# Distance and Displacement

## Practice

### practice problem 1

#### solution

Add vectors in the same direction with "ordinary" addition.

x = 11,200 m + 8,400 mx = 19,600 my = 3200 m + 1700 my = 4900 m |

Add vectors at right angles with a combination of pythagorean theorem for magnitude…

r = √(x^{2} + y^{2})r = √[(19,600 m)^{2} + (4,900 m)^{2}]r = 20,200 m |

and tangent for direction.

tan θ = | y |
= | 4,900 m | |

x |
19,600 m | |||

θ = | 14.0° | |||

Don't forget to answer the question.

Camp B is 20,200 m away from base camp at an angle of elevation of 14.0°.

Oh yeah, and don't forget to make a drawing. I probably should have told you to do that earlier.

### practice problem 2

#### solution

Vectors that are not at nice angles need to be dealt with. Break them up into their components.

x_{1} = |
r_{1} cos θ_{1} |

x_{1} = |
(11,648 m)cos(15.95°) |

x_{1} = |
11,200 m |

y_{1} = |
r_{1} sin θ_{1} |

y_{1} = |
(11,648 m)sin(15.95°) |

y_{1} = |
3200 m |

x_{2} = |
r_{2} cos θ_{2} |

x_{2} = |
(8,570 m)cos(11.44°) |

x_{2} = |
8400 m |

y_{2} = |
r_{2} sin θ_{2} |

y_{2} = |
(8,570 m)sin(11.44°) |

y_{2} = |
1700 m |

Add vectors in the same direction with "ordinary" addition.

x = |
11,200 m + 8,400 m |

x = |
19,600 m |

y = |
3200 m + 1700 m |

y = |
4900 m |

Add vectors at right angles with a combination of pythagorean theorem for magnitude…

r = |
√(x^{2} + y^{2}) |

r = |
√[(19,600 m)^{2} + (4,900 m)^{2}] |

r = |
20,200 m |

and tangent for direction.

tan θ = | y |
= | 4,900 m | |

x |
19,600 m | |||

θ = | 14.04° | |||

Don't forget to answer the question.

The target of the laser beam is 20,200 m away at an angle of elevation of 14.04°.

Oh yeah, and don't forget to make a drawing. I probably should have told you to do that earlier. I feel bad that I've done that twice in the sections on vectors. I should set a better example. Let me make it up to you by giving you an animated drawing.

If you have a feeling of *déjà vu*, do not be alarmed. The Matrix is fine. I recycled the solution to this problem from an earlier one. The idea was to show a common problem solving method used in physics. Whenever possible, take a difficult problem that you haven't solved and reduce it one that you have solved.

How does one add parallel vectors? Simple — add them. How does one add antiparallel vectors? Also simple — subtract them. How does one add vectors at right angles. Reasonably simple — use pythagorean theorem and tangent. How does one add vectors that aren't at 0°, 180°, or 90°? Brutally simple — resolve them into components. Don't let the vectors make you work harder. Make them point in a direction that's convenient for you. Make them in simpler vectors.

And then the students learned that there really was no such thing as a "bad" vector and everyone lived happily ever after. The End.

### practice problem 3

departure | arrival | distance | bearing |
---|---|---|---|

ATL | ORD | 975 km | 342.8° |

ORD | DFW | 1291 km | 221.6° |

DFW | ATL | ||

round trip | n/a |

#### solution

Start with a diagram. It helps to know a little geography, but it isn't essential. Starting from Atlanta we go north and west to Chicago, then south and west to Dallas. We definitely need to go east to return, but without an accurate drawing, it's hard to decide if the last leg runs north or south.

Whenever you see vectors at odd angles, you should immediately think of components. Since bearings are measured from north (the +*y* axis) and not east (the +*x* axis), sine and cosine are swapped from their "normal" assignments — sine goes with the *x* component and cosine with the *y* component. Do this once for each of the first two stages of the journey.

x_{1} = |
r_{1} sin θ_{1} |

x_{1} = |
(975 km)sin(342.8°) |

x_{1} = |
−288 km |

y_{1} = |
r_{1} cos θ_{1} |

y_{1} = |
(975 km)cos(342.8°) |

y_{1} = |
+931 km |

x_{2} = |
r_{2} sin θ_{2} |

x_{2} = |
(1291 km)sin(221.6°) |

x_{2} = |
−857 km |

y_{2} = |
r_{2} cos θ_{2} |

y_{2} = |
(1291 km)cos(221.6°) |

y_{2} = |
−965 km |

The net displacement of a round trip is zero. In order to get back to Atlanta, the third stage should cancel the first two. It needs to do this in both the *x* (east-west) and *y* (north-south) directions.

x_{1} |
+ | x_{2} |
+ | x_{3} |
= | 0 |

(−288 km) | + | (−857 km) | + | x_{3} |
= | 0 km |

x_{3} |
= | +1145 km | ||||

y_{1} |
+ | y_{2} |
+ | y_{3} |
= | 0 |

(+931 km) | + | (−965 km) | + | y_{3} |
= | 0 km |

y_{3} |
= | +34 km |

Use pythagorean theorem for the magnitude…

r_{3} = √[(+1145 km)^{2} + (−34 km)^{2}]r_{3} = 1146 km |

and tangent for the direction of stage three. We're dealing with angles measured from the +*y* axis, so tangent is ∆*x*/∆*y* not the "standard" ∆*y*/∆*x*.

tan θ_{3} = |
x_{3} |
= | +1145 km | |

y_{3} |
+34 km | |||

To get to Atlanta from Dallas, one must head very much east and only slightly north.

Add all three magnitudes together to get the round trip distance.

∆s = 975 km + 1291 km + 1146 km∆ s = 3412 km |

We have all the information we need to complete the table.

departure | arrival | distance | heading |
---|---|---|---|

ATL | ORD | 975 km | 342.8° |

ORD | DFW | 1291 km | 221.6° |

DFW | ATL | 1146 km | 88.3° |

round trip | 3412 km | n/a |

Final Note: These calculations assume the Earth is flat, which is not quite correct. Using equations derived from spherical geometry, the displacement of the final stage is 1177 km at 82.5°. Please consult a professional pilot before attempting any air travel.

### practice problem 4

- one complete orbit around the Sun
- one-half orbit around the Sun
- one-fourth orbit around the Sun
- What relationship between distance and displacement does this illustrate?

#### solution

Start by drawing a happy orange and yellow sun. Then draw a dark blue Earth in orbit. Then, since it's hard to get the proportions right on the Sun's face, write "not to scale" in great big letters

The problem didn't specify a unit, so I recommend the astronomical unit [au]. An astronomical unit is almost the same as the mean distance from the Earth to the Sun. Mean is not the opposite of happy, so the units do not contradict the drawing. (The opposite of happy is standard deviation, by the way.)

After a lap around the Sun, the Earth has traveled a distance of one circumference…

∆

*s*=*C*= 2π*r*= 2π(1 au)

∆*s*= 6.28 aubut it's right back where it started from, so it's displacement is…

*r*= 0.00 auAfter half a lap around the Sun, the Earth has traveled a distance of half a circumference.

∆

*s*= ½*C*= ½(2π*r*) = 1π(1 au)

∆*s*= 3.14 auBut it's one diameter away from where it started, so it's displacement is…

*r*= 2.00 auAfter one-quarter lap around the Sun, the Earth has traveled a distance of one-quarter circumference.

∆

*s*= ¼*C*= ¼(2π*r*) = ½π(1 au)

∆*s*= 1.57 auDo you see the special triangle in the diagram when the Earth is in this position? Two radii at right angles create a 45°, 45°, 90° triangle, the sides of which are in the ratio 1:1:√2.

*r*= √2 (1 au)*r*= 1.41 auDisplacement approaches distance as time approaches zero. I added a few extra calculated values to show the trend.

segment | time | distance | displacement | ∆ |
---|---|---|---|---|

a. ^{1}1 orbit |
12 months | 6.28 au | 0.00 au | 6.28 au |

b. ^{1}^{1}_{2} orbit |
06 months | 3.14 au | 2.00 au | 1.14 au |

c. ^{1}^{1}_{4} orbit |
03 months | 1.57 au | 1.41 au | 0.77 au |

^{1}d. ^{1}_{8} orbit |
1^{1}_{2} months |
0.79 au | 0.77 au | 0.02 au |

d. ^{1}_{16} orbit |
0^{3}_{4} months |
0.39 au | 0.39 au | 0.00 au |

⋮ | ⋮ | ⋮ | ⋮ | ⋮ |

infinitesimal | 0 months | 0 au | 0 au | 0 au |