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The next several examples consider the motion of the subway train shown in [link] . In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.

In part (a), a subway train moves from left to right from an initial position of x equals 4 point 7 kilometers to a final position of x equals 6 point 7 kilometers, with a displacement of 2 point 0 kilometers. In part (b), the train moves toward the left, from an initial position of 5 point 25 kilometers to a final position of 3 point 75 kilometers.
One-dimensional motion of a subway train considered in [link] , [link] , [link] , [link] , [link] , and [link] . Here we have chosen the x -axis so that + means to the right and means to the left for displacements, velocities, and accelerations. (a) The subway train moves to the right from x 0 to x f . Its displacement Δ x is +2.0 km. (b) The train moves to the left from x 0 to x f size 12{ { {x}} sup { ' } rSub { size 8{f} } } {} . Its displacement Δ x size 12{Δx'} {} is 1 .5 km . (Note that the prime symbol (′) is used simply to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.)

Calculating displacement: a subway train

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of [link] ?

Strategy

A drawing with a coordinate system is already provided, so we don’t need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation Δ x = x f x 0 size 12{Δx=x rSub { size 8{f} } - x rSub { size 8{0} } } {} . This is straightforward since the initial and final positions are given.

Solution

1. Identify the knowns. In the figure we see that x f = 6.70 km and x 0 = 4.70 km for part (a), and x f = 3 .75 km and x 0 = 5 .25 km for part (b).

2. Solve for displacement in part (a).

Δ x = x f x 0 = 6 . 70 km 4 . 70 km = + 2 . 00 km size 12{Δx=x rSub { size 8{f} } - x rSub { size 8{0} } =6 "." "70 km" - 4 "." "70 km""=+"2 "." "00 km"} {}

3. Solve for displacement in part (b).

Δ x = x f x 0 = 3.75 km 5.25 km = 1.50 km size 12{Δx'= { {x}} sup { ' } rSub { size 8{f} } - { {x}} sup { ' } rSub { size 8{0} } =3 "." "75 km" - 5 "." "25 km"= - 1 "." "50 km"} {}

Discussion

The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.

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Comparing distance traveled with displacement: a subway train

What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in [link] ?

Strategy

To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in [link] . Distance traveled is the total length of the path traveled between the two positions. (See Displacement .) In the case of the subway train shown in [link] , the distance traveled is the same as the distance between the initial and final positions of the train.

Solution

1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km.

2. The displacement for part (b) was −1.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.

Discussion

Distance is a scalar. It has magnitude but no sign to indicate direction.

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Calculating acceleration: a subway train speeding up

Suppose the train in [link] (a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

Strategy

It is worth it at this point to make a simple sketch:

A point represents the initial velocity of 0 kilometers per second. Below the point is a velocity vector arrow pointing to the right, representing the final velocity of thirty point zero kilometers per hour. Below the velocity vector is an acceleration vector arrow labeled a equals question mark.

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

Solution

1. Identify the knowns. v 0 = 0 size 12{v rSub { size 8{0} } =0} {} (the trains starts at rest), v f = 30 . 0 km/h size 12{v rSub { size 8{f} } ="30" "." "0 km/h"} {} , and Δ t = 20 . 0 s size 12{Δt="20" "." "0 s"} {} .

2. Calculate Δ v size 12{Δv} {} . Since the train starts from rest, its change in velocity is Δ v = + 30.0 km/h size 12{Δv"=+""30" "." 0`"km/h"} {} , where the plus sign means velocity to the right.

3. Plug in known values and solve for the unknown, a - size 12{ { bar {a}}} {} .

a - = Δ v Δ t = + 30.0 km/h 20 . 0 s size 12{ { bar {a}}= { {Δv} over {Δt} } = { {+"30" "." 0`"km/h"} over {"20" "." 0`s} } } {}

4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.)

a - = + 30 km/h 20.0 s 10 3 m 1 km 1 h 3600 s = 0 . 417 m/s 2 size 12{ { bar {a}}= left ( { {+"30 km/h"} over {"20" "." "0 s"} } right ) left ( { {"10" rSup { size 8{3} } " m"} over {"1 km"} } right ) left ( { {"1 h"} over {"3600 s"} } right )=0 "." "417 m/s" rSup { size 8{2} } } {}

Discussion

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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