(a) The greatest speed is reached after accelerating for 5 seconds at 0.4 m/s². Using the formula:

\(v = u + at\)

where \(u = 0\), \(a = 0.4\), \(t = 5\):

\(v = 0 + 0.4 \times 5 = 2 \text{ m/s}\)

The velocity-time graph is a trapezium with a height of 2 m/s and time intervals of 5 s, 30 s, and 40 s.

(b) The total distance is the area under the velocity-time graph, which is a trapezium:

\(\text{Distance} = \frac{1}{2} \times (25 + 5 + 25) \times 2 = 65 \text{ m}\)

(c) Using Newton's second law during deceleration:

\(12250 - 1200g - mg = (1200 + m)(-0.2)\)

Solving for \(m\):

\(12250 = 1200 \times 9.8 + m \times 9.8 + 240 + 0.2m\)

\(12250 = 11760 + 240 + 9.8m + 0.2m\)

\(250 = 10m\)

\(m = 50\)

(d) During acceleration, using Newton's second law for the crate:

\(R - 50g = 50 \times 0.4\)

\(R = 50 \times 9.8 + 20\)

\(R = 520 \text{ N, upwards}\)

(a) The acceleration during the first 1.5 s is calculated using the formula:

\(a = \frac{v - u}{t} = \frac{2 - 0}{1.5} = \frac{4}{3} \text{ m/s}^2\)

(b) The total distance traveled upwards must equal the total distance traveled downwards. The area under the velocity-time graph represents the distance.

Upward distance: \(\frac{1}{2} \times (1.5 + 4.5) \times 2 = 6 \text{ m}\)

Downward distance: \(\frac{1}{2} \times (8.5 + 5) \times V = 6 \text{ m}\)

Equating the two distances gives:

\(\frac{1}{2} \times 13.5 \times V = 6\)

\(V = \frac{12}{13.5} = 1.70 \text{ m/s}\)

(c) Using Newton's second law, the tension \(T\) in the cable when decelerating is given by:

\(T - 1500g = 1500 \times (-2)\)

\(T = 1500g - 3000\)

\(T = 1500 \times 9.8 - 3000 = 12,000 \text{ N}\)

(i) Using energy conservation, the potential energy lost is equal to the kinetic energy gained:

\(0.4g \times 1.8 = \frac{1}{2} \times 0.4 \times v^2\)

Solving for \(v\):

\(v = 6 \text{ m/s}\)

(ii) Applying Newton's second law in the water:

\(0.4g - 5.6 = 0.4a\)

Solving for \(a\):

\(a = -4 \text{ m/s}^2\)

Using the equation \(v = u + at\) with \(u = 6 \text{ m/s}\) and \(v = 0 \text{ m/s}\):

\(0 = 6 - 4t\)

Solving for \(t\):

\(t = 1.5 \text{ s}\)

(iii) The velocity-time graph consists of two straight lines:

• First line: Starts at (0,0) with a positive gradient, reaching maximum velocity \(v = 6 \text{ m/s}\) at \(t = 0.6 \text{ s}\).
• Second line: Starts at the end of the first line with a negative gradient, ending with \(v = 0 \text{ m/s}\) at \(t = 2.1 \text{ s}\).

(i) Using Newton's second law, the net force on the particle while moving in the liquid is given by:

\(3g - R = 3 imes 5.5\)

Solving for the resistance \(R\):

\(R = 3g - 3 imes 5.5 = 3 imes 9.8 - 16.5 = 29.4 - 16.5 = 13.5 \text{ N}\)

(ii) To find the velocity at the surface, use the equation:

\(v_s^2 = u^2 + 2as\)

where \(u = 0\), \(a = 9.8 \text{ m/s}^2\), and \(s = 5 \text{ m}\):

\(v_s^2 = 2 imes 9.8 imes 5 = 98\)

\(v_s = \sqrt{98} = 10 \text{ m/s}\)

For the velocity at the bottom, use:

\(v_B^2 = v_T^2 + 2 imes 5.5 imes 4\)

where \(v_T = 10 \text{ m/s}\):

\(v_B^2 = 10^2 + 2 imes 5.5 imes 4 = 100 + 44 = 144\)

\(v_B = \sqrt{144} = 12 \text{ m/s}\)

For the time at the surface, use:

\(v = u + at\)

\(10 = 0 + 9.8t_1\)

\(t_1 = \frac{10}{9.8} \approx 1 \text{ s}\)

For the time at the bottom:

\(12 = 10 + 5.5(t_2 - t_1)\)

\(2 = 5.5(t_2 - 1)\)

\(t_2 - 1 = \frac{2}{5.5}\)

\(t_2 = 1 + \frac{2}{5.5} \approx 1.36 \text{ s}\)

(i) The distance travelled by the elevator is the area under the velocity-time graph. The graph consists of a trapezium with a base of 24 s and a height of 0.4 m/s. The area is calculated as:

\(s = \frac{1}{2} \times 5 \times 0.4 + 19 \times 0.4 + \frac{1}{2} \times 4 \times 0.4 = 9.4 \text{ m}\)

(ii) The acceleration during the first stage is the slope of the line from 0 to 5 s:

\(a = \frac{0.4}{5} = 0.08 \text{ m/s}^2\)

The deceleration during the third stage is the slope of the line from 24 to 28 s:

\(a = \frac{0.4}{4} = 0.1 \text{ m/s}^2\)

(iii) Using Newton's second law for the elevator and box system:

\(T - (800 + 100)g = (800 + 100)a\)

\(T - 900g = 900a\)

For the first stage, \(a = 0.08 \text{ m/s}^2\):

\(T = 900 \times 9.8 + 900 \times 0.08 = 9072 \text{ N}\)

For the second stage, \(a = 0\):

\(T = 900 \times 9.8 = 9000 \text{ N}\)

For the third stage, \(a = -0.1 \text{ m/s}^2\):

\(T = 900 \times 9.8 - 900 \times 0.1 = 8910 \text{ N}\)

(iv) For the force on the box, using \(R - 100g = 100a\):

Greatest force (first stage):

\(R = 100 \times 9.8 + 100 \times 0.08 = 1008 \text{ N}\)

Least force (third stage):

\(R = 100 \times 9.8 - 100 \times 0.1 = 990 \text{ N}\)