\({Screech!}\) A bus driver sees a reckless driver going the wrong way, and SLAMS on the bus brakes. The reckless driver is headed directly toward the bus full of passengers, and is speeding along the suburban road at 14 m/s (while the bus travels at a mere 8 m/s ). If the brake system on the bus slows down the speed of the bus with a constant acceleration of \(10 \,\text{m}/\text{s}^2\), and the reckless driver slams on the car brakes (resulting in a constant acceleration of \(5 \,\text{m}/\text{s}^2\)), the bus and the car fortunately avoid a collision and come to a stop 2 meters apart. The bus driver reports the incident to their supervisor, and the incident is then reported to the company’s automobile insurance and to the police. A police officer investigating the incident is tasked with calculating the distance between the car and the bus at the moment when the bus driver began to press on the brakes; this is necessary to determine whether any wrongdoing occurred and to determine who should be held liable. Based on the information above, calculate the distance between the car and the bus at the moment when the drivers both began slamming on their brakes.
Place a coordinate system in one of the two objects such that at least one of the initial positions is zero. Notice that at the end, their final positions are separated by 2 meters. Also, for both objects, you must use the equations of motion of an object moving with constant acceleration. Finally, notice that one acceleration will be positive and the other one will be negative.
Placing a coordinate system with the origin at the initial location of the bus, and with the X axis pointing along its direction of motion, the bus is initially at \(x_{iB}=0\) meters and moves in the positive X direction. According to the system, the car initially moves in the negative X direction, and we can call \(x_{iC}\) the initial position (which is unknown). Indeed, what we want to find is precisely \(x_{iC}\) which corresponds to the initial distance of the car respect to the bus.
If we use \(x_{fB}\) to refer to the bus’ final position and \(x_{fC}\) to the car’s final position, then it is clear that
\begin{equation*}
x_{fC}=x_{fB}+2 \, \text{m}.
\end{equation*}
Since we know the final speed, the initial speed, the acceleration, and initial position of the bus and the car, it will be very convenient to use the following equation that does not involve time:
\begin{equation*}
v_{f}^2 = v_{i}^2 + 2 a (x_{f} – x_{i}),
\end{equation*}
where solving for \(x_f\) we get:
\begin{equation*}
\frac{v_{f}^2 – v_{i}^2}{2 a} + x_{i} = x_{f}.
\end{equation*}
Replacing in the first equation for both the car and the bus, and then solving for \(x_{iC}\) we get:
\begin{equation*}
x_{iC}=\frac{v_{fB}^2 – v_{iB}^2}{2 a_B} + x_{iB} – \frac{v_{fC}^2 – v_{iC}^2}{2 a_C} + 2\;\text{m},
\end{equation*}
where using numerical values we obtained:
\begin{equation*}
x_{iC} = 24.8 \; \text{m}.
\end{equation*}
That is, the car was initially 24.8 meters apart from the bus.
For a more detailed explanation of any of these steps, click on “Detailed Solution”.
Let’s start by placing a coordinate system with the origin at the initial location of the bus, and with the X axis pointing along its direction of motion (see figure 1).
Figure 1: We place the coordinate system on the ground in front of the bus just as it starts to push the brakes.
According to this system, the bus is initially at \(x_{iB}=0\) meters and moves in the positive X direction. Also, according to the system, the car initially moves in the negative X direction, and we can call \(x_{iC}\) the initial position (which is unknown). Indeed, what we want to find is precisely \(x_{iC}\), which corresponds to the initial distance between the car respect to the bus (we used that \(x_{iB}=0\)).
If we use \(x_{fB}\) to refer to the bus’ final position and \(x_{fC}\) to the car’s final position, then it is clear from our coordinate system that
\begin{equation}
\label{BusCarro_posiciones}
x_{fC}=x_{fB}+2 \, \text{m},
\end{equation}
since in the end, the car and the bus are two meters apart, as we can see in figure 2.
Figure 2: Final positions for the bus and the car. When they stop they are 2 meters apart.
Since we know the final speed, the initial speed, the acceleration, and the bus’ initial position, it will be very convenient to use the following equation that does not involve time (since we do not know it):
\begin{equation}
v_{fB}^2 = v_{iB}^2 + 2 a_B (x_{fB} – x_{iB}),
\label{BusCarro_ecuacionBus}
\end{equation}
where \(v_{fB}\) is the bus’ final speed (it is zero, since at the end it stops), \(v_{iB}\) is the bus’ initial position, \(a_B\) is the acceleration and where \(x_{fB} – x_{iB}\) is the difference between the bus’ final position and the initial one (if we had used \(\vec{x_f}=\vec{x_i}+\vec{v_i}t+1/2at^2\), then we would have had to find the time first). It is important to stress that the sign of the acceleration in this equation depends on the direction of the acceleration (the acceleration is a vector, and so it has a direction).
The direction of the bus’ acceleration is negative. This is because the final velocity is zero and the initial velocity is positive. This means that the term \((\vec{v_f}-\vec{v_i})/t\), which corresponds to the acceleration, will be negative.
Like in the bus’ case, it will be convenient to use the following equation for the car:
\begin{equation}
v_{fC}^2 = v_{fC}^2 + 2 a_B (x_{fC} – x_{iC}),
\label{BusCarro_ecuacionCarro}
\end{equation}
where \(v_{fC}\) is the car’s final speed (which is zero, since at the end it would have stopped), \(v_{fC}\) is the initial speed, \(a_B\) is the acceleration and \(x_{fC} – x_{iC}\) is the difference between the final and the initial speed.
Since initially the car’s velocity is negative (it is moving in the negative X direction) and it is zero at the end, then we know it is increasing in X, which means that the acceleration must be positive in X. That is, for the car, \((\vec{v_f}-\vec{v_i})/t\) is positive (be careful: as the car brakes, the speed is decreasing but the velocity is increasing due to the coordinate system that we chose).
Hence, both from equation \eqref{BusCarro_ecuacionBus} and from equation \eqref{BusCarro_ecuacionCarro} we need to solve for the final position. Take equation \eqref{BusCarro_ecuacionBus} and move the initial velocity term to the left side:
\begin{equation}
v_{fB}^2 – v_{iB}^2 = 2 a_B (x_{fB} – x_{iB}).
\end{equation}
Divide now both sides by \(2 a_B\):
\begin{equation}
\frac{v_{fB}^2 – v_{iB}^2}{2 a_B} = x_{fB} – x_{iB}.
\end{equation}
Finally, move \(x_{iB}\) to the left side:
\begin{equation}
\frac{v_{fB}^2 – v_{iB}^2}{2 a_B} + x_{iB} = x_{fB}.
\label{BusCarro_posicionFinalBus}
\end{equation}
If we follow the same process for the car’s equation, we find
\begin{equation}
\frac{v_{fC}^2 – v_{iC}^2}{2 a_C} + x_{iC} = x_{fC}.
\label{BusCarro_posicionFinalCarro}
\end{equation}
If we use equation \eqref{BusCarro_posicionFinalBus} and equation \eqref{BusCarro_posicionFinalCarro} in equation \eqref{BusCarro_posiciones}, we get:
\begin{equation}
\; {\frac{v_{fC}^2 – v_{iC}^2}{2 a_C} + x_{iC}} \; ={\frac{v_{fB}^2 – v_{iB}^2}{2 a_B} + x_{iB}} \; + \;
2\;\text{m}.
\label{BusCarro_posicionesConVariables}
\end{equation}
We can find \(x_{iC}\) from this equation if we move the term \(\frac{v_{fC}^2 – v_{iC}^2}{2 a_C}\) to the other side:
\begin{equation}
x_{iC}=\frac{v_{fB}^2 – v_{iB}^2}{2 a_B} + x_{iB} – \frac{v_{fC}^2 – v_{iC}^2}{2 a_C} + 2\;\text{m}.
\end{equation}
Finally, let’s insert the numerical values for the different quantities:
\begin{equation}
\frac{ ({0 \; \text{m/s}})^2 – ({8 \; \text{m/s}})^2}{2 {(-10 \; \text{m/s}^2)} } + {(0 \; \text{m})} – \frac{ ({0 \; \text{m/s}})^2 – ({-14 \; \text{m/s}})^2}{2 {(5 \; \text{m/s}^2)} } + {(2 \; \text{m})} = x_{iC},
\label{BusCarro_reemplazandoValores}
\end{equation}
where we used that the car’s acceleration is positive, the bus’ is negative and car’s initial velocity is negative. The final result is
\begin{equation}
x_{iC} = 24.8 \; \text{m}.
\end{equation}
That is, the car was initially 24.8 meters apart from the bus.
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