Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).

The ratio of maximum height \( H \) to horizontal range \( R \) is given by \( \frac{H}{R} = \frac{\tan \theta}{4} \). For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \), leading to \( \frac{H}{R} = \frac{1}{4} \).

The distance travelled in the \( n^{\text{th}} \) second is \( s_n = u + \frac{a}{2}(2n - 1) \). Substituting \( u = 0 \), \( a = 2 \, \text{m/s}^2 \), and \( n = 4 \) yields \( s_4 = 0 + \frac{2}{2}(2(4) - 1) = 7 \, \text{m} \).

Displacement is \( S = \int v \, dt = \int_0^t (6 - 3t) \, dt = 6t - 1.5t^2 \). Returning to the starting point means \( S = 0 ⇒ 6t - 1.5t^2 = 0 ⇒ t = 4\text{ s} \).

Relative velocity required: \( v_{\text{rel}} = \frac{\text{distance}}{\text{time}} = \frac{1500\text{ m}}{60\text{ s}} = 25\text{ m/s} \). Since \( v_{\text{rel}} = v_s - v_c \), we get \( v_s = v_c + v_{\text{rel}} = 20 + 25 = 45\text{ m/s} \).

Maximum height \( H \propto u^2 \) and Time of flight \( T \propto u \) for a given angle. Since particle B has a larger initial velocity, both parameters are greater for B.

Concept: Average speed is total distance divided by total time. Let distance from X to Y be \( D \). Time taken to go to Y: \( t_u = D/v_u \). Time taken to return to X: \( t_d = D/v_d \). Total distance \( = 2D \). Total time \( = t_u + t_d = D/v_u + D/v_d = D \frac{v_u + v_d}{v_u v_d} \). Average speed \( = \frac{2D}{D \frac{v_u + v_d}{v_u v_d}} = \frac{2v_u v_d}{v_u + v_d} \).

Concept: Velocity is the time derivative of displacement. First, express \( x \) as a function of \( t \): from \( t = \sqrt{x} + 3 \), we get \( \sqrt{x} = t - 3 \), so \( x = (t-3)^2 \). Then find velocity \( v = \frac{dx}{dt} = 2t-6 \). Set \( v=0 \) to find when it is at rest: \( 2t-6=0 \) implies \( t=3 \text{ s} \). Substitute \( t=3 \text{ s} \) back into the displacement equation: \( x(3) = (3-3)^2 = 0 \text{ m} \).

Concept: Average speed is total distance over total time. Let total distance be \( D \). The time taken for each one-third distance is \( t_1 = \frac{D/3}{10} = \frac{D}{30} \), \( t_2 = \frac{D/3}{20} = \frac{D}{60} \), \( t_3 = \frac{D/3}{60} = \frac{D}{180} \). Total time \( T = t_1+t_2+t_3 = D \left( \frac{6+3+1}{180} \right) = \frac{D}{18} \). Average speed \( = \frac{D}{T} = \frac{D}{D/18} = 18 \text{ km/h} \).

Concept: Average speed for two equal distance segments is \( v_{avg} = \frac{2v_1 v_2}{v_1 + v_2} \). Given \( v_{avg} = 48 \text{ km/h} \), \( v_1 = 40 \text{ km/h} \), and \( v_2 = v \). Substitute these values: \( 48 = \frac{2 \times 40 \times v}{40 + v} \). Simplify and solve for \( v \): \( 48(40+v) = 80v \) leads to \( 32v = 1920 \), so \( v = 60 \text{ km/h} \).