Question 2.7: THE ACELA: THE PORSCHE OF AMERICAN TRAINS GOAL Find accelera...
THE ACELA: THE PORSCHE OF AMERICAN TRAINS
GOAL Find accelerations and displacements from a velocity vs. time graph.
PROBLEM The sleek high-speed electric train known as the Acela (pronounced “ah cel’ la”) is currently in service on the Washington-New York-Boston run. The Acela consists of two power cars and six coaches and can carry 304 passengers at speeds up to 170 \mathrm{mi} / \mathrm{h}. To negotiate curves comfortably at high speeds, the train carriages tilt as much as 6^{\circ} from the vertical, preventing passengers from being pushed to the side. A velocity vs. time graph for the Acela is shown in Figure 2.19a. (a) Describe the motion of the Acela. (b) Find the peak acceleration of the Acela in miles per hour per second ((\mathrm{mi} / \mathrm{h}) / \mathrm{s}) as the train speeds up from 45 \mathrm{mi} / \mathrm{h} to 170 \mathrm{mi} / \mathrm{h}. (c) Find the train’s displacement in miles between t=0 and t=200 \mathrm{~s}. (d) Find the average acceleration of the Acela and its displacement in miles in the interval from 200 \mathrm{~s} to 300 \mathrm{~s}. (The train has regenerative braking, which means that it feeds energy back into the utility lines each time it stops!) (e) Find the total displacement in the interval from 0 to 400 \mathrm{~s}. Note: Assume that all given quantities and estimates are good to two significant figures. (Estimates by different individuals may vary and result in slightly different answers.)
STRATEGY For part (a), remember that the slope of the tangent line at any point of the velocity vs. time graph gives the acceleration at that time. To find the peak acceleration in part (b), study the graph and locate the point at which the slope is steepest. In parts (c) through (e), estimating the area under the curve gives the displacement during a given period, with areas below the time axis, as in part (e), subtracted from the total. The average acceleration in part (d) can be obtained by substituting numbers taken from the graph into the definition of average acceleration, \bar{a}=\Delta v / \Delta t.
(a) Describe the motion.
From about −50 \mathrm{~s} to 50 \mathrm{~s}, the Acela cruises at a constant velocity in the +x-direction. Then the train accelerates in the +x-direction from 50 \mathrm{~s} to 200 \mathrm{~s}, reaching a top speed of about 170 \mathrm{mi} / \mathrm{h}, whereupon it brakes to rest at 350 \mathrm{~s} and reverses, steadily gaining speed in the -x-direction.
(b) Find the peak acceleration.
Calculate the slope of the steepest tangent line, which connects the points (50 \mathrm{~s}, 50 \mathrm{mi} / \mathrm{h}) and (100 \mathrm{~s}, 150 \mathrm{mi} / \mathrm{h} ) (the light blue line in Figure 2.19 \mathrm{~b} ) :
\begin{aligned}a &=\text { slope }=\frac{\Delta v}{\Delta t}=\frac{\left(15 \times 10^{2} \quad 50 \times 10^{1}\right) \mathrm{mi} / \mathrm{h}}{\left(10 \times 10^{2} \quad 50 \times 10^{1}\right) \mathrm{s}} \\&=2.0(\mathrm{mi} / \mathrm{h}) / \mathrm{s}\end{aligned}
(c) Find the displacement between 0 \mathrm{~s} and 200 \mathrm{~s}.
Using triangles and rectangles, approximate the area in Figure 2.19c:
\begin{aligned}\Delta x_{0 \rightarrow 200 \mathrm{~s}}=& \text { area }_{1}+\text { area }_{2}+\text { area }_{3}+\text { area }_{4}+\text { area }_{5} \\& \approx\left(5.0 \times 10^{1} \mathrm{mi} / \mathrm{h}\right)\left(5.0 \times 10^{1} \mathrm{~s}\right) \\&+\left(5.0 \times 10^{1} \mathrm{mi} / \mathrm{h}\right)\left(5.0 \times 10^{1} \mathrm{~s}\right) \\&+\left(1.6 \times 10^{2} \mathrm{mi} / \mathrm{h}\right)\left(1.0 \times 10^{2} \mathrm{~s}\right) \\&+\frac{1}{2}\left(5.0 \times 10^{1} \mathrm{~s}\right)\left(1.0 \times 10^{2} \mathrm{mi} / \mathrm{h}\right) \\&+{ }_{2}^{1}\left(1.0 \times 10^{2} \mathrm{~s}\right)\left(1.7 \times 10^{2} \mathrm{mi} / \mathrm{h} \quad 1.6 \times 10^{2} \mathrm{mi} / \mathrm{h}\right) \\=& 2.4 \times 10^{4}(\mathrm{mi} / \mathrm{h}) \mathrm{s}\end{aligned}
Convert units to miles by converting hours to seconds:
\Delta x_{0 \rightarrow 200 \mathrm{~s}} \approx 24 \times 10^{4} \frac{\mathrm{mi} \cdot \mathrm{s}}{\mathrm{h}}\left(\frac{1 \mathrm{~h}} {3600 \mathrm{~s}}\right)=6.7 \mathrm{mi}
(d) Find the average acceleration from 200 \mathrm{~s} to 300 \mathrm{~s}, and find the displacement.
The slope of the green line is the average acceleration from 200 \mathrm{~s} to 300 \mathrm{~s} (Fig. 2.19b):
\begin{aligned}& \bar{a}=\operatorname{slop} \mathrm{e}=\frac{\Delta v}{\Delta t}=\frac{\left(1.0 \times 10^{1} – 1.7 \times 10^{2}\right) \mathrm{mi} / \mathrm{h}}{1.0 \times 10^{2} \mathrm{~s}} \\& =-1.6(\mathrm{mi} / \mathrm{h}) / \mathrm{s} \end{aligned}
The displacement from 200 \mathrm{~s} to 300 \mathrm{~s} is equal to area _{6}, which is the area of a triangle plus the area of a very narrow rectangle beneath the triangle:
\begin{aligned} \Delta x_{200} \rightarrow 300 \mathrm{~s} &\approx{ }_{2}^{1}\left(1.0 \times 10^{2} \mathrm{~s}\right)\left(17 \times 10^{2} – 1.0 \times 10^{1}\right) \mathrm{mi} / \mathrm{h} \\& +\left(1.0 \times 10^{1} \mathrm{mi} / \mathrm{h}\right)\left(1.0 \times 10^{2} \mathrm{~s}\right) \\& =9.0 \times 10^{3}(\mathrm{mi} / \mathrm{h})(\mathrm{s})=2.5 \mathrm{mi}\end{aligned}
(e) Find the total displacement from 0 s to 400 \mathrm{~s}.
The total displacement is the sum of all the individual displacements. We still need to calculate the displacements for the time intervals from 300 \mathrm{~s} to 350 \mathrm{~s} and from 350 \mathrm{~s} to 400 \mathrm{~s}. The latter is negative because it’s below the time axis.
\begin{aligned}\Delta x_{300 \rightarrow 350 \mathrm{~s}} &\approx \frac{1}{2}\left(5.0 \times 10^{1} \mathrm{~s}\right)\left(10 \times 10^{1} \mathrm{mi} / \mathrm{h}\right) \\&=2.5 \times 10^{2}(\mathrm{mi} / \mathrm{h})(\mathrm{s}) \\\Delta x_{350 \rightarrow 400 \mathrm{~s}} &\approx \frac{1}{2}\left(5.0 \times 10^{1} \mathrm{~s}\right)\left(-5.0 \times 10^{1} \mathrm{mi} / \mathrm{h}\right) \\=& -1.3 \times 10^{3}(\mathrm{mi} / \mathrm{h})(\mathrm{s}) \end{aligned}
Find the total displacement by summing the parts:
\begin{aligned}\Delta x_{0 \rightarrow 400 \mathrm{~s}} \approx &\left(2.4 \times 10^{4}+9.0 \times 10^{3}+2.5 \times 10^{2}\right.\\&\left.-1.3 \times 10^{3}\right)(\mathrm{mi} / \mathrm{h})(\mathrm{s})=8.9 \mathrm{mi}\end{aligned}
REMARKS There are a number of ways to find the approximate area under a graph. Choice of technique is a personal preference.