A question on uniformly accelerated motion In my book on uniformly accelerated motion there is the following exercise:
A body, initially stationary, moves with constant acceleration and travels $20$ m in $5$ s.
What space does it travel in $20$ s?
A:$1600\, m.$
B:$320\, m.$
C:$400\, m.$
D: cannot be answered because the acceleration
of the motion.

The book's answer is B.
My attempts: If I apply the formula $$x=x_0+\left(\frac{v+v_0}{2}\right)t \tag 1$$
I obtain $v=\frac{20 m}{5 s}= 4\frac ms$ and $\Delta x=x-x_0=\left(\frac{v+v_0}{2}\right)t=40 m$ and it not possible. If I do a proportion
$$20 m : 5 s = x: 20 s \implies x=80 \,m$$
but this not is the solution. How should one proceed because I don't have any additional information? The question is taken from a second-year high school physics textbook.
 A: Using $x=\frac{1}{2}at^2$ and given $x_1=20,t_1=5,t_2=20$, we can say
$x_1=\frac{1}{2}at_1^2$, and so $a=\frac{2x_1}{t_1^2}$
Then $x_2=\frac{1}{2}\frac{2x_1}{t_1^2}t_2^2=x_1\frac{t_2^2}{t_1^2}=x_1\left(\frac{t_2}{t_1}\right)^2=20\cdot16=320$.
A: We start from rest and move with constant acceleration $a$ which gives us $$\frac{dv}{dt}=a\implies v = at + v_0\implies v=at$$ because our initial speed is zero.
We now have $$\frac{dx}{dt}=at\implies x=\frac{1}{2}at^2+x_0$$ and since we are allowed to choose our starting position we take $x_0=0$ to get $$x=\frac{1}{2}at^2$$
We are told we travel $20$m in $5$ seconds which gives us $$\frac{1}{2}a(5)^2=20\implies a=\frac{8}{5}$$ meters per second squared. Plugging $a$ back into our equation of motion we get $$x=\frac{4}{5}t^2$$
Therefore, $x(20)=\frac{4}{5}(400)=320$
A: Peraphs I have understood.
$$x-x_0=\left(\frac{v+v_0}{2}\right)t \iff \frac{2(x-x_0)}{t}=v \tag 2$$
being $v_0=0$.
Then
$$\frac{2\cdot 20m}{5}=v=8 m/s$$
$$a=\frac{8m/s}{5s}=1.6 m/s^2$$
$$v=at=1.6\cdot 20=32 m/s$$
$$\Delta x=\frac 12 a t^2=.5\cdot 1.6 \cdot 400=320 m.$$
