# Question 1ad1b

Jul 21, 2017

$4 {v}_{1}$

#### Explanation:

Let use setup some variables to describe the problem:

 { (d, "total distance (or displacement) travelled"), (t_1, "time taken for first part of the journey"), (t_2, "time taken for second part of the journey"), (v_2, "velocity for second part of the journey") :} 

For the first part of the Journey:

 { (s, = 1/3d), (u, = v_1), (t, = t_1) :} 

Applying $s = u t$ we have:

$\frac{1}{3} d = {v}_{1} {t}_{1}$ ..... [A]

For the second part of the Journey:

 { (s, = 2/3d), (u, = v_2), (t, = t_2) :} 

Applying $s = u t$ we have:

$\frac{2}{3} d = {v}_{2} {t}_{2}$ ..... [B]

For the overall Journey:

 { (s, = d), (u, = 2v_1), (t, = t_1+t_2) :} #

Applying $s = u t$ we have:

$d = 2 {v}_{1} \left({t}_{1} + {t}_{2}\right)$ ..... [C]

From [A] and [C] we have:

$d = 3 {v}_{1} {t}_{1} = 2 {v}_{1} \left({t}_{1} + {t}_{2}\right)$

$\therefore 3 {t}_{1} = 2 {t}_{1} + 2 {t}_{2}$

$\therefore {t}_{1} = 2 {t}_{2}$

So then from [C] we get:

$d = 2 {v}_{1} \left(2 {t}_{2} + {t}_{2}\right) = 6 {v}_{1} {t}_{2}$

Inserting into [B] we then have:

$\frac{2}{3} \cdot 6 {v}_{1} {t}_{2} = {v}_{2} {t}_{2}$

$\therefore {v}_{2} = \frac{2}{3} \cdot 6 {v}_{1} = 4 {v}_{1}$