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

Question

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 [latex]170 \mathrm{mi} / \mathrm{h}[/latex]. To negotiate curves comfortably at high speeds, the train carriages tilt as much as [latex]6^{\circ}[/latex] 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 [latex]((\mathrm{mi} / \mathrm{h}) / \mathrm{s})[/latex] as the train speeds up from [latex]45 \mathrm{mi} / \mathrm{h}[/latex] to [latex]170 \mathrm{mi} / \mathrm{h}[/latex]. (c) Find the train's displacement in miles between [latex]t=0[/latex] and [latex]t=200 \mathrm{~s}[/latex]. (d) Find the average acceleration of the Acela and its displacement in miles in the interval from [latex]200 \mathrm{~s}[/latex] to [latex]300 \mathrm{~s}[/latex]. (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 [latex]400 \mathrm{~s}[/latex]. 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, [latex]\bar{a}=\Delta v / \Delta t[/latex].

Accepted Answer

(a) Describe the motion.

From about [latex]−50 \mathrm{~s}[/latex] to [latex]50 \mathrm{~s}[/latex], the Acela cruises at a constant velocity in the [latex]+x[/latex]-direction. Then the train accelerates in the [latex]+x[/latex]-direction from [latex]50 \mathrm{~s}[/latex] to [latex]200 \mathrm{~s}[/latex], reaching a top speed of about [latex]170 \mathrm{mi} / \mathrm{h}[/latex], whereupon it brakes to rest at [latex]350 \mathrm{~s}[/latex] and reverses, steadily gaining speed in the [latex]-x[/latex]-direction.

(b) Find the peak acceleration.

Calculate the slope of the steepest tangent line, which connects the points [latex](50 \mathrm{~s}, 50 \mathrm{mi} / \mathrm{h})[/latex] and [latex](100 \mathrm{~s}[/latex], [latex]150 \mathrm{mi} / \mathrm{h}[/latex] ) (the light blue line in Figure [latex]2.19 \mathrm{~b}[/latex] ) :

[latex]\begin{aligned}a &= ext { slope }=\frac{\Delta v}{\Delta t}=\frac{\left(15 imes 10^{2} \quad 50 imes 10^{1} ight) \mathrm{mi} / \mathrm{h}}{\left(10 imes 10^{2} \quad 50 imes 10^{1} ight) \mathrm{s}} \&=2.0(\mathrm{mi} / \mathrm{h}) / \mathrm{s}\end{aligned}[/latex]

(c) Find the displacement between [latex]0 \mathrm{~s}[/latex] and [latex]200 \mathrm{~s}[/latex].

Using triangles and rectangles, approximate the area in Figure 2.19c:

[latex]\begin{aligned}\Delta x_{0 ightarrow 200 \mathrm{~s}}=& ext { area }_{1}+ ext { area }_{2}+ ext { area }_{3}+ ext { area }_{4}+ ext { area }_{5} \& \approx\left(5.0 imes 10^{1} \mathrm{mi} / \mathrm{h} ight)\left(5.0 imes 10^{1} \mathrm{~s} ight) \&+\left(5.0 imes 10^{1} \mathrm{mi} / \mathrm{h} ight)\left(5.0 imes 10^{1} \mathrm{~s} ight) \&+\left(1.6 imes 10^{2} \mathrm{mi} / \mathrm{h} ight)\left(1.0 imes 10^{2} \mathrm{~s} ight) \&+\frac{1}{2}\left(5.0 imes 10^{1} \mathrm{~s} ight)\left(1.0 imes 10^{2} \mathrm{mi} / \mathrm{h} ight) \&+{ }_{2}^{1}\left(1.0 imes 10^{2} \mathrm{~s} ight)\left(1.7 imes 10^{2} \mathrm{mi} / \mathrm{h} \quad 1.6 imes 10^{2} \mathrm{mi} / \mathrm{h} ight) \=& 2.4 imes 10^{4}(\mathrm{mi} / \mathrm{h}) \mathrm{s}\end{aligned}[/latex]

Convert units to miles by converting hours to seconds:

[latex]\Delta x_{0 ightarrow 200 \mathrm{~s}} \approx 24 imes 10^{4} \frac{\mathrm{mi} \cdot \mathrm{s}}{\mathrm{h}}\left(\frac{1 \mathrm{~h}} {3600 \mathrm{~s}} ight)=6.7 \mathrm{mi}[/latex]

(d) Find the average acceleration from [latex]200 \mathrm{~s}[/latex] to [latex]300 \mathrm{~s}[/latex], and find the displacement.

The slope of the green line is the average acceleration from [latex]200 \mathrm{~s}[/latex] to [latex]300 \mathrm{~s}[/latex] (Fig. 2.19b):

[latex]\begin{aligned}& \bar{a}=\operatorname{slop} \mathrm{e}=\frac{\Delta v}{\Delta t}=\frac{\left(1.0 imes 10^{1} - 1.7 imes 10^{2} ight) \mathrm{mi} / \mathrm{h}}{1.0 imes 10^{2} \mathrm{~s}} \& =-1.6(\mathrm{mi} / \mathrm{h}) / \mathrm{s} \end{aligned}[/latex]

The displacement from [latex]200 \mathrm{~s}[/latex] to [latex]300 \mathrm{~s}[/latex] is equal to area [latex]_{6}[/latex], which is the area of a triangle plus the area of a very narrow rectangle beneath the triangle:

[latex]\begin{aligned} \Delta x_{200} ightarrow 300 \mathrm{~s} &\approx{ }_{2}^{1}\left(1.0 imes 10^{2} \mathrm{~s} ight)\left(17 imes 10^{2} - 1.0 imes 10^{1} ight) \mathrm{mi} / \mathrm{h} \& +\left(1.0 imes 10^{1} \mathrm{mi} / \mathrm{h} ight)\left(1.0 imes 10^{2} \mathrm{~s} ight) \& =9.0 imes 10^{3}(\mathrm{mi} / \mathrm{h})(\mathrm{s})=2.5 \mathrm{mi}\end{aligned}[/latex]

(e) Find the total displacement from 0 s to [latex]400 \mathrm{~s}[/latex].

The total displacement is the sum of all the individual displacements. We still need to calculate the displacements for the time intervals from [latex]300 \mathrm{~s}[/latex] to [latex]350 \mathrm{~s}[/latex] and from [latex]350 \mathrm{~s}[/latex] to [latex]400 \mathrm{~s}[/latex]. The latter is negative because it's below the time axis.

[latex]\begin{aligned}\Delta x_{300 ightarrow 350 \mathrm{~s}} &\approx \frac{1}{2}\left(5.0 imes 10^{1} \mathrm{~s} ight)\left(10 imes 10^{1} \mathrm{mi} / \mathrm{h} ight) \&=2.5 imes 10^{2}(\mathrm{mi} / \mathrm{h})(\mathrm{s}) \\Delta x_{350 ightarrow 400 \mathrm{~s}} &\approx \frac{1}{2}\left(5.0 imes 10^{1} \mathrm{~s} ight)\left(-5.0 imes 10^{1} \mathrm{mi} / \mathrm{h} ight) \=& -1.3 imes 10^{3}(\mathrm{mi} / \mathrm{h})(\mathrm{s}) \end{aligned}[/latex]

Find the total displacement by summing the parts:

[latex]\begin{aligned}\Delta x_{0 ightarrow 400 \mathrm{~s}} \approx &\left(2.4 imes 10^{4}+9.0 imes 10^{3}+2.5 imes 10^{2} ight.\&\left.-1.3 imes 10^{3} ight)(\mathrm{mi} / \mathrm{h})(\mathrm{s})=8.9 \mathrm{mi}\end{aligned}[/latex]

REMARKS There are a number of ways to find the approximate area under a graph. Choice of technique is a personal preference.

Question 2.7: THE ACELA: THE PORSCHE OF AMERICAN TRAINS GOAL Find accelera...

Related Answered Questions

Verified Answer:

Verified Answer:

CATCHING A FLY BALL GOAL Apply the definition of instantaneous acceleration. PROBLEM A baseball player moves in a straight line path in order to catch a fly ball hit to the outfield. His velocity as a func tion of time is shown in Figure 2.11a. Find his instantaneous acceleration at points ...

Verified Answer:

A ROCKET GOES BALLISTIC GOAL Solve a problem involving a powered ascent followed by free-fall motion. PROBLEM A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s². It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height ...

Verified Answer:

Verified Answer:

NOT A BAD THROW FOR A ROOKIE! GOAL Apply the kinematic equations to a freely falling object with a nonzero initial velocity. PROBLEM A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer: