Vectors

Let $v_m$ be the velocity at $C$, the midpoint of $A$ and $B$.
Let distance $AB = l$, so distance $AC = \frac{l}{2}$ and $CB = \frac{l}{2}$.
For part (a), using the relation $v^2 = u^2 + 2as$:
For path $AC$: $v_m^2 = u^2 + 2a\left(\frac{l}{2}\right) = u^2 + al$ ... (1)
For path $CB$: $v^2 = v_m^2 + 2a\left(\frac{l}{2}\right) = v_m^2 + al$ ... (2)
Subtracting (1) from (2): $v^2 - v_m^2 = v_m^2 - u^2$
$2v_m^2 = u^2 + v^2$
$v_m = \sqrt{\frac{u^2 + v^2}{2}}$
For part (b), let time taken for $CB$ be $t$, then time taken for $AC$ is $2t$.
Using average velocity relation $s = \left(\frac{u_1 + u_2}{2}\right)t$:
For $AC$: $\frac{l}{2} = \left(\frac{u + v_m}{2}\right)2t$ ... (4)
For $CB$: $\frac{l}{2} = \left(\frac{v_m + v}{2}\right)t$ ... (5)
Equating equations (4) and (5):
$\left(\frac{u + v_m}{2}\right)2t = \left(\frac{v_m + v}{2}\right)t$
$2u + 2v_m = \frac{v_m + v}{2}$
$4u + 4v_m = v_m + v$
$v_m = \frac{v - 4u}{3}$ (or expanding via substitution of $v_m = \sqrt{\frac{u^2 + v^2}{2}}$ as shown in text: $v^2 - 8uv + 7u^2 = 0$)
Solving the quadratic equation $(v - u)(v - 7u) = 0$.
Since the motion is accelerated, $v - u \neq 0$, therefore $v = 7u$.
Thus, the ratio $\frac{v}{u} = 7$.
Step Solution:
Step 1: Apply the equation $v^2 = u^2 + 2as$ for segments $AC$ and $CB$ to derive the velocity at the midpoint $v_m = \sqrt{\frac{u^2 + v^2}{2}}$.
Step 2: Express the displacement over intervals $AC$ and $CB$ using average velocity multiplied by time intervals $2t$ and $t$ respectively.
Step 3: Equate the expressions for distance $AC$ and $CB$ to establish a relation between velocities $u$, $v$, and $v_m$.
Step 4: Substitute $v_m = \sqrt{\frac{u^2 + v^2}{2}}$ into the derived time-displacement relation to form a quadratic equation in terms of $v$ and $u$.
Step 5: Factor the quadratic equation into $(v - u)(v - 7u) = 0$ and solve for the ratio $\frac{v}{u}$, yielding $7$.