Concept: Distance covered in the \(n^{text{th}}\) second for uniformly accelerated motion.
Formula: \(S_n = u + \frac{a}{2}(2n - 1)\). Since it starts from rest, \(u=0\).
Solution: \(S_4 = \frac{a}{2}(2(4) - 1) = \frac{7a}{2}\), \(S_3 = \frac{a}{2}(2(3) - 1) = \frac{5a}{2}\). Ratio \(S_4:S_3 = 7a/2 : 5a/2 = 7:5\).

For equal distances, average speed \(v_{avg} = \frac{2v_1 v_2}{v_1 + v_2}\). Given \(v_1 = 40 \text{ km/h}\) and \(v_2 = 60 \text{ km/h}\). So, \(v_{avg} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h}\).

Given (u=0), \(v = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}\), \(t = 20 \text{ s}\). Using \(S = \frac{u+v}{2}t), we get \(S = \frac{0+40}{2} \times 20 = 20 \times 20 = 400 \text{ m}\).

Given \(x = at^2 - bt^3\). Velocity \(v = \frac{dx}{dt} = 2at - 3bt^2\). Acceleration \(a_c = \frac{dv}{dt} = 2a - 6bt\). For zero acceleration, \(2a - 6bt = 0 \Rightarrow 2a = 6bt \Rightarrow t = \frac{2a}{6b} = \frac{a}{3b}\).

Concept: Free fall under gravity.
Formula: Distance \(h = \frac{1}{2}gt^2\) ⇒ time \(t = \sqrt{\frac{2h}{g}}\), so \(t \propto \sqrt{h}\)
For body A, \(h_A = 16\text{ m}\); for body B, \(h_B = 25\text{ m}\).
Ratio: \(t_A/t_B = \sqrt{h_A/h_B} = \sqrt{16/25} = 4/5\).

Concept: Distance covered in the \(n^{\text{th}}\)) second of free fall from rest.
Formula: \(h_n = u + \frac{g}{2}(2n-1)\). Since \(u=0\), \(h_n = \frac{g}{2}(2n-1)\).
Distance in 4th second (\(n=4\)): \(h_4 = \frac{g}{2}(2 times 4 - 1) = \frac{7g}{2}\).
Distance in 5th second (\(n=5\)): \(h_5 = \frac{g}{2}(2 times 5 - 1) = \frac{9g}{2}\).
Ratio: \(h_4 : h_5 = \frac{7g}{2} : \frac{9g}{2} = 7:9\).

Initial velocity \(\vec{v}_i = 30\hat{i}\). Final velocity \(\vec{v}_f = 40\hat{j}\). Time interval \(\Delta t = 10\text{ s}\). Average acceleration \(\vec{a}_{av} = \frac{\vec{v}_f - \vec{v}_i}{\Delta t} = \frac{40\hat{j} - 30\hat{i}}{10} = -3\hat{i} + 4\hat{j}\). The magnitude of average acceleration is \(|\vec{a}_{av}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\text{ m/s}^2\).

The cross product of any vector with the zero vector is the zero vector. Therefore, \(\vec{A} \times \vec{0} = \vec{0}\).

Given \(|\vec{A}|=3, |\vec{B}|=4, |\vec{C}|=5\) and \(\vec{A} + \vec{B} = \vec{C}\). Squaring both sides: \(|\vec{A} + \vec{B}|^2 = |\vec{C}|^2\). \(|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|cos\theta = |\vec{C}|^2\). Substituting values: \(3^2 + 4^2 + 2(3)(4)cos\theta = 5^2\) which gives \(9 + 16 + 24cos\theta = 25\). Thus, \(24cos\theta = 0\) and \(\theta = \pi/2\).

Velocity \(\vec{v}(t) = \frac{d\vec{r}}{dt} = 4\hat{i} + 4\that{j}\). At \(t = 1 \text{s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2} \text{m/s}\). The angle with the x-axis is \(tan\theta = \frac{v_y}{v_x} = \frac{4}{4} = 1 ⇒
\theta = 45^\circ\).