(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, \(d_1 = 1.6d_2\).