Lesson 1.1 · Unit 1 · CED topic 1.1
Scalars and vectors in one dimension
Two drivers leave the same town at the same speed of 60 miles per hour. One ends up in the city to the north and the other in the city to the south. A speedometer cannot tell them apart, yet they are in very different places. The reason is that “60 mph” is only half a description of their motion: it says how fast but not which way. Physics sorts every measurable quantity into one of two families according to whether direction matters, and the whole of mechanics is written in that language. This lesson sets the vocabulary and the sign rules you will use in every one of the eight units.
Describe a scalar or vector quantity using magnitude and direction, as appropriate. Describe a vector sum in one dimension.
- Scalars are quantities described by magnitude only; vectors are quantities described by both magnitude and direction.
- Vectors can be visually modeled as arrows with appropriate direction and lengths proportional to their magnitude.
- Distance and speed are examples of scalar quantities, while position, displacement, velocity, and acceleration are examples of vector quantities.
- Vectors are notated with an arrow above the symbol for that quantity. Vector notation is not required for vector components along an axis. In one dimension, the sign of the component completely describes the direction of that component.
- When determining a vector sum in a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.
Magnitude and direction
A scalar is a quantity fully described by a number and a unit: a mass of 2.0 kg, a time of 7 s, a distance of 300 m, a speed of 15 m/s. A vector needs a number, a unit and a direction: a displacement of 300 m north, a velocity of 15 m/s to the right. The number and unit together are the vector’s magnitude, which is always zero or positive, because it measures size only.
A vector is drawn as an arrow. The arrow points in the direction of the vector and its length is proportional to the magnitude, so an arrow twice as long represents a vector twice as large. In print a vector is named with an arrow over its symbol, \(\vec{v}\), \(\vec{a}\), \(\Delta\vec{x}\); the magnitude is written without the arrow, \(v\), or with bars, \(|\vec{v}|\).
| Scalars (magnitude only) | Vectors (magnitude and direction) |
|---|---|
| distance \(d\), the total path length | position \(x\) and displacement \(\Delta x\) |
| speed, the magnitude of velocity | velocity \(v\) and acceleration \(a\) |
| time, mass, energy (Unit 3) | force (Unit 2), momentum (Unit 4) |
One dimension: the sign is the direction
When all motion lies along a single straight line, you do not need arrows over your symbols. You choose a coordinate axis, say \(x\), and decide which way is positive. Every vector component along that axis is then a signed number: a plus sign means “along the positive direction” and a minus sign means “along the negative direction.” A velocity of \(v_x = -3\ \text{m/s}\) is a motion of 3 m/s in the negative direction. The sign is part of the answer.
Two habits prevent the most common errors. First, state the axis before you calculate: write “right is positive” or “up is positive” at the top of your work. Second, remember that the minus sign says nothing about size. A velocity of \(-8\ \text{m/s}\) has a greater magnitude than \(+3\ \text{m/s}\): it is the faster motion, only in the opposite direction. “Negative” does not mean “smaller” and it does not mean “slowing down.”
The choice of axis is yours, and the physics is the same for either choice, as Figure 1.1b shows. Flipping the axis flips every sign in the problem, but the magnitudes, the times and the physical conclusions do not change. What you may not do is mix: if right is positive for displacement, it must be positive for velocity and acceleration in the same problem.
Position, displacement and distance
Position \(x\) tells where an object is relative to a chosen origin, and it is a vector: \(x = +12\ \text{m}\) and \(x = -12\ \text{m}\) are different places. Displacement is the change in position, \(\Delta x = x_f - x_i\), the straight-line “how far and which way from the start to the finish.” It can be positive, negative or zero. Distance is the total length of the path actually traveled, which is why it is a scalar and can never be negative or decrease. Any trip that ends where it started has zero displacement but a large distance.
Adding vectors in one dimension
In one dimension the vector sum is an ordinary sum of signed numbers. Vectors pointing the same way reinforce; vectors pointing in opposite directions partly cancel, because they carry opposite signs. If a plane flies at \(+200\ \text{m/s}\) relative to the air and the wind adds \(-30\ \text{m/s}\) (a headwind), the resulting velocity relative to the ground is \(+200 + (-30) = +170\ \text{m/s}\).
The same rule explains the first equation of motion on the AP reference sheet. An object’s final velocity is its initial velocity plus the change in velocity, and the change in velocity is the acceleration times the time: \[ \vec{v} = \vec{v}_0 + \vec{a}\,t \qquad\text{which in one dimension reads}\qquad v_x = v_{x0} + a_x t. \] Each term has its own sign. You will use this equation constantly, and it is nothing more than a vector sum.
A hiker walks 40 m east, turns back and walks 25 m west, then walks 10 m east again, in a total time of 50 s. Take east as positive. Find (a) the distance, (b) the displacement, (c) the average speed and (d) the magnitude and direction of the average velocity.
Step 1, signed legs. The displacements of the three legs are \(+40\ \text{m}\), \(-25\ \text{m}\) and \(+10\ \text{m}\).
Step 2, distance. Distance adds lengths and ignores direction: \(40 + 25 + 10 = 75\ \text{m}\).
Step 3, displacement. Displacement adds the signed values: \(\Delta x = +40 - 25 + 10 = +25\ \text{m}\), which is 25 m east of the start.
Step 4, average speed. Speed uses distance: \(\dfrac{75\ \text{m}}{50\ \text{s}} = 1.5\ \text{m/s}\).
Step 5, average velocity. Velocity uses displacement: \(\dfrac{+25\ \text{m}}{50\ \text{s}} = +0.50\ \text{m/s}\), that is, 0.50 m/s east.
Check. The average speed (1.5 m/s) is larger than the magnitude of the average velocity (0.50 m/s). That is always true unless the path is a straight line with no backtracking.
A cart rolls to the left at 6.0 m/s along a track, and a steady push gives it an acceleration of \(+2.0\ \text{m/s}^2\) (to the right) for 4.0 s. Take right as positive. What is its velocity at the end?
Step 1, signs. The initial velocity is to the left, so \(v_{x0} = -6.0\ \text{m/s}\); the acceleration is to the right, \(a_x = +2.0\ \text{m/s}^2\).
Step 2, equation. \(v_x = v_{x0} + a_x t = -6.0 + (2.0)(4.0) = -6.0 + 8.0 = +2.0\ \text{m/s}\).
Interpretation. The cart now moves to the right at 2.0 m/s. It slowed down, stopped for an instant after 3.0 s, reversed direction, and sped up the other way. The sign change shows the reversal; no “turning around” step had to be added by hand.
A student rolls a ball from a mark at \(x = +2.0\ \text{m}\) to a mark at \(x = +7.0\ \text{m}\). Describe the displacement if (a) right is positive and the marks are as given, and (b) the student instead calls left positive.
(a) \(\Delta x = x_f - x_i = 7.0 - 2.0 = +5.0\ \text{m}\). (b) With the axis reversed, the same two places have coordinates \(-2.0\ \text{m}\) and \(-7.0\ \text{m}\), so \(\Delta x = -7.0 - (-2.0) = -5.0\ \text{m}\). The ball moved 5.0 m in both descriptions; only the sign changed, because the positive direction changed.
- Name the axis. Free-response graders reward “take up as positive” because every later sign depends on it.
- Keep the sign in your answer. “\(-4\ \text{m/s}\)” and “\(4\ \text{m/s}\)” are different answers. When a question asks for a magnitude, drop the sign; when it asks for a velocity, keep it.
- Distance is not displacement. Average speed uses distance and average velocity uses displacement; confusing them is the single most common slip in Unit 1.
- Do not say “negative means slowing.” A negative sign gives direction only. Whether something slows down depends on how its velocity and acceleration compare (lesson 1.2).
- Vector sums are signed sums. In one dimension there is no need for a triangle or a protractor; opposite directions are opposite signs.
- Classify each as scalar or vector: temperature of 20 \(^\circ\)C, a force of 30 N downward, a time of 12 s, an acceleration of \(-3.0\ \text{m/s}^2\), a speed of 25 m/s.
Show answer
Scalars: temperature, time, speed (they have magnitude only). Vectors: the 30 N downward force and the acceleration of \(-3.0\ \text{m/s}^2\), since each has a direction (given in words in the first case and by the sign in the second).
- A runner goes once around a 400 m track and finishes where she started in 80 s. Find her distance, displacement, average speed and average velocity.
Show answer
Distance 400 m. Displacement 0 m, since she ends at the start. Average speed \(400/80 = 5.0\ \text{m/s}\). Average velocity \(0/80 = 0\ \text{m/s}\).
- With right positive, a car moves at \(-18\ \text{m/s}\) and a bicycle at \(+9\ \text{m/s}\). Which is moving faster? Which way is each moving?
Show answer
The car is faster: its magnitude, 18 m/s, exceeds the bicycle’s 9 m/s. The car moves to the left (negative direction) and the bicycle to the right.
- A student walks 30 m north, then 50 m south, then 10 m north. Take north as positive. Find the displacement and the distance.
Show answer
\(\Delta x = +30 - 50 + 10 = -10\ \text{m}\), which is 10 m south of the start. The distance is \(30 + 50 + 10 = 90\ \text{m}\).
- A skater glides at \(+5.0\ \text{m/s}\) and a gust gives her \(a_x = -1.5\ \text{m/s}^2\) for 2.0 s. Find her final velocity.
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\(v_x = v_{x0} + a_x t = 5.0 + (-1.5)(2.0) = 5.0 - 3.0 = +2.0\ \text{m/s}\). She is still moving in the positive direction, only more slowly.
- Explain, using the idea of a sign convention, why a problem solved with “down is positive” and the same problem solved with “up is positive” give different signs but the same physical result.
Show answer
The positive direction is an arbitrary choice. Reversing it multiplies every vector component by \(-1\) (displacement, velocity and acceleration all flip sign together), but the magnitudes, the times and the real motion are unchanged. As long as the same choice is used throughout one solution, both versions describe the same event.