(a) The distance covered in the first 8 seconds is the area of the triangle under the velocity-time graph. The formula for the area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). Here, the base is 8 seconds and the height is 12.6 m/s.

\(\text{Distance} = \frac{1}{2} \times 8 \times 12.6 = 50.4 \text{ m}\)

(b) To find \(a\), use the equations of motion for the deceleration phase. From 48 to 62 seconds, the velocity decreases from 12.6 m/s to \(v_1\).

\(v_1 = 12.6 - (62 - 48)a\)

From 62 to 70 seconds, the velocity decreases from \(v_1\) to 0.

\(0 = v_1 - 2a \times (70 - 62)\)

Substituting \(v_1\) from the first equation into the second gives:

\(0 = (12.6 - 14a) - 16a\)

\(0 = 12.6 - 30a\)

\(a = \frac{12.6}{30} = 0.42\)

(c) The average speed is the total distance divided by the total time. Calculate the distance for each segment:

\(s_1 = 50.4 \text{ m}\) (from part a)

\(s_2 = (48 - 8) \times 12.6 = 504 \text{ m}\)

\(s_3 = 0.5 \times (12.6 + 6.72) \times (62 - 48) = 135.24 \text{ m}\)

\(s_4 = 0.5 \times 6.72 \times (70 - 62) = 26.88 \text{ m}\)

Total distance \(s = s_1 + s_2 + s_3 + s_4 = 50.4 + 504 + 135.24 + 26.88 = 716.52 \text{ m}\)

\(Total time = 70 seconds\)

Average speed = \(\frac{716.52}{70} = 10.236 \text{ m/s}\)

(i) The displacement of P from O at t = 10 is the area under the velocity-time graph from t = 0 to t = 10. This area is a trapezium with bases 10 and 4, and height 6. The area is given by:

\(\text{Area} = \frac{1}{2} \times (10 + 4) \times 6 = 42 \text{ m}\)

(ii) To find the velocity of P at t = 12, we use the gradient of the line segment from t = 10 to t = 12. The change in velocity is from 6 to 0 over 4 seconds, so:

\(\frac{v}{2} = \frac{6}{4}\)

Solving gives \(v = 3 \text{ m s}^{-1}\).

(iii) The total distance covered by P is 49.5 m. The distance from t = 0 to t = 10 is 42 m. The remaining distance is covered from t = 10 to t = T at a constant velocity of 3 m s^-1:

\(42 + \frac{1}{2} (T - 10) \times 3 = 49.5\)

Solving for \(T\) gives \(T = 15 \text{ s}\).

(iv) The acceleration of Q from t = 2 to t = 6 is 1.75 m s^-2. The change in velocity over this time is:

\(V = 1.75 \times 4 = 7 \text{ m s}^{-1}\)

The distance traveled by Q is the area under its velocity-time graph:

\(\text{Distance} = \frac{1}{2} (13 + 6) \times 7 = 66.5 \text{ m}\)

The distance between P and Q when they both come to rest is:

\(66.5 + 42 - 7.5 = 101 \text{ m}\)

(i) The acceleration between \(t = 3.5\) and \(t = 6\) is given by the change in velocity divided by the time interval. The velocity changes from \(10 \text{ m s}^{-1}\) to \(-15 \text{ m s}^{-1}\) over \(2.5 \text{ s}\). Thus, the acceleration is:

\(a = \frac{-15 - 10}{6 - 3.5} = \frac{-25}{2.5} = -10 \text{ m s}^{-2}\)

(ii) The acceleration between \(t = 6\) and \(t = 10\) is \(7.5 \text{ m s}^{-2}\). The initial velocity at \(t = 6\) is \(-15 \text{ m s}^{-1}\). The velocity at \(t = 10\) is \(V \text{ m s}^{-1}\). Using the formula:

\(V = -15 + 7.5 \times 4\)

\(V = 15 \text{ m s}^{-1}\)

(iii) The particle comes to rest at \(t = T\). The total distance travelled is \(100 \text{ m}\). The areas under the velocity-time graph represent the distance travelled. The areas are calculated as follows:

Area from \(t = 0\) to \(t = 4.5\):

\(\frac{1}{2} \times (4.5 + 2) \times 10\)

Area from \(t = 4.5\) to \(t = 8\):

\(\frac{1}{2} \times (8 - 4.5) \times 15\)

Area from \(t = 8\) to \(t = T\):

\(\frac{1}{2} \times (T - 8) \times 15\)

Setting the total distance to \(100 \text{ m}\):

\(\frac{1}{2} \times (4.5 + 2) \times 10 + \frac{1}{2} \times (8 - 4.5) \times 15 + \frac{1}{2} \times (T - 8) \times 15 = 100\)

Solving for \(T\):

\(T = 13.5 \text{ s}\)

(i) The distance covered in the first 42 s is the area under the velocity-time graph from 0 to 42 s. This consists of a triangle and a rectangle. The area of the triangle is \(\frac{1}{2} \times 6 \times 8.2 = 24.6\) m. The area of the rectangle is \(36 \times 8.2 = 295.2\) m. Therefore, the total distance is \(24.6 + 295.2 = 319.8\) m.

(ii) The total distance for the race is 400 m. The distance covered in the last 10 s is \(400 - 319.8 = 80.2\) m. The area under the graph from 42 s to 52 s is a trapezium with parallel sides 8.2 m/s and V m/s, and height 10 s. The area is \(\frac{1}{2} \times (8.2 + V) \times 10 = 80.2\). Solving for V gives \(V = 7.84\).

(iii) The deceleration is the change in velocity over time. The change in velocity is \(8.2 - 7.84 = 0.36\) m/s over 10 s. Therefore, the deceleration is \(\frac{0.36}{10} = 0.036\) m/s².

(i) To find the acceleration, we use the formula for acceleration as the gradient of the velocity-time graph.

For 0 < t < 30, the velocity changes from 1.5 m s^-1 to 2.1 m s^-1 over 30 seconds. Thus, the acceleration is:

\(a = \frac{2.1 - 1.5}{30} = 0.02 \text{ m s}^{-2}\)

For 30 < t < 40, the velocity changes from 2.1 m s^-1 to 0 m s^-1 over 10 seconds. Thus, the acceleration is:

\(a = \frac{0 - 2.1}{10} = -0.21 \text{ m s}^{-2}\)

\((ii) The distance AB is the area under the velocity-time graph from t = 0 to t = 60.\)

Calculate the area of each segment:

• From t = 0 to t = 30: Trapezium with bases 1.5 and 2.1, height 30.
• Area = \(\frac{1}{2} (1.5 + 2.1) \times 30 = 54 \text{ m}\)
• From t = 30 to t = 40: Triangle with base 10, height 2.1.
• Area = \(\frac{1}{2} \times 2.1 \times 10 = 10.5 \text{ m}\)
• From t = 40 to t = 52: Triangle with base 12, height 2.2.
• Area = \(\frac{1}{2} \times 2.2 \times 12 = 13.2 \text{ m}\)
• From t = 52 to t = 60: Triangle with base 8, height 2.2.
• Area = \(\frac{1}{2} \times 2.2 \times 8 = 8.8 \text{ m}\)

\(Total displacement = 54 + 10.5 - 13.2 - 8.8 = 42.5 m\)

(iii) The total distance walked is the sum of the absolute values of the areas:

\(Total distance = 54 + 10.5 + 13.2 + 8.8 = 86.5 m\)