(a) Use the formula for distance traveled in the nth second:

\(s = u + \frac{1}{2}a(2n-1)\).

For the 2nd second:

\(52 = u + \frac{1}{2}a(3)\).

For the 4th second:

\(64 = u + \frac{1}{2}a(7)\).

These simplify to:

\(52 = u + 1.5a\)

\(64 = u + 3.5a\).

Subtract the first equation from the second:

\(12 = 2a\)

\(a = 6 \text{ m/s}^2\).

Substitute \(a = 6\) into \(52 = u + 1.5a\):

\(52 = u + 9\)

\(u = 43 \text{ m/s}\).

(b) Use the formula for total distance traveled:

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

Substitute \(u = 43\), \(a = 6\), and \(t = 10\):

\(s = 43 \times 10 + \frac{1}{2} \times 6 \times 10^2\)

\(s = 430 + 300\)

\(s = 730 \text{ m}\).

To solve this problem, we can use the equations of motion and the concept of areas under a velocity-time graph.

Let the constant speed be denoted by \(V\).

1. During the first 12 seconds, the bus accelerates from rest to speed \(V\). The distance covered is given by:

\(\text{Distance} = \frac{1}{2} \times 12 \times V = 6V\)

2. During the next 30 seconds, the bus travels at constant speed \(V\). The distance covered is:

\(\text{Distance} = 30V\)

3. During the final 6 seconds, the bus decelerates to rest. The distance covered is:

\(\text{Distance} = \frac{1}{2} \times 6 \times V = 3V\)

The total distance is the sum of these distances:

\(6V + 30V + 3V = 585\)

\(39V = 585\)

\(V = \frac{585}{39} = 15 \text{ m/s}\)

Thus, the constant speed of the bus is 15 m/s.

4. To find the magnitude of the deceleration, we use the equation:

\(V = u + at\)

where \(u = 15 \text{ m/s}\), \(V = 0 \text{ m/s}\), and \(t = 6 \text{ s}\).

\(0 = 15 + a \times 6\)

\(a = -\frac{15}{6} = -2.5 \text{ m/s}^2\)

The magnitude of the deceleration is 2.5 m/s².

(a) Use the constant acceleration equations to find expressions for \(s_{AB}\) and \(s_{BC}\):

\(s_{AB} = 2 \times 4.5 - \frac{1}{2} a \times 2^2\)

\(s_{BC} = 2 \times 4.5 + \frac{1}{2} a \times 2^2\)

Given \(s_{AB} = \frac{4}{5} s_{BC}\), substitute:

\([2 \times 4.5 - \frac{1}{2} a \times 2^2] = \frac{4}{5} [2 \times 4.5 + \frac{1}{2} a \times 2^2]\)

Solve for \(a\):

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

(b) Find \(AC\):

\(s_{AB} = 8 \text{ m}\) and \(s_{BC} = 10 \text{ m}\)

\(s_{AC} = s_{AB} + s_{BC} = 8 + 10 = 18 \text{ m}\)

(i) Using the equation of motion: \(s = ut + \frac{1}{2}at^2\).

For PQ: \(s_{PQ} = 20 \times 10 - 0.5 \times a \times 10^2\).

For QR: \(s_{QR} = 20 \times 10 + 0.5 \times a \times 10^2\).

Given \(s_{QR} = 1.5 \times s_{PQ}\), we have:

\(1.5(200 - 50a) = 200 + 50a\).

Simplifying gives \(100 = 125a\) leading to \(a = 0.8 \text{ m s}^{-2}\).

(ii) For distance QS, use \(s = ut + \frac{1}{2}at^2\).

\(s_{QS} = 20 \times 20 + \frac{1}{2} \times 0.8 \times 20^2\).

\(s_{QS} = 400 + 160 = 560 \text{ m}\).

Average speed between Q and S is \(\frac{560}{20} = 28 \text{ m s}^{-1}\).

We use the equation for motion under constant acceleration: \(s = ut + \frac{1}{2}at^2\).

For the first 80 m in 5 s:

\(80 = 5u + \frac{1}{2}a \times 5^2\)

\(80 = 5u + 12.5a\)

For the first 160 m in 8 s:

\(160 = 8u + \frac{1}{2}a \times 8^2\)

\(160 = 8u + 32a\)

We now have two simultaneous equations:

1. \(5u + 12.5a = 80\)
2. \(8u + 32a = 160\)

Solving these equations, we first multiply the first equation by 8 and the second by 5 to eliminate \(u\):

\(40u + 100a = 640\)

\(40u + 160a = 800\)

Subtract the first from the second:

\(60a = 160\)

\(a = \frac{8}{3}\)

Substitute \(a = \frac{8}{3}\) back into the first equation:

\(5u + 12.5 \times \frac{8}{3} = 80\)

\(5u + \frac{100}{3} = 80\)

\(5u = 80 - \frac{100}{3}\)

\(5u = \frac{240}{3} - \frac{100}{3}\)

\(5u = \frac{140}{3}\)

\(u = \frac{28}{3}\)

Thus, \(a = \frac{8}{3} \text{ m s}^{-2}\) and \(u = \frac{28}{3} \text{ m s}^{-1}\).