$$B$$ catches $$C$$ in time $$t$$ then $$t = \frac{d}{{u - 10}}.$$
Separation by this time has increased by $$'d'$$ between $$A$$ and $$C$$ hence
$$\left( {10 - 5} \right) \times \frac{d}{{\left( {u - 10} \right)}} = d,u = 15\,m/s$$

The tangential acceleration is given by
$${a_T} = r\alpha \,......\left( {\text{i}} \right)$$
From 2nd equation of motion for rotational motion,
$${\omega ^2} = \omega _0^2 + 2\alpha \theta $$
Here, initial angular velocity, $${\omega _0} = 0,$$
Final angular velocity,
$$\eqalign{ & \omega = \frac{v}{I} = \frac{{80}}{{\frac{{20}}{\pi }}} \cr & = 4\pi \,rad/s \cr & \theta = 2 \times 2\pi \,rad \cr} $$
So, angular acceleration
$$\alpha = \frac{{{\omega ^2}}}{{2\theta }} = \frac{{{{\left( {4\pi } \right)}^2}}}{{2 \times \left( {2 \times 2\pi } \right)}} = \frac{{16{\pi ^2}}}{{8\pi }} = 2\pi $$
Hence, from Eq. (i), we have
$${a_T} = r\alpha = \frac{{20}}{\pi } \times 2\pi = 40\;m/{s^2}$$
Alternative
Initial velocity, $$u = 0$$
Final velocity, $$v = 80\,m/s$$
Radius of circle $$r = \left( {\frac{{20}}{{11}}} \right)m$$
Distance travelled, $$S = 2 \times \left( {2\pi r} \right) = 2 \times \left( {2\pi \times \frac{{20}}{\pi }} \right)$$
$$ = 80\,m$$
Now, by applying third equation of motion,
$$\eqalign{ & {v^2} = {u^2} + 2as \cr & {\left( {80} \right)^2} = 0 + 2 \times {a_f} \times 80 \cr & {a_f} = 40\,m/{s^2} \cr} $$

From displacement-time graph, it is clear that in equal intervals of time displacements are not equal infact, decreases and after $$40\,s$$  displacement constant i.e. the particle stops.

The area of the triangle
$$A = \frac{1}{2}\left| {\left( {\hat i + \hat j + \hat k} \right) \times \left( {3\hat i} \right)} \right| = \frac{3}{{\sqrt 2 }}.$$

$$R$$ is same for both $$\theta $$ and $$\left( {{{90}^ \circ } - \theta } \right).$$   If angle w.r.t. vertical is $${40^ \circ }$$ then w.r.t. horizontal direction it will be $${90^ \circ } - {40^ \circ } = {50^ \circ }.$$

$$\eqalign{ & {\text{Speed,}}\,\,u = 60 \times \frac{5}{{18}}\,m/s = \frac{{50}}{3}\,m/s \cr & d = 20\,m,\,\,u' = 120 \times \frac{5}{{18}} = \frac{{100}}{3}\,m/s \cr & {\text{Let declearation be a then }}{\left( 0 \right)^2} - {u^2} = - 2ad \cr & {\text{or, }}{\left( 0 \right)^2} - {u^2} = - 2ad\,\,\,\,{\text{or}},\,\,{u^2} = 2ad\,.....(1) \cr & {\text{and }}{\left( 0 \right)^2} - u{'^2} = - 2ad'\,\,\,\,\,{\text{or}},\,\,u{'^2} = 2ad'.....\left( 2 \right) \cr & (2)\;{\text{divided }}(1)\,{\text{gives,}} \cr & 4 = \frac{{d'}}{d} \Rightarrow d' = 4 \times 20 = 80\,m \cr} $$

$$\eqalign{ & {v_A} = \tan {30^ \circ }\,\,{\text{and}}\,\,{v_B} = \tan {60^ \circ } \cr & \therefore \frac{{{v_A}}}{{{v_b}}} = \frac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} = \frac{{\frac{1}{{\sqrt 3 }}}}{{\sqrt 3 }} = \frac{1}{3} \cr} $$

$$\eqalign{ & \frac{{dv}}{{dt}} = - 2.5\sqrt v \Rightarrow \frac{{dv}}{{\sqrt v }} = - 2.5\,dt \cr & {\text{Integratini g,}}\,\int\limits_{6.25}^0 {{v^{ - \frac{1}{2}}}dv} = - 2.5\int\limits_0^t {dt} \cr & \Rightarrow \left[ {\frac{{{v^{ + \frac{1}{2}}}}}{{\left( {\frac{1}{2}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t \cr & \Rightarrow - 2{\left( {6.25} \right)^{\frac{1}{2}}} = - 2.5t \Rightarrow t = 2\,\sec \cr} $$

Velocity of each car is given by
$$\eqalign{ & {V_P} = \frac{{d{x_P}\left( t \right)}}{{dt}} = a + 2bt \cr & {\text{and}}\,{V_Q} = \frac{{d{x_Q}\left( t \right)}}{{dt}} = f - 2t \cr} $$
It is given that $${V_P} = {V_Q}$$
$$\eqalign{ & \Rightarrow a + 2bt = f - 2t \cr & \Rightarrow t = \frac{{f - a}}{{2\left( {b + 1} \right)}} \cr} $$

Relative speed of each train with respect to each other be, $$v = 10 + 15 = 25\,m/s$$
Here distance covered by each train = sum of their lengths $$= 50 + 50 = 100\,m$$
$$\therefore {\text{Required time}} = \frac{{100}}{{25}} = 4\,\sec $$