Solution to Question 1:

1. Explanation: Distance is a scalar quantity measuring the actual path length traveled (the 1.5 laps), while displacement is a vector quantity measuring the shortest straight-line distance between the initial position (A) and final position (B). Because the track is curved, the actual path is much longer than the straight-line shortcut across the field.
2. Calculation of Displacement:

   First, find the radius (\( \displaystyle R \)) of the track. One full lap (circumference) is \( \displaystyle 400\text{m} \).

   \[ \displaystyle 2 \times \pi \times R = 400 \]

   \[ \displaystyle 2 \times \left(\frac{22}{7}\right) \times R = 400 \]

   \[ \displaystyle R = \frac{400 \times 7}{44} \approx 63.63\text{ m} \]

   Since the runner completes 1.5 laps, they end up exactly diametrically opposite to their starting point. Therefore, the displacement is the diameter of the track:

   \[ \displaystyle \text{Displacement} = 2R = 2 \times 63.63 = 127.26\text{ m} \]

3. Speed and Velocity:

   \[ \displaystyle \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{600\text{ m}}{100\text{ s}} = 6.0\text{ m/s} \]

   \[ \displaystyle \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{127.26\text{ m}}{100\text{ s}} \approx 1.27\text{ m/s} \]

   The average velocity is lesser because it relies on displacement (the shortest path), which is inherently shorter than or equal to the actual curved path taken.

Solution to Question 2:

1. Graph Description: The v-t graph forms a trapezium composed of three parts:
   • A right-angled triangle (\( \displaystyle 0 \) to \( \displaystyle 5\text{s} \)) representing uniform acceleration.
   • A rectangle (\( \displaystyle 5\text{s} \) to \( \displaystyle 15\text{s} \)) representing constant velocity.
   • A right-angled triangle (\( \displaystyle 15\text{s} \) to \( \displaystyle 20\text{s} \)) representing uniform deceleration.
2. Acceleration Calculation:

   \[ \displaystyle \text{Acceleration } (a) = \frac{v – u}{t} \]

   First 5 seconds: \[ \displaystyle a = \frac{15 – 0}{5} = +3\text{ m/s}^2 \]

   Final 5 seconds: \[ \displaystyle a = \frac{0 – 15}{5} = -3\text{ m/s}^2 \]

   (Retardation is \( \displaystyle 3\text{ m/s}^2 \))

3. Maximum Height Calculation:

   The distance (height) is the area of the v-t graph (Trapezium).

   \[ \displaystyle \text{Area} = \text{Area of Triangle 1} + \text{Area of Rectangle} + \text{Area of Triangle 2} \]

   \[ \displaystyle \text{Area} = \left(\frac{1}{2} \times 5 \times 15\right) + (10 \times 15) + \left(\frac{1}{2} \times 5 \times 15\right) \]

   \[ \displaystyle \text{Area} = 37.5 + 150 + 37.5 = 225\text{ m} \]

   The drone reached a maximum height of \( \displaystyle 225 \) meters.

Solution to Question 3:

1. Unit Conversion:

   Initial velocity, \( \displaystyle u = 90\text{ km/h} \)

   \[ \displaystyle u = 90 \times \left(\frac{5}{18}\right) = 25\text{ m/s} \]

2. Stopping Distance:

   Final velocity, \( \displaystyle v = 0\text{ m/s} \) (since it stops)

   Acceleration, \( \displaystyle a = -5\text{ m/s}^2 \)

   Using the third equation of motion: \[ \displaystyle v^2 – u^2 = 2as \]

   \[ \displaystyle (0)^2 – (25)^2 = 2 \times (-5) \times s \]

   \[ \displaystyle -625 = -10 \times s \]

   \[ \displaystyle s = 62.5\text{ m} \]

   Since the stopping distance (\( \displaystyle 62.5\text{ m} \)) is greater than the distance to the barricade (\( \displaystyle 60\text{ m} \)), the car WILL hit the barricade.

3. Wet Road Scenario:

   New acceleration, \( \displaystyle a = -3\text{ m/s}^2 \)

   \[ \displaystyle (0)^2 – (25)^2 = 2 \times (-3) \times s \]

   \[ \displaystyle -625 = -6 \times s \]

   \[ \displaystyle s = \frac{625}{6} \approx 104.16\text{ m} \]

   Implication: A lower magnitude of deceleration (due to reduced friction on a wet road) drastically increases the stopping distance from \( \displaystyle 62.5\text{ m} \) to \( \displaystyle 104.16\text{ m} \), highlighting the scientific necessity for reduced speeds during rain.