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 substitution: $x(t) = 3t^2 + 2t + 10$. 3. Calculate positions: - $x(0) = 3(0)^2 + 2(0) + 10 = 10$ m - $x(1) = 3(1)^2 + 2(1) + 10 = 3 + 2 + 10 = 15$ m - $x(2) = 3(2)^2 + 2(2) + 10 = 12 + 4 + 10 = 26$ m - $x(3) = 3(3)^2 + 2(3) + 10 = 27 + 6 + 10 = 43$ m - $x(4) = 3(4)^2 + 2(4) + 10 = 48 + 8 + 10 = 66$ m - $x(5) = 3(5)^2 + 2(5) + 10 = 75 + 10 + 10 = 95$ m 4. Average speed between $t=0$ and $t=2$ is given by $$\text{Average speed} = \frac{x(2) - x(0)}{2 - 0} = \frac{26 - 10}{2} = \frac{16}{2} = 8\ \text{m/s}$$ --- 5. **Problem 3.2:** Given speed $v = 3t^2 + 6t + 4$ m/s, find instantaneous acceleration at $t=0,1,2,3,4,5$. 6. Acceleration is derivative of velocity: $$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 + 6t + 4) = 6t + 6$$ 7. Calculate acceleration values: - $a(0) = 6(0) + 6 = 6$ m/s² - $a(1) = 6(1) + 6 = 12$ m/s² - $a(2) = 6(2) + 6 = 18$ m/s² - $a(3) = 6(3) + 6 = 24$ m/s² - $a(4) = 6(4) + 6 = 30$ m/s² - $a(5) = 6(5) + 6 = 36$ m/s² --- 8. **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 $t=6$: $$V(3) = 0.1(3) + 0.02(3)^2 = 0.3 + 0.18 = 0.48\ \text{m/s}$$ $$V(6) = 0.1(6) + 0.02(6)^2 = 0.6 + 0.72 = 1.32\ \text{m/s}$$ $$\Delta V = V(6) - V(3) = 1.32 - 0.48 = 0.84\ \text{m/s}$$ (ii) Average acceleration: $$a_{avg} = \frac{\Delta V}{\Delta t} = \frac{0.84}{6 - 3} = \frac{0.84}{3} = 0.28\ \text{m/s}^2$$ --- 9. **Problem 3.4:** Car 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/s². 10. Use formula: $$v = u + at \Rightarrow t = \frac{v - u}{a} = \frac{33.33 - 16.67}{2} = 8.33\ \text{s}$$ 11. Distance traveled: $$s = ut + \frac{1}{2}at^2 = 16.67(8.33) + \frac{1}{2}(2)(8.33)^2 = 138.89 + 69.44 = 208.33\ \text{m}$$ --- 12. **Problem 3.5:** Ball thrown at $20$ m/s at $30^\circ$ from $50$ m height. (i) Time of flight: Vertical velocity component: $$v_y = 20 \sin 30^\circ = 20 \times 0.5 = 10\ \text{m/s}$$ Use equation: $$y = v_yt + \frac{1}{2}(-g)t^2$$ $$-50 = 10t - 5t^2$$ Rearranged: $$5t^2 - 10t - 50 = 0$$ Divide by 5: $$\cancel{5}t^2 - \cancel{10}t - \cancel{50} = 0 \Rightarrow t^2 - 2t - 10 = 0$$ Solve quadratic: $$t = \frac{2 \pm \sqrt{4 + 40}}{2} = \frac{2 \pm \sqrt{44}}{2} = 1 \pm 3.32$$ Positive root: $$t = 4.32\ \text{s}$$ (ii) Horizontal range: $$v_x = 20 \cos 30^\circ = 20 \times 0.866 = 17.32\ \text{m/s}$$ $$R = v_x t = 17.32 \times 4.32 = 74.8\ \text{m}$$ (iii) Velocity on impact: Vertical velocity at impact: $$v_{y_f} = v_y - gt = 10 - 10 \times 4.32 = 10 - 43.2 = -33.2\ \text{m/s}$$ Magnitude of velocity: $$v = \sqrt{v_x^2 + v_{y_f}^2} = \sqrt{17.32^2 + (-33.2)^2} = \sqrt{299.9 + 1102.2} = \sqrt{1402.1} = 37.44\ \text{m/s}$$ --- 13. **Problem 3.6:** Iron ball and feather fall from $10$ m, $g=10$ m/s². 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 in same time and reach same velocity ignoring air resistance. --- 14. **Problem 3.7:** Position $x = 2 - 5t + 6t^2$. Initial velocity is derivative of position at $t=0$: $$v = \frac{dx}{dt} = -5 + 12t$$ $$v(0) = -5\ \text{m/s}$$ --- 15. **Problem 3.8:** Train speed $54$ km/h = $15$ m/s, stops over $225$ m. Use formula: $$v^2 = u^2 + 2as \Rightarrow 0 = 15^2 + 2a(225)$$ $$0 = 225 + 450a \Rightarrow a = -\frac{225}{450} = -0.5\ \text{m/s}^2$$ Retardation is $0.5$ m/s². --- 16. **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: $$t = \frac{90}{15} = 6\ \text{s}$$ --- 17. **Problem 3.10:** Position vector $\mathbf{r} = 3t^2 \hat{i} + 5t \hat{j} + 4 \hat{k}$. (a) Velocity: $$\mathbf{v} = \frac{d\mathbf{r}}{dt} = 6t \hat{i} + 5 \hat{j} + 0 \hat{k}$$ $$\mathbf{v}(3) = 18 \hat{i} + 5 \hat{j}$$ (b) Speed: $$|\mathbf{v}| = \sqrt{18^2 + 5^2} = \sqrt{324 + 25} = \sqrt{349} = 18.68\ \text{m/s}$$ (c) Acceleration: $$\mathbf{a} = \frac{d\mathbf{v}}{dt} = 6 \hat{i} + 0 \hat{j} + 0 \hat{k} = 6 \hat{i}$$ --- Final answers summarized in steps above.