(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 \(u = \frac{28}{3} \text{ m s}^{-1}\).

(i) Using the equation of motion:

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

For AB:

\(100 = 4u + 8a\)

For BC:

\(148 = 4v + 8a\)

Using the equation \(v = u + at\) for AB to BC:

\(v = u + 4a\)

Substitute \(v = u + 4a\) into the equation for BC:

\(148 = 4(u + 4a) + 8a\)

Simplifying gives:

\(148 = 4u + 24a\)

Now solve the system of equations:

\(100 = 4u + 8a\)

\(148 = 4u + 24a\)

Subtract the first from the second:

\(48 = 16a\)

Thus, \(a = 3\) m s^-2.

Substitute \(a = 3\) into \(100 = 4u + 8a\):

\(100 = 4u + 24\)

\(76 = 4u\)

\(u = 19\) m s^-1.

(ii) Using the equation of motion:

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

For AD:

\(61^2 = 19^2 + 2 \times 3 \times s\)

\(3721 = 361 + 6s\)

\(3360 = 6s\)

\(s = 560\)

Distance CD is:

\(CD = 560 - 248 = 312\) m.

(i) For the section AB, using the equation of motion:

\(s_{AB} = ut + \frac{1}{2} a t^2\)

where \(u = 14\) m s^-1, \(t = 5\) s, and \(a\) is the acceleration.

\(s_{AB} = 14 \times 5 + \frac{1}{2} a \times 5^2\)

\(s_{AB} = 70 + \frac{25}{2} a\)

For the section AC, the time is \(5 + 3 = 8\) s:

\(s_{AC} = 14 \times 8 + \frac{1}{2} a \times 8^2\)

\(s_{AC} = 112 + 32a\)

Since \(AC = 2AB\), we have:

\(112 + 32a = 2(70 + 12.5a)\)

\(112 + 32a = 140 + 25a\)

\(7a = 28\)

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

(ii) To find the speed at point C, use the equation:

\(v = u + at\)

where \(u = 14\) m s^-1, \(a = 4\) m s^-2, and \(t = 8\) s:

\(v = 14 + 4 \times 8\)

\(v = 46 \text{ m s}^{-1}\)

(i) To find the ratio of distances AB : BC, we use the equation of motion:

\(v^2 = u^2 + 2(-a)s\)

For AB: \(12^2 = 20^2 - 2a \times AB\)

\(144 = 400 - 2a \times AB\)

\(2a \times AB = 256\)

\(AB = \frac{128}{a}\)

For BC: \(6^2 = 12^2 - 2a \times BC\)

\(36 = 144 - 2a \times BC\)

\(2a \times BC = 108\)

\(BC = \frac{54}{a}\)

The ratio \(AB : BC = \frac{128/a}{54/a} = \frac{128}{54} = \frac{64}{27}\).

(ii) To find the distance BC, we first find the deceleration \(a\) using the entire distance AD:

\(0 = 20^2 - 2a \times 80\)

\(0 = 400 - 160a\)

\(a = 2.5\)

Now, substitute \(a\) back to find BC:

\(BC = \frac{54}{2.5} = 21.6 \text{ m}\).

To solve this problem, we use the equation of motion:

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

where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance.

1. (a) For the distance \(d_1\) between points \(A\) and \(B\):

   Initial speed \(u = 5 \text{ m s}^{-1}\), final speed \(v = 7 \text{ m s}^{-1}\).

   \(7^2 = 5^2 + 2ad_1\)

   \(49 = 25 + 2ad_1\)

   \(24 = 2ad_1\)

   \(12 = ad_1\)

2. (b) For the distance \(d_2\) between points \(B\) and \(C\):

   Initial speed \(u = 7 \text{ m s}^{-1}\), final speed \(v = 8 \text{ m s}^{-1}\).

   \(8^2 = 7^2 + 2ad_2\)

   \(64 = 49 + 2ad_2\)

   \(15 = 2ad_2\)

   \(7.5 = ad_2\)

Now, to find \(d_1\) in terms of \(d_2\):

From \(ad_1 = 12\) and \(ad_2 = 7.5\), we have:

\(\frac{ad_1}{ad_2} = \frac{12}{7.5}\)

\(\frac{d_1}{d_2} = \frac{24}{15}\)

\(\frac{d_1}{d_2} = 1.6\)

Thus, \(d_1 = 1.6d_2\).