Solved Problem on One-dimensional Motion
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A fox, chased by a greyhound, has 63 front jumps on the dog. The greyhound gives 3 jumps while the fox gives 4 jumps, but 6 jumps from the greyhound are worth 10 jumps of the fox. How many jumps the greyhound should give to reach the fox?
Problem data:
- Advantage of the fox on the greyhound: 63 fox jumps;
- Ratio between the greyhound and fox jumps: \( \dfrac{6\;\text{greyhound jumps}}{10\;\text{fox jumps}} \);
- Greyhound jumps per unit of time: 3 greyhound jumps/ time unit;
- Fox jumps per unit of time: 4 fox pulls/time unit.
As 6 greyhound jumps (gj) worth 10 fox jumps (fj), this represents a conversion factor (Figure 1)
\[
\begin{gather}
6\;\text{pg}=10\;\text{pr}
\end{gather}
\]
Note: This is a conversion factor as in the case, we say 1 foot = 0.3046 meters
\[
\begin{gather}
1\;\text{ft}=0.3048\;\text{m}
\end{gather}
\]
But in this case, the measurement units are greyhound and fox.
At the same time interval, the greyhound gives 3 jumps the fox gives 4 jumps (Figure 2). Let's take this time interval, which is the same for both, as the unit of time (ut). The speed of the greyhound will be
\[
\begin{gather}
v_{g}=3\;\text{pg/ut}
\end{gather}
\]
the speed of the fox will be
\[
\begin{gather}
v_{r}=4\;\text{pr/ut}
\end{gather}
\]
Note: These two values represent speeds in different units, as if we had two bodies with
speeds, for example
\[
\begin{gather}
v_{1}=5\;\text{ft/s}
\end{gather}
\]
\[
\begin{gather}
v_{2}=8\;\text{m/s}
\end{gather}
\]
The fox has 63 jumps (of the fox) of advantage, this is the initial position of the fox.
\[
\begin{gather}
S_{0r}=63\;\text{pr}
\end{gather}
\]
As we want the number of the greyhound, we will convert the fox speed and the initial position of the fox
to the jumps of the greyhound.
\[
\begin{gather}
S_{0r}=63\;\cancel{\text{pr}}\times \frac{6\;\text{pg}}{10\;\cancel{\text{pr}}}=\frac{378}{10}\;\text{pg}\\[10pt]
v_{r}=4\;\frac{\cancel{\text{pr}}}{\text{ut}}\times \frac{6\;\text{pg}}{10\;\cancel{\text{pr}}}=\frac{24}{10}\;\text{pg/ut}
\end{gather}
\]
So we have the following diagram of the problem (Figure 3)
As the two animals have constant speeds, the equation for displacement as a function of time with constant speed is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_{0}+vt}
\end{gather}
\]
We choose the greyhound position at the origin, S0g=0, applying the expression
above for the two animals.
\[
\begin{gather}
S_{g}=S_{0g}+v_{g}t\\[5pt]
S_{g}=3t
\end{gather}
\]
\[
\begin{gather}
S_{r}=S_{0r}+v_{r}t\\[5pt]
S_{r}=\frac{378}{10}+\frac{24}{10}t
\end{gather}
\]
Equating the two expressions, we have the instant in which the greyhound reaches the fox.
\[
\begin{gather}
S_{g}=S_{r}\\[5pt]
3t=\frac{378}{10}+\frac{24}{10}t\\[5pt]
3t-\frac{24}{10}t=\frac{378}{10}
\end{gather}
\]
multiplying both sides of the expression by 10
\[
\begin{gather}
\qquad\qquad\quad 3t-\frac{24}{10}t=\frac{378}{10} \qquad \times{(10)}\\[5pt]
10\times 3t-\cancel{10}\times \frac{24}{\cancel{10}}t=\cancel{10}\times \frac{378}{\cancel{10}}\\[5pt]
30t-24t=378\\[5pt]
6t=378\\[5pt]
t=\frac{378}{6}\\[5pt]
t=63\;\text{ut}
\end{gather}
\]
substituindo este valor na expressão do galgo
\[
\begin{gather}
S_{g}=3\times 63
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S_{g}=189\;\text{pg}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .