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

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.

Substituting the components into the vector equation:

$$(V_{gx}, 0) = (90 \sin\alpha, 90 \cos\alpha) + (0, 20)$$

Equating the y-components:

$$0 = 90 \cos\alpha + 20$$ $$90 \cos\alpha = -20$$ $$\cos\alpha = -\frac{20}{90} = -\frac{2}{9}$$

$$\alpha = \arccos\left(-\frac{2}{9}\right)$$ $$\alpha \approx 102.89^\circ$$

The driver must steer at a bearing of $102.89^\circ$ true.

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

2.1.2 Calculate how long the journey will take.

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.

$$\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(-\frac{2}{9}\right)^2} = \sqrt{1 - \frac{4}{81}} = \sqrt{\frac{77}{81}} = \frac{\sqrt{77}}{9}$$ $$V_{gx} = 90 \times \frac{\sqrt{77}}{9} = 10\sqrt{77} \text{ km/h}$$ $$V_{gx} \approx 10 \times 8.775 \approx 87.75 \text{ km/h}$$

The distance to the warehouse is $D = 120 \text{ km}$.

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$ $$T = \frac{120 \text{ km}}{87.75 \text{ km/h}}$$ $$T \approx 1.3675 \text{ hours}$$

Converting to minutes:

$$T \approx 1.3675 \times 60 \text{ minutes} \approx 82.05 \text{ minutes}$$

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

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 $g = 9.8 \text{ m/s}^2$ (downwards).

$$u_x = u \cos\theta = 18 \cos(25^\circ) \approx 18 \times 0.9063 = 16.313 \text{ m/s}$$ $$u_y = u \sin\theta = 18 \sin(25^\circ) \approx 18 \times 0.4226 = 7.607 \text{ m/s}$$

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

At maximum height, the vertical velocity $v_y = 0$. Using the kinematic equation $v_y^2 = u_y^2 + 2a_y \Delta y$:

$$0^2 = (7.607)^2 + 2(-9.8) H_{rel}$$ $$0 = 57.866 - 19.6 H_{rel}$$ $$H_{rel} = \frac{57.866}{19.6} \approx 2.952 \text{ m}$$

$$H_{max} = h_0 + H_{rel} = 48 \text{ m} + 2.952 \text{ m} = 50.952 \text{ m}$$

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

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

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:

$$t = \frac{7.607 + 31.602}{9.8} = \frac{39.209}{9.8} \approx 4.001 \text{ s}$$

The horizontal distance $x$ is given by $x = u_x t$:

$$x = 16.313 \text{ m/s} \times 4.001 \text{ s} \approx 65.268 \text{ m}$$

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

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.

$$t \approx 4.001 \text{ s}$$

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

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 $t_3 = 60 \text{ s}$.

$$72 \text{ km/h} = 72 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 20 \text{ m/s}$$

Convert time for Phase 2 to seconds:

$$t_2 = 3 \text{ minutes} = 3 \times 60 \text{ s} = 180 \text{ s}$$

2.3.1 Calculate the acceleration of the train.

Using the kinematic equation $v_1 = u_1 + a t_1$:

$$20 \text{ m/s} = 0 + a (45 \text{ s})$$ $$a = \frac{20}{45} \text{ m/s}^2 = \frac{4}{9} \text{ m/s}^2$$ $$a \approx 0.444 \text{ m/s}^2$$

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

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

Using $s_1 = u_1 t_1 + \frac{1}{2} a t_1^2$:

$$s_1 = (0)(45) + \frac{1}{2} \left(\frac{4}{9}\right) (45)^2$$ $$s_1 = \frac{2}{9} \times 2025 = 2 \times 225 = 450 \text{ m}$$

First, find the deceleration $a_3$. Using $v_3 = u_3 + a_3 t_3$:

$$0 = 20 + a_3 (60)$$ $$a_3 = -\frac{20}{60} = -\frac{1}{3} \text{ m/s}^2$$

Now, calculate the distance $s_3$ using $s_3 = u_3 t_3 + \frac{1}{2} a_3 t_3^2$:

$$s_3 = (20)(60) + \frac{1}{2} \left(-\frac{1}{3}\right) (60)^2$$ $$s_3 = 1200 - \frac{1}{6} (3600) = 1200 - 600 = 600 \text{ m}$$

$$S_{total} = s_1 + s_2 + s_3 = 450 \text{ m} + 3600 \text{ m} + 600 \text{ m}$$ $$S_{total} = 4650 \text{ m}$$

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