(a) To find the value of T, equate the accelerations during the two acceleration phases. The first acceleration phase is from 0 to 20 m s^-1 over 5 s, giving an acceleration of \(\frac{20}{5} = 4\) m s^-2. The second acceleration phase is from 6 to 20 m s^-1 over (50 - T) s, giving an acceleration of \(\frac{20 - 6}{50 - T} = \frac{14}{50 - T}\) m s^-2. Equating these gives:

\(\frac{14}{50 - T} = 4\)

Solving for T:

\(14 = 4(50 - T)\)

\(14 = 200 - 4T\)

\(4T = 186\)

\(T = 46.5\)

(b) To find the total distance travelled, calculate the area under the velocity-time graph. The areas are:

• Triangle from 0 to 5 s: \(\frac{1}{2} \times 5 \times 20 = 50\)
• Rectangle from 5 to 25 s: \(20 \times 20 = 400\)
• Trapezium from 25 to 30 s: \(\frac{1}{2} \times 5 \times (20 + 6) = 65\)
• Rectangle from 30 to T s: \(6 \times (T - 30) = 6(T - 30)\)
• Trapezium from T to 50 s: \(\frac{1}{2} \times (50 - T) \times (20 + 6) = \frac{1}{2} \times (50 - T) \times 26\)
• Triangle from 50 to 60 s: \(\frac{1}{2} \times 10 \times 20 = 100\)

Substitute \(T = 46.5\) into the expression for the total distance:

\(50 + 400 + 65 + 6(46.5 - 30) + \frac{1}{2} \times (50 - 46.5) \times 26 + 100\)

\(= 50 + 400 + 65 + 99 + 45.5 + 100\)

\(= 759.5 \text{ m}\)