# Lily's question at Yahoo! Answers regarding kinematics

#### MarkFL

Staff member
Here is the question:

AP physics/math question?

A mouse is moving with a constant acceleration along a straight ditch. It starts its stopwatch as it passes a fence post and notes that it takes it 10s to reach a pine tree 10m farther along the ditch. As it passes the pine tree, its speed is 1.2 m/s. How far was it from the fence post when it started from rest?
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello Lily,

Let's first draw a diagram:

We may use the kinematics relations for constant acceleration:

$$\displaystyle \tag{1}v_f=at+v_i$$

$$\displaystyle \tag{2}v_f=\sqrt{2ad+v_i^2}$$

where:

$$\displaystyle v_f=\text{final velocity}$$

$$\displaystyle a=\text{acceleration}$$

$$\displaystyle d=\text{displacement}$$

$$\displaystyle v_i=\text{initial velocity}$$

Now, when the mouse starts, he begins from rest so his velocity is $$\displaystyle 0\frac{\text{m}}{\text{s}}$$,when the mouse passes the fence post, his velocity is $v_0$, and when he passes the pine tree, his velocity is $$\displaystyle 1.2\frac{\text{m}}{\text{s}}$$.

We also know that average velocity is displacement per time:

$$\displaystyle \overline{v}=\frac{d}{t}$$

For the interval between the fence post and the pine tree, we then find:

$$\displaystyle \overline{v}=\frac{10\text{m}}{10\text{s}}=1\frac{\text{m}}{\text{s}}$$

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:

$$\displaystyle 5a+v_0=1$$

And using (2), we have:

$$\displaystyle \frac{36}{25}=20a+v_0^2$$

Solving both equations for $a$, we then obtain:

$$\displaystyle a=\frac{1-v_0}{5}=\frac{36-25v_0^2}{500}$$

Which leads to the quadratic in $v_0$:

$$\displaystyle 25v_0^2-100v_0+64=0$$

Factoring, we find:

$$\displaystyle \left(5v_0-4 \right)\left(5v_0-16 \right)=0$$

Since we must have $$\displaystyle v_0<\frac{6}{5}$$, we take the root:

$$\displaystyle v_0=\frac{4}{5}\frac{\text{m}}{\text{s}}$$

From this, we find from either of our two equations above:

$$\displaystyle a=\frac{1}{25}\frac{\text{m}}{\text{s}^2}$$

Now, on the interval between the start and the fence post, using (2) we find:

$$\displaystyle x=\frac{v_0^2}{2a}=\frac{ \left(\dfrac{4}{5}\dfrac{ \text{m}}{ \text{s}} \right)^2}{2\left(\dfrac{1}{25}\dfrac{ \text{m}}{ \text{s}^2} \right)}=8 \text{ m}$$