Lesson 1.3 · Unit 1 · HS-PS2-1
How fast, and how fast in which direction
Speed and velocity are treated as synonyms in ordinary language and are not synonyms in physics. The difference is the same one as between distance and displacement, and it has the same consequence: two correct answers to a journey that differ by a large factor.
- Average velocity is displacement over time: \( \bar{v} = \dfrac{\Delta x}{\Delta t} \). It is a vector.
- Average speed is distance over time. It is a scalar, and it is not the magnitude of the average velocity.
- Instantaneous velocity is the velocity at one moment, found by shrinking the time interval toward zero.
- On a position-time graph, velocity is the slope. Steeper means faster.
- A negative slope means motion in the negative direction, and a horizontal line means at rest.
- A curved position-time graph means the velocity is changing, which is lesson 1.4's subject.
- The instantaneous speed is the magnitude of the instantaneous velocity. For instants, unlike averages, the two do correspond.
Where students lose marks: computing an average speed by averaging two speeds. A trip at 60 km/h out and 30 km/h back does not average 45 km/h, because more time is spent at the slower speed. Average speed is always total distance divided by total time, and nothing else.
The problem. Using the runner from lesson 1.2: she runs 240 m east in 30 s, then 180 m west in 20 s. (a) Find her average velocity. (b) Find her average speed. (c) Explain why the two differ so much. (d) Describe the position-time graph of her run, and say what its slope shows.
Step one: collect what is needed for (a). Average velocity needs displacement and total time. From lesson 1.2, \( \Delta x = +60\ \text{m} \) with east positive. Total time: \( 30 + 20 = 50\ \text{s} \).
Step two: compute (a). \( \bar{v} = \dfrac{+60\ \text{m}}{50\ \text{s}} = +1.2\ \text{m/s} \), that is 1.2 m/s east. The direction is part of the answer, so the sign or the word "east" must appear.
Step three: compute (b). Average speed needs distance and total time. Distance was 420 m. \( \text{average speed} = \dfrac{420\ \text{m}}{50\ \text{s}} = 8.4\ \text{m/s} \).
Step four: answer (c). The two differ by a factor of seven because the reversal makes the path far longer than the net change in position. Neither figure is a lie about the run. At no point was she moving at 1.2 m/s; that number describes the net result of 50 seconds of running, not her pace. At no point was she moving at exactly 8.4 m/s either; that is the average of two different paces. Averages describe intervals, not moments.
Step five: find the pace on each leg. Out: \( \dfrac{240}{30} = 8.0\ \text{m/s} \) east. Back: \( \dfrac{180}{20} = 9.0\ \text{m/s} \) west. Check the average speed against these. The overall 8.4 m/s lies between 8.0 and 9.0, closer to 8.0, which fits because more time was spent on the outward leg. Averaging the two paces would have given 8.5, which is wrong for exactly that reason.
Step six: begin (d). With east positive and the marker at the origin, the graph rises from \( (0, 0) \) to \( (30\ \text{s}, 240\ \text{m}) \) in a straight line, because her pace on that leg was steady.
Step seven: the second leg. From \( (30, 240) \) the line falls to \( (50\ \text{s}, 60\ \text{m}) \), again straight. Slope of the first segment: \( \dfrac{240 - 0}{30 - 0} = +8.0\ \text{m/s} \) ✓ Slope of the second: \( \dfrac{60 - 240}{50 - 30} = \dfrac{-180}{20} = -9.0\ \text{m/s} \) ✓ The slopes reproduce the velocities including their signs, which is what makes the graph worth drawing.
Step eight: read the average velocity off the graph. Draw the straight line from the first point \( (0, 0) \) to the last \( (50, 60) \). Its slope is \( \dfrac{60}{50} = +1.2\ \text{m/s} \), matching part (a). That is the general rule. Average velocity over an interval is the slope of the straight line connecting the endpoints, and instantaneous velocity is the slope of the graph at a single point. When the graph is straight the two coincide; when it curves they do not. What the graph cannot show. Average speed. The graph plots position, so its slopes give velocities, and recovering the distance requires adding the sizes of the rises and falls separately. That is why 8.4 m/s does not appear anywhere on the picture as a slope.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.
- Write the definition of average velocity.
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Displacement divided by time
- What does the slope of a position-time graph give?
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Velocity
- An object moves \( +40\ \text{m} \) in 8.0 s. Find its average velocity.
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\( +5.0\ \text{m/s} \)
- What does a horizontal line on a position-time graph mean?
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Zero slope. The object is at rest
- A car covers 150 m in 10 s. Find its average speed.
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15 m/s
- A runner completes one lap of a 400 m track in 50 s. Find the average speed and the
average velocity.
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Average speed: \( \dfrac{400}{50} = 8.0\ \text{m/s} \). Average velocity: the start and finish are the same point, so the displacement is zero and \( \bar{v} = \dfrac{0}{50} = 0 \). Both are correct. She was certainly moving, but she got nowhere, and velocity measures getting somewhere. Speed 8.0 m/s, velocity zero
- A car travels 60 km at 60 km/h, then 60 km at 30 km/h. Find the average speed for the
whole trip.
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Do not average the speeds. Find the times first. Leg 1: \( \dfrac{60\ \text{km}}{60\ \text{km/h}} = 1.0\ \text{h} \). Leg 2: \( \dfrac{60\ \text{km}}{30\ \text{km/h}} = 2.0\ \text{h} \). Total distance 120 km, total time 3.0 h. \( \dfrac{120}{3.0} = 40\ \text{km/h} \). Not 45. Twice as much time was spent at the slower speed, so the average is pulled toward it. 40 km/h
- A position-time graph is a curve that gets steeper as time increases. Describe the
motion.
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Slope is velocity, and the slope is increasing, so the velocity is increasing. The object is moving in the positive direction and speeding up. Contrast the other curvatures. A curve flattening toward horizontal means slowing down. A curve bending below the horizontal means it reversed direction at the turning point. Moving in the positive direction and accelerating
- Explain why average speed is not the magnitude of average velocity.
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Because they are built from different numerators: one from distance, the other from displacement, and those agree only when the motion never reverses. The definitions. Average speed is \( \dfrac{\text{distance}}{t} \); average velocity is \( \dfrac{\text{displacement}}{t} \). The denominators match, so any difference comes entirely from the top. Why the tops differ. Lesson 1.2 established that distance is at least the magnitude of displacement, and strictly greater whenever the motion reverses. So average speed is at least the magnitude of the average velocity, and usually more. The extreme case. A closed loop has zero displacement, so the average velocity is zero while the average speed is not. The two cannot be the same quantity if one can be zero while the other is 8 m/s. When they do agree. Motion in a straight line without reversing. On a one-way trip the two are numerically equal, which is why the distinction is easy to miss until a problem involves a return leg. For instants the situation is different. Instantaneous speed *is* the magnitude of instantaneous velocity, because over a vanishingly small interval there is no opportunity to reverse. The mismatch is a property of averages only. Distance exceeds displacement whenever motion reverses, so the two averages come apart
- A train accelerates from rest, travels at a steady speed, then brakes to a stop,
covering 2400 m in 120 s altogether. Sketch the shape of its position-time graph, and
explain why its average velocity of 20 m/s is not its speed during the middle phase.
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Compute the average velocity. \( \bar{v} = \dfrac{2400\ \text{m}}{120\ \text{s}} = 20\ \text{m/s} \), in the direction of travel. Describe the graph in three parts. Phase 1, accelerating. The graph starts at the origin with zero slope, because the train starts at rest, and curves upward with increasing steepness. Phase 2, steady speed. The graph becomes a straight line, because constant velocity means constant slope. This is the steepest part. Phase 3, braking. The graph curves again, flattening until its slope reaches zero at the final position, where the train stops. The overall shape is a stretched letter S: shallow, steep, shallow, rising throughout because the train never reverses. Why 20 m/s is not the cruising speed. The average is the total displacement over the total time, and part of that time was spent moving more slowly than the cruise, during acceleration and braking. To average 20 m/s while spending time below it, the train must exceed 20 m/s during the middle phase. Put a number on it. Suppose the acceleration and braking take 20 s each and the cruise takes 80 s. Using the result from lesson 1.4 that a constant acceleration from rest covers half the distance a constant top speed would, accelerating to speed \( v \) over 20 s covers \( 10v \) meters, and the same for braking, while cruising covers \( 80v \). Total: \( 10v + 80v + 10v = 100v = 2400 \), so \( v = 24\ \text{m/s} \). Check. The cruise speed 24 m/s is above the 20 m/s average, as argued ✓ And the average sits between zero and 24, as any average must. The general reading. An average velocity is a single number standing in for a whole interval. It is exactly right about the net displacement and says nothing reliable about any particular instant. Whenever a problem gives you an average and asks about a moment, that gap is usually where the question lives. An S-shaped rising curve; the 20 m/s average is below the cruising speed of about 24 m/s