Define the following terms: Resultant force, Elastic limit, Strain, Hydrostatic pressure, Second moment of area.

Here are the solutions to the kinematics problems.

SECTION B

QUESTION 2: KINEMATICS

2.1

• Given: Truck's speed relative to air $V_t = 90 \text{ km/h}$. Wind velocity $V_w = 20 \text{ km/h}$ from the South (i.e., blowing North). Warehouse is $120 \text{ km}$ due East.

2.1.1 Determine the true course the driver must steer to reach the warehouse.

Step 1: Define the velocity vectors. Let the resultant ground velocity be $\vec{V}_g$, the truck's velocity relative to air be $\vec{V}_t$, and the wind velocity be $\vec{V}_w$. The destination is due East, so the resultant ground velocity must be purely in the East direction. The wind blows North. We have the vector equation: $\vec{V}_g = \vec{V}_t + \vec{V}_w$ Let East be the positive x-direction and North be the positive y-direction. The wind velocity is $\vec{V}_w = (0, 20) \text{ km/h}$. The magnitude of the truck's velocity relative to air is $|\vec{V}_t| = 90 \text{ km/h}$. Let the driver steer at a bearing $\alpha$ (clockwise from North). So, $\vec{V}_t = (90 \sin\alpha, 90 \cos\alpha)$. The resultant ground velocity is $\vec{V}_g = (V_{gx}, 0)$, where $V_{gx}$ is the ground speed towards East.

Step 2: Set up and solve the component equations. Substituting the components into the vector equation: $(V_{gx}, 0) = (90 \sin\alpha, 90 \cos\alpha) + (0, 20)$ Equating the y-components: $0 = 90 \cos\alpha + 20$ $90 \cos\alpha = -20$ $\cos\alpha = -\frac{20}{90} = -\frac{2}{9}$ Step 3: Calculate the bearing. $\alpha = \arccos\left(-\frac{2}{9}\right)$ $\alpha \approx 102.89^\circ$ The driver must steer at a bearing of $102.89^\circ$ true.

The true course the driver must steer is $\boxed{102.89^\circ true}$.

2.1.2 Calculate how long the journey will take.

Step 1: Calculate the magnitude of the resultant ground velocity. From the x-components: $V_{gx} = 90 \sin\alpha$ Using $\cos\alpha = -\frac{2}{9}$, we can find $\sin\alpha$. Since $\alpha$ is in the second quadrant ($102.89^\circ$), $\sin\alpha$ is positive. $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(-\frac{2}{9}\right)^2} = \sqrt{1 - \frac{4}{81}} = \sqrt{\frac{77}{81}} = \frac{\sqrt{77}}{9}$ $V_{gx} = 90 \times \frac{\sqrt{77}}{9} = 10\sqrt{77} km/h$ $V_{gx} \approx 10 \times 8.775 \approx 87.75 km/h$ Step 2: Calculate the time taken for the journey. The distance to the warehouse is $D = 120 \text{ km}$. $Time = \frac{Distance}{Speed}$ $T = \frac{120 km}{87.75 km/h}$ $T \approx 1.3675 hours$ Converting to minutes: $T \approx 1.3675 \times 60 minutes \approx 82.05 minutes$

The journey will take approximately $\boxed{82.05 minutes}$.

2.2

• Given: Initial height $h_0 = 48 \text{ m}$. Initial velocity $u = 18 \text{ m/s}$. Angle $\theta = 25^\circ$ above horizontal. Acceleration due to gravity $g = 9.8 m/s^2$ (downwards).

Step 1: Resolve initial velocity into components. $u_x = u \cos\theta = 18 \cos(25^\circ) \approx 18 \times 0.9063 = 16.313 m/s$ $u_y = u \sin\theta = 18 \sin(25^\circ) \approx 18 \times 0.4226 = 7.607 m/s$

2.2.1 The maximum height above the ground reached by the ball.

Step 2: Calculate the vertical displacement from the projection point to the maximum height. At maximum height, the vertical velocity $v_y = 0$. Using the kinematic equation $v_y^2 = u_y^2 + 2a_y \Delta y$: $0^2 = (7.607)^2 + 2(-9.8) H_{rel}$ $0 = 57.866 - 19.6 H_{rel}$ $H_{rel} = \frac{57.866}{19.6} \approx 2.952 m$ Step 3: Calculate the maximum height above the ground. $H_{max} = h_0 + H_{rel} = 48 m + 2.952 m = 50.952 m$

The maximum height above the ground reached by the ball is $\boxed{50.95 m}$.

2.2.2 The horizontal distance from the foot of the building where the ball hits the ground.

Step 1: Calculate the total time of flight. The ball starts at $y_0 = 48 \text{ m}$ and hits the ground at $y = 0 \text{ m}$. Using the kinematic equation $y = y_0 + u_y t + \frac{1}{2} a_y t^2$: $0 = 48 + 7.607 t + \frac{1}{2}(-9.8) t^2$ $0 = 48 + 7.607 t - 4.9 t^2$ Rearranging into a quadratic equation: $4.9 t^2 - 7.607 t - 48 = 0$ Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-(-7.607) \pm \sqrt{(-7.607)^2 - 4(4.9)(-48)}}{2(4.9)}$ $t = \frac{7.607 \pm \sqrt{57.866 + 940.8}}{9.8}$ $t = \frac{7.607 \pm \sqrt{998.666}}{9.8}$ $t = \frac{7.607 \pm 31.602}{9.8}$ We take the positive root for time: $t = \frac{7.607 + 31.602}{9.8} = \frac{39.209}{9.8} \approx 4.001 s$ Step 2: Calculate the horizontal distance. The horizontal distance $x$ is given by $x = u_x t$: $x = 16.313 m/s \times 4.001 s \approx 65.268 m$

The horizontal distance from the foot of the building where the ball hits the ground is $\boxed{65.27 m}$.

2.2.3 The time taken for the ball to hit the ground.

The time taken for the ball to hit the ground was calculated in 2.2.2. $t \approx 4.001 s$

The time taken for the ball to hit the ground is $\boxed{4.00 s}$.

2.3

• Given:
  • Phase 1: Accelerates from rest ($u_1 = 0$) to $v_1 = 72 \text{ km/h}$ in $t_1 = 45 \text{ s}$.
  • Phase 2: Travels at constant speed ($v_2 = 72 \text{ km/h}$) for $t_2 = 3 \text{ minutes}$.
  • Phase 3: Brakes uniformly to rest ($v_3 = 0$) in $t_3 = 60 \text{ s}$.

Step 1: Convert velocities to SI units. $72 km/h = 72 \times \frac{1000 m}{3600 s} = 20 m/s$ Convert time for Phase 2 to seconds: $t_2 = 3 minutes = 3 \times 60 s = 180 s$

2.3.1 Calculate the acceleration of the train.

Step 2: Calculate acceleration during Phase 1. Using the kinematic equation $v_1 = u_1 + a t_1$: $20 m/s = 0 + a (45 s)$ $a = \frac{20}{45} m/s^2 = \frac{4}{9} m/s^2$ $a \approx 0.444 m/s^2$

The acceleration of the train is $\boxed{0.444 m/s^2}$.

2.3.2 Determine the total distance travelled by the train from the start until it comes to rest.

Step 1: Calculate distance for Phase 1 (acceleration). Using $s_1 = u_1 t_1 + \frac{1}{2} a t_1^2$: $s_1 = (0)(45) + \frac{1}{2} \left(\frac{4}{9}\right) (45)^2$ $s_1 = \frac{2}{9} \times 2025 = 2 \times 225 = 450 m$ Step 2: Calculate distance for Phase 2 (constant speed). $s_2 = v_2 t_2 = (20 m/s)(180 s) = 3600 m$ Step 3: Calculate distance for Phase 3 (deceleration). First, find the deceleration $a_3$. Using $v_3 = u_3 + a_3 t_3$: $0 = 20 + a_3 (60)$ $a_3 = -\frac{20}{60} = -\frac{1}{3} m/s^2$ Now, calculate the distance $s_3$ using $s_3 = u_3 t_3 + \frac{1}{2} a_3 t_3^2$: $s_3 = (20)(60) + \frac{1}{2} \left(-\frac{1}{3}\right) (60)^2$ $s_3 = 1200 - \frac{1}{6} (3600) = 1200 - 600 = 600 m$ Step 4: Calculate the total distance. $S_{total} = s_1 + s_2 + s_3 = 450 m + 3600 m + 600 m$ $S_{total} = 4650 m$

The total distance travelled by the train is $\boxed{4650 m}$.

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