26 - 10}{2} = \frac{16}{2} = 8 \text{ m/s}$$ --- 4. **Problem 3.2:** Given speed $v = 3t^2 + 6t + 4$ m/s, find instantaneous acceleration at $t=0,1,2,3,4,5$. 5. Acceleration is derivative of velocity: $$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 + 6t + 4) = 6t + 6$$ Calculate: - $a(0) = 6(0) + 6 = 6$ - $a(1) = 6(1) + 6 = 12$ - $a(2) = 6(2) + 6 = 18$ - $a(3) = 6(3) + 6 = 24$ - $a(4) = 6(4) + 6 = 30$ - $a(5) = 6(5) + 6 = 36$ --- 6. **Problem 3.3:** Velocity $V = Ct + Dt^2$ with $C=0.1$, $D=0.02$ m/s. (i) Change in velocity between $t=3$ and $t=6$: $$V(3) = 0.1(3) + 0.02(3)^2 = 0.3 + 0.18 = 0.48$$ $$V(6) = 0.1(6) + 0.02(6)^2 = 0.6 + 0.72 = 1.32$$ Change: $$\Delta V = V(6) - V(3) = 1.32 - 0.48 = 0.84$$ (ii) Average acceleration: $$a_{avg} = \frac{\Delta V}{\Delta t} = \frac{0.84}{6 - 3} = \frac{0.84}{3} = 0.28 \text{ m/s}^2$$ --- 7. **Problem 3.4:** Initial velocity $u = 60$ km/h = $\frac{60 \times 1000}{3600} = 16.67$ m/s, final velocity $v = 120$ km/h = $33.33$ m/s, acceleration $a=2$ m/s$^2$. Use formula: $$v = u + at \Rightarrow t = \frac{v - u}{a} = \frac{33.33 - 16.67}{2} = 8.33 \text{ s}$$ Distance traveled: $$s = ut + \frac{1}{2}at^2 = 16.67(8.33) + \frac{1}{2}(2)(8.33)^2 = 138.89 + 69.44 = 208.33 \text{ m}$$ --- 8. **Problem 3.5:** Ball thrown at $20$ m/s at $30^\circ$ from height $50$ m. (i) Time of flight: Vertical velocity component: $$v_y = 20 \sin 30^\circ = 10 \text{ m/s}$$ Use equation: $$y = v_yt + \frac{1}{2}(-g)t^2$$ $$-50 = 10t - 5t^2$$ Rearranged: $$5t^2 - 10t - 50 = 0$$ Divide by 5: $$t^2 - 2t - 10 = 0$$ Use quadratic formula: $$t = \frac{2 \pm \sqrt{4 + 40}}{2} = \frac{2 \pm \sqrt{44}}{2} = 1 \pm 3.32$$ Positive root: $$t = 4.32 \text{ s}$$ (ii) Horizontal range: $$v_x = 20 \cos 30^\circ = 17.32 \text{ m/s}$$ $$R = v_x t = 17.32 \times 4.32 = 74.8 \text{ m}$$ (iii) Velocity on hitting ground: Vertical velocity at impact: $$v_{y_f} = v_y - gt = 10 - 10 \times 4.32 = -33.2 \text{ m/s}$$ Magnitude: $$v = \sqrt{v_x^2 + v_{y_f}^2} = \sqrt{17.32^2 + (-33.2)^2} = \sqrt{300 + 1102} = \sqrt{1402} = 37.45 \text{ m/s}$$ --- 9. **Problem 3.6:** Iron ball and feather fall from $10$ m, $g=10$ m/s$^2$. Time to fall: $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 10}{10}} = \sqrt{2} = 1.414 \text{ s}$$ Velocity on impact: $$v = gt = 10 \times 1.414 = 14.14 \text{ m/s}$$ Both fall with same time and velocity ignoring air resistance. --- 10. **Problem 3.7:** Position $x = 2 - 5t + 6t^2$. Velocity is derivative: $$v = \frac{dx}{dt} = -5 + 12t$$ Initial velocity at $t=0$: $$v(0) = -5 + 12(0) = -5 \text{ m/s}$$ --- 11. **Problem 3.8:** Train speed $54$ km/h = $15$ m/s, stops over $225$ m. Use: $$v^2 = u^2 + 2as$$ $$0 = 15^2 + 2a(225)$$ $$2a(225) = -225$$ $$a = \frac{-225}{450} = -0.5 \text{ m/s}^2$$ Retardation is $0.5$ m/s$^2$. --- 12. **Problem 3.9:** Train speeds $36$ km/h = $10$ m/s and $18$ km/h = $5$ m/s opposite directions, length $90$ m. Relative speed: $$v_r = 10 + 5 = 15 \text{ m/s}$$ Time to pass: $$t = \frac{90}{15} = 6 \text{ s}$$ --- 13. **Problem 3.10:** Position vector: $$\vec{r} = 3t^2 \hat{i} + 5t \hat{j} + 4 \hat{k}$$ Velocity: $$\vec{v} = \frac{d\vec{r}}{dt} = 6t \hat{i} + 5 \hat{j} + 0 \hat{k}$$ At $t=3$: $$\vec{v} = 18 \hat{i} + 5 \hat{j}$$ Speed: $$|\vec{v}| = \sqrt{18^2 + 5^2} = \sqrt{324 + 25} = \sqrt{349} = 18.68 \text{ m/s}$$ Acceleration: $$\vec{a} = \frac{d\vec{v}}{dt} = 6 \hat{i} + 0 \hat{j} + 0 \hat{k} = 6 \hat{i}$$ --- Final answers summarized: - Positions (3.1): $10, 15, 26, 43, 66, 95$ m - Average speed (3.1): $8$ m/s - Accelerations (3.2): $6, 12, 18, 24, 30, 36$ m/s$^2$ - Velocity change (3.3): $0.84$ m/s - Average acceleration (3.3): $0.28$ m/s$^2$ - Time (3.4): $8.33$ s, Distance: $208.33$ m - Time of flight (3.5): $4.32$ s, Range: $74.8$ m, Impact velocity: $37.45$ m/s - Fall time (3.6): $1.414$ s, Impact velocity: $14.14$ m/s - Initial velocity (3.7): $-5$ m/s - Retardation (3.8): $0.5$ m/s$^2$ - Passing time (3.9): $6$ s - Velocity (3.10): $18 \hat{i} + 5 \hat{j}$ m/s, Speed: $18.68$ m/s, Acceleration: $6 \hat{i}$ m/s$^2$