It’s a hot summer day, and a conservation biologist in India is observing the behavior of the local wildlife. The scientist notices that an elephant begins moving its trunk in a way to signal to the other elephants that the herd should find water. The elephant herd is initially 5 km away from the nearest lake, and they are moving towards the lake at a constant speed of 8 km/hr. Half an hour later, the lake is in sight, and the Elephant Scout gazes longingly at the lake and decides to speed up; the elephant accelerates at a constant rate and reaches the lake 5 minutes later. The elephant takes a nice, long, and refreshing drink.
(a) What was the acceleration of the elephant during the final mad dash to the lake?
(b) If the elephant had begun accelerating 5 minutes earlier, how much time it would have saved?
(a) Split the distance traveled in two stages: first a stage where the elephant moves with constant velocity, and then a stage where the elephant moves with constant acceleration. Once you find the distance traveled in the first stage, use that distance in the equation of motion of the second stage.
(b) Same hint as before. Remember to use a different value of time.
(a) The equation that describes the motion of the elephant in the first stage (constant speed) is
\begin{equation*}
\vec{x}_{f1} = \vec{v} t_1\,
\end{equation*}
where using explicitly the given numerical values for each variable, we have:
\begin{equation*}
x_{f1} = 4 \, \text{km},
\end{equation*}
After the first stage, the elephant will continue with a constant acceleration motion for \(5\) minutes. The equation that describes its motion in this second stage is:
\begin{equation*}
\vec{x}_{f2} = \vec{x}_{i2} + \vec{v}_i t_2 + \frac{1}{2} \vec{a} t_2^2,
\end{equation*}
where \(x_{f2}\) is the final position of the whole motion ( \( 5 \, \text{km} \) ) and \(x_{i2}\) is the final position of the first stage ( \( 4 \, \text{km} \) ).
Solving for \(a\) we get:
\begin{equation*}
a = \frac{2(x_{f2} – x_{i2} – v_i t_2)}{t_2^2}.
\end{equation*}
The final result is:
\begin{equation*}
a=96 \, \frac{\text{km}}{\text{h}^2}.
\end{equation*}
(b) The first stage will be 25 minutes. Proceeding exactly as before, we get the final’s first stage position:
\begin{equation*}
x_{f1} = \frac{10}{3} \, \text{km} \approx 3.33 \, \text{km}.
\end{equation*}
Now, using the same equation of motion for the second stage where now the acceleration is the same as found in the previous numeral and solving for \(t\), which is the unknown variable, we get:
\begin{equation*}
0 = \left( \frac{1}{2} a \right) t_2^2 + (v_i) t_2 + (x_{i2} – x_{f2} ),
\end{equation*}
Taking the positive root, we have:
\begin{equation*}
t_{2} = \frac{-1 + \sqrt{6}}{12} \, \text{h} \approx 0.12 \, \text{h} \approx 7.25 \, \text{min}.
\end{equation*}
For a more detailed explanation of any of these steps, click on “Detailed Solution”.
Figure 1: We place our coordinate system on the ground in front of the elephant as it starts its journey.
The problem states that the elephant will travel a total distance of 5km towards the lake in two stages: in the first stage, it will move with a constant speed for 30 minutes and then, in the second stage, it will move with constant acceleration for 5 minutes. Using this information, we must find the acceleration of the elephant in the second stage.
The equation that describes the motion of the elephant in the first stage (constant speed) is
\begin{equation}
x_{f1} \;\hat{\textbf{i}} = x_{i1}\; \hat{\textbf{i}} + v t_1\; \hat{\textbf{i}},
\label{elefanteXCte}
\end{equation}
where \(x_{i1}\) is the initial position (which is zero with respect to our coordinate system), \(x_{f1}\) is the final position of this stage, \(v\;\hat{\textbf{i}}\) is the elephant’s velocity (which is positive according to our coordinate system) and \(t_1\) is the time expended on this first stage (\(\hat{\textbf{i}}\) is the unitary vector in the X axis).
After the first stage, the elephant will continue with a constant acceleration motion for \(5\) minutes. The equation that describes its motion in this second stage is
\begin{equation}
x_{f2} \hat{\textbf{i}} = x_{i2} \hat{\textbf{i}} + v_i t_2 \hat{\textbf{i}} + \frac{1}{2} a t_2^2 \hat{\textbf{i}},
\label{elefanteXAcelerado}
\end{equation}
where \(x_{f2}\) is the final position of the whole motion, which is the lake’s position; \(x_{i2}\) is the initial position of the second stage, which is the same as the final position of the first stage \(x_{f1}\) (since the second stage begins where the first stage ends); \(v_i\) is the initial speed of the second stage (which is the same as the constant speed of the first stage); \(a\) is the acceleration that we must solve for; and \(t_2\) is the time expended on this second stage. Notice that in order to find \(a\) from equation \eqref{elefanteXAcelerado}, we must first find \(x_{i2}\) (which is the same as \(x_{f1}\)). To find \(x_{f1}\) we must first use equation \eqref{elefanteXCte}.
Both \eqref{elefanteXCte} and \eqref{elefanteXAcelerado} are written in vector form. It follows that if two vectors are equal, their magnitudes (norms) are also equal. Thus, equations \eqref{elefanteXCte} and \eqref{elefanteXAcelerado} are valid with just their norms, that is, without the unitary vector \(\hat{\textbf{i}}\) in each term. Using this in equation \eqref{elefanteXCte},
we obtain
\begin{equation}
\label{primera}
x_{f1} = x_{i1} + v t_1.
\end{equation}
Using explicitly the given numerical values for each variable, we have:
\begin{equation}
x_{f1} = 0 + {\left( 8\,\frac{\text{km}}{\text{h}} \right)} \cdot {\left(\frac{1}{2} \, \text{h}\right)}= 4 \, \text{km},
\label{elefanteX1Resultado}
\end{equation}
which means that in the first stage the elephant traveled \(4\) kilometers.
Therefore, we know that in the second stage the elephant will have an initial position of 4 km with respect to the origin, thus \(x_{i2} = 4\) km. Using equation \eqref{elefanteXAcelerado} with just the norms of the vectors, we may find \(a\):
\begin{equation}
x_{f2} = x_{i2} + v_i t_2 + \frac{1}{2} a t_2^2.
\label{segunda}
\end{equation}
Let’s now leave the acceleration term in the right side of the equation and take the other terms to the left side of the equation
\begin{equation}
x_{f2} – x_{i2} – v_i t_2 = \frac{1}{2} a t_2^2.
\end{equation}
Multiplying and dividing both sides by \(2\) and \(t_2^2\) respectively, we obtain
\begin{equation}
\frac{2(x_{f2} – x_{i2} – v_i t_2)}{t_2^2} = a.
\label{elefanteADespejada}
\end{equation}
Finally, using the numerical values for the known variables, we can conclude that
\begin{eqnarray}
a = \frac{2\left({(5\, \text{km})} – {(4\, \text{km})} – {\left( 8\,\frac{\text{km}}{\text{h}} \right)} \cdot {\left(\frac{1}{12} \, \text{h}\right)} \right)}{{\left( \frac{1}{12} \,\text{h} \right)^2}}.
\label{elefanteAResultado1}
\end{eqnarray}
The final result is:
\begin{equation}
a=96 \, \frac{\text{km}}{\text{h}^2}
\label{elefanteAResultado}
\end{equation}
Since all this time we were dealing with the vector norms, the result above is the norm of the acceleration. Its direction will be in the positive X axis, so we can write the acceleration more precisely as 96 \( \text{km}\)/\(\text{h}^2 \;\hat{\textbf{i}} \).
(b) If the elephant begins to accelerate 5 minutes earlier, it means that the time expended on the first stage will be 25 minutes. Now we want to know what will be the time expended on the second stage with the acceleration obtained in \eqref{elefanteAResultado}.
We proceed in a similar fashion as before, with equation \eqref{primera} and using the following numerical values: \(x_{i1} = 0\), \(v=8\; \text{km}/\text{h}\) and \(t=25 \; \text{min} = \frac{5}{12}\; \text{h}\). We get:
\begin{equation}
x_{f1} = 0 + {\left( 8\,\frac{\text{km}}{\text{h}} \right)} \cdot {\left(\frac{5}{12} \, \text{h}\right)},
\end{equation}
and so we obtain
\begin{equation}
x_{f1} = \frac{10}{3} \, \text{km} \approx 3.33 \, \text{km}.
\label{elefanteX1ResultadoB}
\end{equation}
Hence, now we know that during the first stage, the elephant traveled \(10/3\) km. Thus, for the second stage, \(x_{i2} = 10/3\) km, \(x_{f2} = 5\) km, \(v_i = 8\) km/h and \(a = 96 \, \text{km}/{\text{h}^2}\). Our unknown quantity is now the time \(t_2\), which we will be able to solve using equation \eqref{segunda}.
Let’s move all the terms in equation \eqref{segunda} to the right side:
\begin{equation}
0 = \left( \frac{1}{2} a \right) t_2^2 + (v_i) t_2 + (x_{i2} – x_{f2} ),
\label{elefanteTOrganizada}
\end{equation}
where we have used a parenthesis to highlight the fact that this is a quadratic equation for time \(t_2\). If we replace the known variables for their numerical values, we obtain the following equation (which is clearly quadratic)
\begin{equation}
0={\left( 48\,\frac{\text{km}}{\text{h}^2} \right)} t_2^2 + {\left( 8\,\frac{\text{km}}{\text{h}} \right)} t_2 + {\left(-\frac{5}{3} \, \text{km} \right)}.
\label{elefanteTReemplazadoTodo}
\end{equation}
The quadratic equation has two solutions (often called roots). We’ll take the positive root, which is the one we are interested in finding out (negative times would correspond to moments before the elephant starts walking). Thus, we have
\begin{equation}
t_{2} = \frac{-1 + \sqrt{6}}{12} \, \text{h} \approx 0.12 \, \text{h} \approx 7.25 \, \text{min}.
\label{elefanteTSolucion}
\end{equation}
So, we know the elephant took 25 minutes in the first stage and 7.25 minutes on the second stage for a total time of 32.25 minutes, saving 2.75 minutes compared to the time expended traveling as described in point a).
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