(i) Use the equation of motion: \(v^2 = u^2 + 2as\).

Substitute \(v = 25 \text{ m s}^{-1}\), \(u = 10 \text{ m s}^{-1}\), and \(s = 525 \text{ m}\):

\(25^2 = 10^2 + 2a \times 525\)

\(625 = 100 + 1050a\)

\(525 = 1050a\)

\(a = \frac{525}{1050} = 0.5 \text{ m s}^{-2}\)

(ii) Apply Newton's second law: \(F = ma\).

The net force \(F\) is given by the driving force minus the resistance: \(900 - R = 1200 \times 0.5\).

\(900 - R = 600\)

\(R = 900 - 600 = 300 \text{ N}\)

(i) Resolve the forces vertically. The normal force \(N\) and the vertical component of the force \(X \cos \theta\) balance the weight of the slab:

\(N + X \cos \theta = mg\)

Given \(\tan \theta = \frac{7}{24}\), we find \(\cos \theta = \frac{24}{25}\).

Substitute \(m = 320 \text{ kg}\) and \(g = 10 \text{ m/s}^2\):

\(N + X \left( \frac{24}{25} \right) = 320 \times 10\)

\(N = 3200 - \frac{24}{25}X\)

(ii) Resolve the forces horizontally. The frictional force \(F\) is given by \(F = X \sin \theta\).

Given \(\sin \theta = \frac{7}{25}\), the frictional force is:

\(F = X \left( \frac{7}{25} \right)\)

The slab is about to slip when \(F = \mu N\), where \(\mu = \frac{3}{8}\):

\(\frac{7}{25}X = \frac{3}{8} \left( 3200 - \frac{24}{25}X \right)\)

Solving for \(X\):

\(\frac{7}{25}X = \frac{3}{8} \times 3200 - \frac{3}{8} \times \frac{24}{25}X\)

\(\frac{7}{25}X + \frac{72}{200}X = 1200\)

\(\frac{175}{625}X + \frac{72}{200}X = 1200\)

\(\frac{187}{625}X = 1200\)

\(X = 1875\)

(i) The frictional force is given by the formula:

\(F = \mu R\)

where \(R = mg\) is the normal reaction force. Thus,

\(F = 0.025 \times 0.15 \times 9.8 = 0.0375 \text{ N}\).

(ii) Using \(F = ma\), we have:

\(-0.0375 = 0.15a\)

Solving for \(a\), we get:

\(a = -0.25 \text{ m/s}^2\).

(iii) Using the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

\(s = 5.5 \times 4 + \frac{1}{2}(-0.25) \times 16\)

\(s = 20 \text{ m}\).

(iv) Using \(v^2 = u^2 + 2as\), we have:

\(v^2 = 3.5^2 - 2 \times 0.25 \times 20\)

\(v = 1.5 \text{ m/s}\).

(v) The return distance is given by:

\(\frac{3.5^2}{2 \times 0.25}\)

\(Total distance = 24.5 + 20 = 44.5 m.\)

Using Newton's second law for the trailer, we have:

\(300 - 200 = m \times 1.25\)

\(100 = m \times 1.25\)

\(m = \frac{100}{1.25} = 80\)

For the car, using Newton's second law:

\(3200 - F - 300 = 1500 \times 1.25\)

\(2900 - F = 1875\)

\(F = 2900 - 1875 = 1025\)

Thus, the mass of the trailer \(m\) is 80 kg and the resistance force \(F\) is 1025 N.

(a) Apply Newton's second law to the van:

\(2500 - 700 - T = 3600a\)

For the trailer:

\(T - 300 = 1200a\)

For the system of van and trailer:

\(2500 - 700 - 300 = (3600 + 1200)a\)

Simplifying gives:

\(1500 = 4800a\)

Thus, the acceleration \(a\) is:

\(a = \frac{5}{16} = 0.3125\)

Substitute \(a\) back into the equation for the trailer:

\(T - 300 = 1200 \times 0.3125\)

\(T - 300 = 375\)

\(T = 675 \text{ N}\)

(b) For the van with braking force \(F\):

\(-F - 700 = 3600a\)

For the trailer:

\(-300 = 1200a\)

For the system:

\(-F - 700 - 300 = (3600 + 1200)a\)

Solving gives:

\(F = 200 \text{ N}\)