Master Manipulation (Kinematics)

The velocity-time graph of Drake will be as shown. The slope of the velocity-time graph gives the acceleration and also the area under it gives the displacement.

The slope for $AB$ $\frac{v_{\text{max}} -v}{T_1} = a \,\, \Rightarrow \,\, T_1 = \frac{v_{\text{max}} -v}{a}.$ The slope for $BC$ $\frac{v_{\text{max}}}{T_2} = a \,\, \Rightarrow \,\, T_2 = \frac{v_{\text{max}}}{a}.$ Area under $AB=L_1$ $L_1 = \frac{1}{2} \cdot (v+v_{\text{max}}) T_1 \,\, \Rightarrow \,\, L_1 = \frac{v_{\text{max}}^2 - v^2} {2a}.$ Area under $BC = L_2$ $L_2 = \frac{1}{2} \cdot v_{\text{max}} T_2 \,\, \Rightarrow \,\, L_2 = \frac{v_{\text{max}}^2}{2a}.$

We know that $\frac{L_2}{L_1}=k$ $k=\frac{v_{\text{max}}^2}{v_{\text{max}}^2 - v^2} \\ \Rightarrow v_{\text{max}} = \boxed{\sqrt{\frac{k}{k-1}}v}.$

We will use the well known equation from kinematics

$\boxed{v^2-u^2=2as}$

where u and v are initial and final velocities, a is a constant acceleration and s is the distance travelled.

Let $V$ be the maximum speed, which will occur after the first leg of the journey. Applying the boxed formula to each leg of the journey gives

$V^2-v^2=2aL_{1}\dots(1)$

$V^2=2aL_{2}$

Use the given relationship between the lengths to write this as

$V^2=k2aL_{1}$

Substitute for $2aL_{1}$ from equation (1) to get

$V^2=k(V^2-v^2)$

Now an easy rearrangement gives

$\boxed{V=\sqrt{\frac{k}{k-1}}v}$

Say that the final velocity which it reaches in the first journey be $v'$ .

So by Equation of Motion

$v'^{2} - v^2 = 2L_1a$ -------(1)

Again, In second journey, Final velocity = 0 and Initial Velocity = $v'$

Hence $0^{2} - v'^{2} = 2L_2(-a)$

or $v'^{2} = 2L_2(a)$

Hence Max velocity in this journey is $v'$ itself.

Hence we get,

$v' = \sqrt{2L_2a}$ -------(2)

Putting $v' = \sqrt{2L_2a}$ in equation 1 we get,

$2L_2a - v^2 = 2L_1a$

hence $v^2 = 2a(L_2 - L_1)$ ----(3)

So a= $\dfrac{v^2}{L_2-L_1}$

Now $\dfrac{L_2}{L_1} = k$

So $L_2 - L_1 = L_1(k-1)$

So we have got everything

Now put in all these to get $v'$ = $v \sqrt{\dfrac{k}{k-1}}$

My solution is the easiest solution ,apart from the $massive$ ones here ( which are no doubt good, but are a bit long).

$2*a*L1$ $=$ $v_2 ^2-v^2$ ... $(1)$ (Here $v_2$ is max velocity)

$2*a*L2$ $=$ $v^2$ ... $(2)$

Divide ( $2$ ) by ( $1$ ) ,cross multiply then factorize ( no its an easy factorization)and we are done.