Hello Lily,
Let's first draw a diagram:
View attachment 1359
We may use the kinematics relations for constant acceleration:
$$\tag{1}v_f=at+v_i$$
$$\tag{2}v_f=\sqrt{2ad+v_i^2}$$
where:
$$v_f=\text{final velocity}$$
$$a=\text{acceleration}$$
$$d=\text{displacement}$$
$$v_i=\text{initial velocity}$$
Now, when the mouse starts, he begins from rest so his velocity is $$0\frac{\text{m}}{\text{s}}$$,when the mouse passes the fence post, his velocity is $v_0$, and when he passes the pine tree, his velocity is $$1.2\frac{\text{m}}{\text{s}}$$.
We also know that average velocity is displacement per time:
$$\overline{v}=\frac{d}{t}$$
For the interval between the fence post and the pine tree, we then find:
$$\overline{v}=\frac{10\text{m}}{10\text{s}}=1\frac{\text{m}}{\text{s}}$$
Since his velocity is increasing linearly, we know his instantaneous velocity in this interval at 5 seconds is equal to the average velocity, and so using (1), we may state:
$$5a+v_0=1$$
And using (2), we have:
$$\frac{36}{25}=20a+v_0^2$$
Solving both equations for $a$, we then obtain:
$$a=\frac{1-v_0}{5}=\frac{36-25v_0^2}{500}$$
Which leads to the quadratic in $v_0$:
$$25v_0^2-100v_0+64=0$$
Factoring, we find:
$$\left(5v_0-4 \right)\left(5v_0-16 \right)=0$$
Since we must have $$v_0<\frac{6}{5}$$, we take the root:
$$v_0=\frac{4}{5}\frac{\text{m}}{\text{s}}$$
From this, we find from either of our two equations above:
$$a=\frac{1}{25}\frac{\text{m}}{\text{s}^2}$$
Now, on the interval between the start and the fence post, using (2) we find:
$$x=\frac{v_0^2}{2a}=\frac{ \left(\dfrac{4}{5}\dfrac{ \text{m}}{ \text{s}} \right)^2}{2\left(\dfrac{1}{25}\dfrac{ \text{m}}{ \text{s}^2} \right)}=8 \text{ m}$$
