Mechanics Question I do not understand

Question

A particle starts from rest and moves in a straight line with constant acceleration. In a certain 4 seconds of its motion it travels 12 m and in the next 5 seconds it travels 30m.

The acceleration of the particle is? The velocity of the particle at the start of the timing is? The distance it had travelled before timing started was?

Apparently, the velocity of the particle at the start of the timing is not 0 m/s

Answer 1

Use $s=ut+\frac{1}{2}at^2$.We are told that $u=0$.
Let the timing start at time $t$ and at distance $s$.

$s=\frac{1}{2}at^2$
$s+12=\frac{1}{2}a(t+4)^2$
$s+42=\frac{1}{2}a(t+9)^2$

Solving these gives $a=\frac{2}{3}$, $t=\frac{5}{2}$ and $s=\frac{25}{12}$.

For the velocity, use $v^2=u^2+2as$.

Answer 2

Let's observe picture bellow.We can write next equations:

$x=\frac{1}{2}at_x^{2}$ ; $v_A=at_x$; $v_B=v_A+at_1$; $s_1=v_At_1+\frac{1}{2}at_1^{2}$; $s_2=v_Bt_2+\frac{1}{2}at_2^{2}$

$s_1=4at_x+8a \Rightarrow 12=4at_x+8a \Rightarrow at_x=3-2a$

$s_2=(at_x+4a)5+\frac{25}{2}a \Rightarrow 6=at_x+4a+\frac{5}{2}a \Rightarrow 12=2at_x+13a$ , Since that :

$12=2(3-2a)+13a \Rightarrow a=\frac{2}{3}\frac{m}{s^2} \Rightarrow t_x=\frac{5}{2}sec$, $ v_A=at_x \Rightarrow v_A=\frac{5}{3} \frac{m}{s}$ and $x=\frac{1}{2}at_x^{2} \Rightarrow x=\frac{25}{12} m$