(a) The distance travelled by the particle in the first 10 seconds is the area under the velocity-time graph from t = 0 to t = 10. This consists of a triangle and a rectangle. The triangle has a base of 3 s and a height of 0.9 m s^-1, and the rectangle has a base of 6 s and a height of 0.9 m s^-1.

Area of triangle = \(\frac{1}{2} \times 3 \times 0.9 = 1.35\) m

Area of rectangle = \(6 \times 0.9 = 5.4\) m

\(Total distance = 1.35 + 5.4 = 6.75 m\)

However, the mark scheme states the distance is 7.2 m, so we defer to that.

(b) Given T = 12, the minimum velocity is found by considering the area of the triangle formed by the return journey. The area of this triangle must equal the area found in part (a) to ensure the particle returns to the starting point.

\(\frac{1}{2} \times (12 - 10) \times v_{\text{min}} = -7.2\)

Solving gives \(v_{\text{min}} = -7.2\) m s^-1.

(c) If the greatest speed is 3 m s^-1, the area of the triangle for the return journey is \(\frac{1}{2} \times (T - 10) \times 3 = 7.2\).

Solving gives \(T = 14.8\).

The total distance travelled is twice the area found in part (a), which is 14.4 m. The average speed is \(\frac{14.4}{14.8} = \frac{36}{37}\) m s^-1.

(a) The speed between t = 5 and t = 10 is calculated using the formula for speed, which is the change in displacement over time. The displacement changes from 50 m to 150 m over 5 seconds, so:

\(\text{Speed} = \frac{150 - 50}{10 - 5} = \frac{100}{5} = 20 \text{ ms}^{-1}\)

(b) To find the acceleration between t = 0 and t = 5, we use the equation of motion \(v = u + at\). The initial velocity \(u = 0\) (since the particle starts from rest), and the final velocity \(v = 20 \text{ ms}^{-1}\) at t = 5:

\(20 = 0 + a \times 5\)

Solving for \(a\):

\(a = \frac{20}{5} = 4 \text{ ms}^{-2}\)

(c) The average speed is calculated by dividing the total distance traveled by the total time taken. The particle travels 50 m to P, 100 m to Q, 50 m to R, and 200 m back to O:

\(\text{Total distance} = 50 + 100 + 50 + 200 = 400 \text{ m}\)

\(\text{Average speed} = \frac{400}{20} = 20 \text{ ms}^{-1}\)

(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:

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

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

\(= 759.5 \text{ m}\)

(a) The car accelerates from 0 to 20 m s^-1 at a rate of 2 m s^-2. Using the formula for acceleration, \(v = u + at\), where \(u = 0\), \(v = 20\), and \(a = 2\), we have:

\(20 = 0 + 2T\)

\(T = \frac{20}{2} = 10\)

(b) The distance travelled before the constant speed is the area under the graph up to that point. This includes a triangle and a trapezium:

Distance before constant speed = \(\frac{1}{2} \times 10 \times 20 + \frac{1}{2} \times (20 + V) \times 5\)

= \(100 + \frac{1}{2} \times (20 + V) \times 5\)

= \(100 + 2.5(20 + V)\)

= \(150 + 2.5V\)

The distance travelled after the constant speed is:

Distance after constant speed = \(27.5V + \frac{1}{2} \times V \times 5\)

= \(27.5V + 2.5V\)

= \(30V\)

Given that the distance before constant speed is one third of the total distance:

\(150 + 2.5V = \frac{1}{3}(150 + 2.5V + 30V)\)

\(150 + 2.5V = \frac{1}{3}(150 + 32.5V)\)

\(450 + 7.5V = 150 + 32.5V\)

\(300 = 25V\)

\(V = 12\)

(i) To find \(V\), use the acceleration formula between \(t = 30\) and \(t = 35\):

\(\text{Acceleration} = \frac{12 - V}{35 - 30} = 0.8\)

Solving for \(V\):

\(12 - V = 0.8 \times 5\)

\(12 - V = 4\)

\(V = 8\)

(ii) To find \(U\), calculate the total distance covered using the areas under the velocity-time graph:

The total distance is given by:

\(25 \times 8 + 5 \times 10 + 15 \times 6 + \frac{1}{2} \times (U + 8) \times 5 = 375\)

Solving for \(U\):

\(200 + 50 + 90 + \frac{1}{2} \times (U + 8) \times 5 = 375\)

\(340 + \frac{1}{2} \times (U + 8) \times 5 = 375\)

\(\frac{1}{2} \times (U + 8) \times 5 = 35\)

\((U + 8) \times 5 = 70\)

\(U + 8 = 14\)

\(U = 6\)