MHB Lily's question at Yahoo Answers regarding kinematics

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The discussion revolves around a physics problem involving a mouse moving with constant acceleration. The mouse measures the time and distance to a pine tree after passing a fence post, noting its final speed. Using kinematic equations, the initial velocity and acceleration are calculated, leading to a quadratic equation for the initial velocity. The solution reveals the initial velocity as 0.8 m/s and the acceleration as 0.04 m/s². Ultimately, the distance from the fence post when the mouse started from rest is determined to be 8 meters.
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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.
 
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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:

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

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