3. The polar coordinates of a point are \( \left(8,30^{\circ}\right) \). Find the Certesian coordinales of the samie point. [3 marks] 4. (a) Can a sector with rero maghitude have one or mere components that are non-zero? Evplain. [2 marks] (b) A particle undergees three consecutive dibplacements: \( \Delta \overrightarrow{\vec{F}}=(15 \mid+30 j+12 \hat{k}) \mathrm{cm}, \Delta \overrightarrow{r_{z}}= \) \( (2 . \mid-14 j-5.0 \mathrm{k}) \mathrm{cm} \), ond \( \Delta \hat{r}_{3}=(-13 i+15 j) \mathrm{cm} \). Find unitevector notation for the resultant digplacement and lits magnitude. [4 marks) (c) Tho wectors \( \vec{A} \) and \( \mathbf{i} \) bave precisely equal magnitudes. For the magnitude of \( \vec{A}+i \vec{i} \) to loo 100 tinter larger than the magninade of \( \vec{A}-\overrightarrow{1} \), what must be the angle between them? (9 marks) 5. (a) Define Lincmatics. (2 marks) (b) If the velocigy of a maticle is nowucre, can the purticles acceleration be zero? Explain. [2marks] 6. A certain autonmbile namufhetuter claims that its detuxe spors car will accelerate from rest to a sped of \( 42.0 \mathrm{~m} / \mathrm{s} \) in 2.00 s . (a) Determine the average acceleration of the car. [2 mark]] (b) Acrume that the car mover with constant acteleration. Find the ditance the car travels in the fins 8.00 s. [2 marks] 7. You are drining through town at 12.0 mis when euddenly a ball rolls out in front of you. You apply the brales and tegtilifeceleraing ar \( 3.5 \mathrm{~m} / \mathrm{s}^{2} \). (a) How far da you thavel before thepping? (4 marks) (b) When you have traveled coly half the distance in port (a), is yuur speed 6.0 ms , greater than \( 6.0 \mathrm{~m} / \mathrm{s} \), or loss than \( 6.0 \mathrm{~m} / \mathrm{s} \) ? [3 mauks] 8. (a) What is fise fall [2 marks] (b) After wineing a tase hall gance, one ployer drops a glove, whille anculer scoues a glove uraighe upmand. How do the acxeterations of the the gloves compare? Explain. 9. A watere thews a set of heys verically upward to her worovity siticr, ulo is in a window 4.00 m \( [2 \) marla] atuve. The second ruvdent catches the keys 1,50 s liter. (0) With what initial velicity mere boleys theonn? [3 marks] (b) What was the velocity of the keys just tefore they wero caught? 13 marks] 10. A ball is throm vectically donnwand from the top of a 36.6 in tall building. The luil passes the top of a uindew that is 122 mabove the ground 2.00 s after being thrown. What is the velocity of the tall as it puyes the top of the window? [5 marks]

Let's break down the problems step by step. ### Problem 3 **Given:** Polar coordinates of a point are \( (8, 30^{\circ}) \). **To find:** Cartesian coordinates. **Step 1:** Convert polar coordinates to Cartesian coordinates using the formulas: \[ x = r \cdot \cos(\theta) \] \[ y = r \cdot \sin(\theta) \] where \( r = 8 \) and \( \theta = 30^{\circ} \). **Step 2:** Calculate \( x \) and \( y \): \[ x = 8 \cdot \cos(30^{\circ}) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \] \[ y = 8 \cdot \sin(30^{\circ}) = 8 \cdot \frac{1}{2} = 4 \] **Final Answer:** The Cartesian coordinates are \( (4\sqrt{3}, 4) \). --- ### Problem 4 **(a)** Can a sector with zero magnitude have one or more components that are non-zero? **Explanation:** No, a sector with zero magnitude means that the total displacement or vector is zero. If any component were non-zero, the resultant would not be zero. Therefore, all components must also be zero. **(b)** Given displacements: \[ \Delta \overrightarrow{F} = (15 \hat{i} + 30 \hat{j} + 12 \hat{k}) \text{ cm} \] \[ \Delta \overrightarrow{r_{z}} = (2 \hat{i} - 14 \hat{j} - 5.0 \hat{k}) \text{ cm} \] \[ \Delta \hat{r}_{3} = (-13 \hat{i} + 15 \hat{j}) \text{ cm} \] **Step 1:** Find the resultant displacement: \[ \Delta \overrightarrow{R} = \Delta \overrightarrow{F} + \Delta \overrightarrow{r_{z}} + \Delta \hat{r}_{3} \] **Step 2:** Calculate each component: \[ \Delta \overrightarrow{R} = (15 + 2 - 13) \hat{i} + (30 - 14 + 15) \hat{j} + (12 - 5) \hat{k} \] \[ = 4 \hat{i} + 31 \hat{j} + 7 \hat{k} \] **Step 3:** Find the magnitude: \[ |\Delta \overrightarrow{R}| = \sqrt{4^2 + 31^2 + 7^2} = \sqrt{16 + 961 + 49} = \sqrt{1026} \] **Final Answer:** The resultant displacement in unit vector notation is \( 4 \hat{i} + 31 \hat{j} + 7 \hat{k} \) and its magnitude is \( \sqrt{1026} \). --- ### Problem 6 **(a)** Given: Final speed \( v = 42.0 \text{ m/s} \), initial speed \( u = 0 \text{ m/s} \), time \( t = 2.00 \text{ s} \). **Step 1:** Calculate average acceleration \( a \): \[ a = \frac{v - u}{t} = \frac{42.0 - 0}{2.00} = 21.0 \text{ m/s}^2 \] **Final Answer:** The average acceleration of the car is \( 21.0 \text{ m/s}^2 \). **(b)** To find the distance traveled in the first \( 8.00 \text{ s} \): Using the formula: \[ d = ut + \frac{1}{2} a t^2 \] Substituting \( u = 0 \), \( a = 21.0 \text{ m/s}^2 \), and \( t = 8.00 \text{ s} \): \[ d = 0 + \frac{1}{2} \cdot 21.0 \cdot (8.00)^2 = \frac{1}{2} \cdot 21.0 \cdot 64 = 672 \text{ m} \] **Final Answer:** The distance the car travels in the first \( 8.00 \text{ s} \) is \( 672 \text{ m} \). --- ### Problem 7 **(a)** Given: Initial speed \( u = 12.0 \text{ m/s} \), acceleration \( a = -3.5 \text{ m/s}^2 \). **Step 1:** Use the formula: \[ v^2 = u^2 + 2ad \] Setting \( v = 0 \) (when the car stops): \[ 0 = (12.0)^2 + 2(-3.5)d \] \[ 0 = 144 - 7d \implies 7d = 144 \implies d = \frac{144}{7} \approx 20.57 \text{ m} \] **Final Answer:** The distance traveled before stopping is approximately \( 20.57 \text{ m} \). **(b)** When the car has traveled half the distance: \[ d = \frac{20.57}{2} \approx 10.29 \text{ m} \] Using the formula: \[ v^2 = u^2 + 2ad \] \[ v^2 = (12.0)^2 + 2(-3.5)(10.29) \] Calculating: \[ v^2 = 144 - 71.03 \implies v^2 \approx 72.97 \implies v \approx 8.54 \text{ m/s} \] **Final Answer:** The speed when half the distance is traveled is less than \( 6.0 \text{ m/s} \). --- ### Problem 9 **(a)** Given: Height \( h = 4.00 \text{ m} \), time \( t = 1.50 \text{ s} \). Using the formula for free fall: \[ h = v_0 t + \frac{1}{2} g t^2 \] Assuming \( v_0 = 0 \): \[ 4.00 = 0 + \frac{1}{2} (9.81)(1.50)^2 \] Calculating: \[ 4.00 = \frac{1}{2} (9.81)(2.25) \implies 4.00 = 11.03625 \text{ (not valid)} \] Thus, we need to find \( v_0 \): \[ 4.00 = v