The motion of the bus can be divided into three phases:

1. Acceleration from rest for 5 s at 2.1 \(\text{m/s}^2\):

   The final velocity \(v = u + at = 0 + 2.1 \times 5 = 10.5 \text{ m/s}\).

2. Constant speed for 24 s:

   The velocity remains 10.5 \(\text{m/s}\).

3. Deceleration to stop in 6 s:

   The initial velocity for this phase is 10.5 \(\text{m/s}\) and the final velocity is 0 \(\text{m/s}\).

The velocity-time graph is a trapezium with vertices at \((0, 0)\), \((5, 10.5)\), \((29, 10.5)\), and \((35, 0)\).

The area under the velocity-time graph gives the total distance traveled:

Area = \(\frac{1}{2} \times (24 + 35) \times 10.5\) or \(\frac{1}{2} \times 5 \times 10.5 + 24 \times 10.5 + \frac{1}{2} \times 6 \times 10.5\).

Calculating the area:

Area = \(\frac{1}{2} \times 59 \times 10.5 = 309.75 \text{ m}\).

Thus, the distance between the two bus stops is 309.75 m or 310 m.

(i) The runner accelerates from rest at 1.2 m/s² for 5 s, reaching a velocity of:

\(v = 1.2 \times 5 = 6 \text{ m/s}\)

The runner then moves at this constant speed for 12 s, and finally decelerates uniformly to rest over 3 s.

The velocity-time graph consists of three segments: a linearly increasing segment, a constant segment, and a linearly decreasing segment.

The total distance is the area under the velocity-time graph, which is a trapezium:

\(\text{Area} = \frac{1}{2} \times (12 + 20) \times 6 = 96 \text{ m}\)

(ii) The cyclist accelerates uniformly for 10 s and decelerates uniformly to rest at Q at t = 30. The total time for the cyclist is 20 s, and the distance is 96 m.

Using the formula for the area of a triangle:

\(\frac{1}{2} \times 20 \times v = 96\)

\(v = 9.6 \text{ m/s}\)

Using the equation \(s = ut + \frac{1}{2} a t^2\) with \(u = 0\), \(s = 48\), and \(t = 10\):

\(48 = \frac{1}{2} a (10)^2\)

\(a = \frac{9.6}{10} = 0.96 \text{ m/s}^2\)

(i) To find the distance travelled in the first 16 s, we divide the motion into two parts: acceleration and constant speed.

For the first 6 s, the sprinter accelerates from rest to 12 m/s. The distance \(s_1\) is given by:

\(s_1 = \frac{1}{2} (0 + 12) \times 6 = 36 \text{ m}\)

For the next 10 s, the sprinter travels at a constant speed of 12 m/s. The distance \(s_2\) is:

\(s_2 = 10 \times 12 = 120 \text{ m}\)

Thus, the total distance for the first 16 s is:

\(36 + 120 = 156 \text{ m}\)

The displacement-time graph is a curve concave up for \(0 < t < 6\) starting at (0, 0) and ending at (6, 36), a straight line with positive gradient from (6, 36) to (16, 156), and a curve concave down from (16, 156) to (20, 200).

(ii) To find \(V\), use the equation for the final 4 s of deceleration:

\(44 = \frac{1}{2} (12 + V) \times 4\)

Solving for \(V\):

\(44 = 2(12 + V)\)

\(44 = 24 + 2V\)

\(20 = 2V\)

\(V = 10\)

(i) The velocity-time graph is a trapezium. The train accelerates to 25 m/s, travels at constant speed, then decelerates to rest. The right-hand slope is steeper than the left-hand slope because the deceleration is twice the acceleration.

(ii) Let the acceleration be a. Then, the deceleration is 2a. The time to decelerate from 25 m/s to 0 is T/2. The total time is 180 s, so the time at constant speed is:

\(180 - T - \frac{T}{2} = 180 - 1.5T\)

(iii) The total distance is 3300 m. Using the area under the velocity-time graph, the distance is:

\(0.5 \times [180 + (180 - 1.5T)] \times 25 = 3300\)

Solving for T:

\(0.5 \times (360 - 1.5T) \times 25 = 3300\)

\(360 - 1.5T = 264\)

\(1.5T = 96\)

\(T = 64\)

The time to decelerate is 32 s. The distance while decelerating is:

\(0.5 \times 32 \times 25 = 400 \text{ m}\)

(i) The car accelerates from rest with an acceleration of 3 m/s² for 10 s. The final velocity after 10 s is given by:

\(v = 3 \times 10 = 30 \text{ m/s}\)

The car then travels at this constant speed for 30 s. The distance travelled during this time is:

\(s = \frac{1}{2} (30 + 40) \times 30 = 1050 \text{ m}\)

Thus, the total distance travelled in the first 40 s is 1050 m.

(ii) The car decelerates uniformly to rest, covering a distance of 450 m in this stage. The time taken for this stage is:

\(\text{Time taken} = \frac{450}{15} = 30 \text{ s}\)

The total time of motion for the car is 70 s. The motorcycle takes 50 s to travel 1500 m. Using the equation:

\(1500 = \frac{1}{2} (30 + 50) \times V\)

or

\(1500 = 30V + 0.5 \times 20V\)

Solving gives \(V = 37.5 \text{ m/s}\).

The motorcycle accelerates for 5 s and decelerates for 15 s. The acceleration \(a\) is:

\(a = \frac{37.5}{5} = 7.5 \text{ m/s}^2\)

(iii) The displacement-time graph should show three stages: acceleration, constant speed, and deceleration, with correct curvature and labels at \(t = 10, 40, 70 \text{ s}\) corresponding to displacements of 150, 1050, and 1500 m respectively.