Lesson 1.2 · Unit 1 · HS-PS2-1
Where something is, and how far it got
Before motion can be described it has to be located, and locating something requires a choice: where zero is and which direction counts as positive. That choice is free, but once made it has to be kept. Two different questions then arise about a journey, and confusing them is the first place physics answers go wrong.
- A position needs a reference frame: an origin, and a direction declared positive.
- The choice is arbitrary but must be fixed. Any consistent choice gives the same physics; changing it midway gives nonsense.
- Displacement is the change in position, \( \Delta x = x_f - x_i \), and it depends only on the endpoints.
- Distance is the length of the path actually traveled, and it depends on the whole journey.
- Displacement is a vector; distance is a scalar. Displacement carries a direction, distance does not.
- In one dimension the sign is the direction. A displacement of \( -8\ \text{m} \) means eight meters in the negative direction.
- Distance is never negative and never less than the size of the displacement. The two are equal only when the motion never reverses.
Where students lose marks: giving a displacement without a direction. "The displacement was 60 m" is incomplete. Write \( +60\ \text{m} \), or 60 m east, or say which way you chose as positive. A vector reported as a bare magnitude has lost half its information.
The problem. A runner starts at a marker, runs 240 m east, then turns and runs 180 m west, ending where she stops. (a) Set up a frame and find her final position. (b) Find her displacement. (c) Find the distance she ran. (d) Explain when distance and displacement agree, and give a case where displacement is zero but distance is large.
Step one: set up the frame for (a). Put the origin at the starting marker, so \( x_i = 0 \), and declare east positive. West is then negative. Everything that follows uses this choice.
Step two: track the position. After the first leg she is at \( +240\ \text{m} \). The second leg is 180 m west, which in this frame is \( -180\ \text{m} \): \( x_f = 240 + (-180) = +60\ \text{m} \). So she finishes 60 m east of the marker.
Step three: find the displacement for (b). \( \Delta x = x_f - x_i = 60 - 0 = +60\ \text{m} \), that is 60 m east. Notice what the calculation used. Only the starting and ending positions. The fact that she went out to 240 m first never entered.
Step four: find the distance for (c). Distance adds the path lengths regardless of direction: \( 240 + 180 = 420\ \text{m} \). The two answers differ by a factor of seven, and both are correct answers to different questions. "How far did she run?" is 420 m. "How far is she from where she started?" is 60 m east.
Step five: check the frame choice does not matter. Redo it with west positive. The first leg is now \( -240\ \text{m} \) and the second is \( +180\ \text{m} \), giving \( x_f = -60\ \text{m} \). That is 60 m in the negative direction, and negative now means east. The same physical answer, described in a different language. What is not allowed is treating east as positive in one line and west as positive in the next, which produces an answer describing no journey at all.
Step six: answer the first half of (d). Distance equals the magnitude of displacement exactly when the motion never reverses direction. If the runner had simply gone 240 m east and stopped, both would be 240 m. Any reversal makes the path longer than the net change, so distance exceeds the displacement's magnitude in every other case.
Step seven: give the extreme case. A runner completes four laps of a 400 m track and stops at the starting line. Distance: \( 4 \times 400 = 1600\ \text{m} \). Displacement: the start and end positions are the same point, so \( \Delta x = 0 \).
Step eight: draw the consequence. Her average velocity over the four laps is exactly zero, while her average speed is clearly not. That is not a paradox; it is the definitions doing what they are for. Why displacement is the useful quantity in physics. Nearly every equation in the next six units involves displacement rather than distance, because the laws of motion relate forces to changes in position, and a change is a difference between two points. The path taken between them does not enter Newton's second law at all. Why distance still matters. It is what the odometer reads, what the fuel is consumed over, and what friction acts along. Lesson 5.6 needs distance rather than displacement for exactly that reason.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.
- Is displacement a vector or a scalar?
Show the full solution
A vector
- Is distance a vector or a scalar?
Show the full solution
A scalar
- An object moves from \( x = 3\ \text{m} \) to \( x = 11\ \text{m} \). Find its
displacement.
Show the full solution
\( 11 - 3 \). \( +8\ \text{m} \)
- An object moves from \( x = 5\ \text{m} \) to \( x = -2\ \text{m} \). Find its
displacement.
Show the full solution
\( -2 - 5 \). \( -7\ \text{m} \)
- Can distance ever be negative?
Show the full solution
It is a path length. No
- A cyclist rides 3.0 km north, then 4.0 km south. Find the distance and the
displacement.
Show the full solution
Take north as positive. Distance: \( 3.0 + 4.0 = 7.0\ \text{km} \). Displacement: \( +3.0 + (-4.0) = -1.0\ \text{km} \), that is 1.0 km south. The sign is the answer's direction, so reporting "1.0 km" alone would be incomplete. Distance 7.0 km, displacement 1.0 km south
- A ball is dropped from a height of 2.0 m, bounces, and is caught at 1.2 m on the way
up. Find the distance traveled and the displacement.
Show the full solution
Take upward positive with the floor at zero. Start at \( +2.0\ \text{m} \), reach \( 0 \), return to \( +1.2\ \text{m} \). Distance: \( 2.0 + 1.2 = 3.2\ \text{m} \). Displacement: \( 1.2 - 2.0 = -0.8\ \text{m} \), that is 0.8 m downward. The ball ends below where it began even though it was last moving upward. Displacement describes the net result, not the final direction of travel. Distance 3.2 m, displacement 0.8 m downward
- A hiker walks 6.0 km east and then 8.0 km north. Find the distance and the magnitude of
the displacement.
Show the full solution
Distance is the path length: \( 6.0 + 8.0 = 14.0\ \text{km} \). The displacement is the straight line from start to finish, and the two legs are perpendicular, so use the Pythagorean theorem: \( \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10.0\ \text{km} \). Direction: \( \tan^{-1}\left( \dfrac{8.0}{6.0} \right) = 53^\circ \) north of east. This is why displacement needs vector addition rather than ordinary addition once the motion leaves a single line. Lesson 1.7 builds on it. Distance 14.0 km, displacement 10.0 km at 53° north of east
- Explain why the choice of positive direction cannot change the physical answer.
Show the full solution
Because the choice is a description of reality, not a property of it, and every quantity in the problem is relabeled together. What changes. The sign attached to each displacement, velocity and acceleration flips. What does not. Which physical direction each sign now names. A displacement of \( +60\ \text{m} \) with east positive and one of \( -60\ \text{m} \) with west positive describe the identical journey, because in the second frame negative means east. Why the flip is consistent. Every vector in the problem is multiplied by \( -1 \) at once, and the equations relating them are unchanged by that, so the relationships survive intact. Where it goes wrong. Not in the choosing but in the forgetting. Taking down as positive for the fall and up as positive for the rebound, inside one problem, produces a calculation that describes no physical situation. Write the choice down at the top of the work and leave it there. A practical tip. Choose the direction of initial motion as positive. It tends to keep the most numbers positive and reduces the chance of a sign slip. The choice relabels every vector together, so the described physics is unchanged
- A delivery van leaves the depot, drives 12 km north, 5.0 km east, then 12 km south, and
stops. Find the distance, the displacement, and explain what the driver's odometer would
read against what a straight-line tracker would report.
Show the full solution
Set up a frame. Origin at the depot, north positive on one axis, east positive on the other. Track the two axes separately, which is the method lesson 1.7 formalizes. North-south axis. \( +12 \) then \( -12 \), so the net is \( 0\ \text{km} \). East-west axis. \( +5.0\ \text{km} \), unchanged by the other two legs. Displacement. The van ends 5.0 km east of the depot and level with it north to south, so the displacement is 5.0 km east. Distance. Add the three path lengths: \( 12 + 5.0 + 12 = 29\ \text{km} \). The odometer. It counts wheel rotations, so it measures path length and reads 29 km. It has no way to know the van turned around. A straight-line tracker. A device comparing the start and end positions reports 5.0 km east, and knows nothing about the route. Neither is wrong. They answer different questions, and both are needed for different purposes: the fuel used depends on the 29 km, while the time saved by a better route depends on how much of that 29 km was wasted relative to the 5.0 km that had to be covered. The general point. A quantity in physics is defined by the question it answers. Asking "which is the real distance?" is the wrong question; asking "which one does this situation depend on?" is the right one. Distance 29 km, displacement 5.0 km east