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Lily's question at Yahoo! Answers regarding kinematics
- Thread starter MarkFL
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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:
\(\displaystyle x=\frac{v_0^2}{2a}=\frac{ \left(\dfrac{4}{5}\dfrac{ \text{m}}{ \text{s}} \right)^2}{2\left(\dfrac{1}{25}\dfrac{ \text{m}}{ \text{s}^2} \right)}=8 \text{ m}\)
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:
\(\displaystyle x=\frac{v_0^2}{2a}=\frac{ \left(\dfrac{4}{5}\dfrac{ \text{m}}{ \text{s}} \right)^2}{2\left(\dfrac{1}{25}\dfrac{ \text{m}}{ \text{s}^2} \right)}=8 \text{ m}\)