(i) Using the equation \(s = \frac{1}{2} (u + v) t\), where \(u = 1.4 \, \text{m/s}\), \(v = 1.1 \, \text{m/s}\), and \(t = 1.2 \, \text{s}\):

\(s = \frac{1}{2} (1.4 + 1.1) \times 1.2 = 1.5 \, \text{m}\)

To find \(d\), use \(v = u + at\):

\(1.1 = 1.4 + (-d) \times 1.2\)

\(d = 0.25 \, \text{m/s}^2\)

(ii) For the motion from \(B\) to \(C\), use \(0 = u^2 + 2(-d)s\):

\(0 = u^2 + 2(-0.25) \times 2\)

\(u = 1 \, \text{m/s}\)

For time \(t\), use \(s = 0 - \frac{1}{2}(-d)t^2\):

\(2 = 0 - \frac{1}{2}(-0.25)t^2\)

\(t = 4 \, \text{s}\)

(iii) The velocity-time graph consists of a line joining \((0, 1.4)\) and \((1.2, 1.1)\), and another line joining \((1.2, -1)\) and \((5.2, 0)\).

(i) The velocity-time graph consists of three straight line segments:

• The first segment has a positive slope, starting from 0 and reaching V at time t₁.
• The second segment is a horizontal line at velocity V from t₁ to t₁ + 600 s.
• The third segment has a negative slope, decreasing from V to 0 at time 1,000 s.

(ii) Using the area under the graph to represent the distance:

For the entire journey: \(\frac{1}{2} (600 + 1000) V = 20000\)

Solving for \(V\):

\(V = 25 \text{ m/s}\)

(iii) Given acceleration \(a = 0.15 \text{ m/s}^2\), find \(t_1\):

\(V = a t_1 \Rightarrow t_1 = \frac{V}{0.15} = \frac{25}{0.15} = 166.6 \text{ s}\)

Time for third stage \(t_3 = 1000 - (t_1 + 600) = 233.3 \text{ s}\)

Distance during third stage \(s_3 = \frac{1}{2} \times 233.3 \times 25 = 2920 \text{ m}\)

(a) To find the time for which the car is accelerating, use the formula for distance under constant acceleration:

\(s = \frac{(u+v)}{2} t\)

where \(s = 200\) m, \(u = 0\) m/s, \(v = 25\) m/s. Solving for \(t\):

\(200 = \frac{(0+25)}{2} t\)

\(200 = 12.5t\)

\(t = 16\) s

(b) The velocity-time graph is a trapezium with vertices at (0, 0), (16, 25), (32, 25), and (37, 0).

(c) The total distance traveled is the sum of the distances during acceleration, constant speed, and deceleration:

Total distance = \(200 + 400 + \frac{1}{2} \times 25 \times 5 = 662.5\) m

The total time is \(16 + 16 + 5 = 37\) s.

The average speed is:

\(\frac{662.5}{37} = 17.9\) m/s (3 s.f.)

(a) The velocity-time graph is a trapezium. The car accelerates from rest to a constant velocity, travels at this constant velocity, and then decelerates to rest. The gradient of the right-hand side is approximately twice that of the left-hand side due to the deceleration being \(2a\).

(b) To find the constant velocity, use the formula: \(v = \frac{500}{25} = 20 \text{ m s}^{-1}\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(v = 20\), and \(s = 50\):

\(20^2 = 0 + 2a \times 50\)

\(400 = 100a\)

\(a = 4\)

(c) Time to accelerate: \(t = \frac{v}{a} = \frac{20}{4} = 5 \text{ s}\).

Deceleration time: \(t = \frac{v}{2a} = \frac{20}{8} = 2.5 \text{ s}\).

Total time = \(5 + 25 + 2.5 = 32.5 \text{ s}\).

(a) The velocity-time graph is a trapezium with the deceleration steeper than the acceleration. The total time from start to stop is 200 s.

(b) The total distance travelled is given by the area under the velocity-time graph, which is a trapezium. The area is \(0.5 \times (170 + 200) \times V = 2775\). Solving for \(V\), we get:

\(0.5 \times 370 \times V = 2775\)

\(185V = 2775\)

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

(c) The acceleration \(a\) can be found using the formula \(a = \frac{V}{t}\), where \(t = 20 \text{ s}\). Thus:

\(a = \frac{15}{20} = 0.75 \text{ m s}^{-2}\)