(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.