Position, displacement and distance from velocity
A speedometer records how fast and in which direction something moves, moment by moment, but not where it is. From the velocity alone, how far did it end up from where it started, and how far did it actually travel?
Where you are 3 in Accumulation and net change
When you reach for it
You reach for this whenever a motion along a line is known through its velocity v(t) or its acceleration a(t) rather than its position s(t): a particle given by a formula, an elevator logged by a sensor, a dropped stone whose only known force is gravity. Three questions come up. Where is it at a later time? How far is it from where it started (the displacement)? How much ground did it cover (the total distance)? Integrals answer all three.
In Position, velocity and acceleration the position was known, and the distance came from adding \lvert s(t_{i+1}) - s(t_i) \rvert between turning points. Here only the rate is known, so the same pieces come from integrals of v.
The procedure
Day 1: displacement, distance and position
Displacement. Position is an antiderivative of velocity, since s' = v. If v is continuous on [a, b], the second fundamental theorem (The second fundamental theorem of calculus) gives
This is the net change theorem (The net change theorem in context) read for motion. Like every integral it is a signed area: time spent moving in the negative direction counts against time spent moving in the positive one.
Position. Rearranged, the same equation finds a later position from an earlier one:
Velocity alone fixes position only up to a constant; one known position fixes the constant.
Total distance. On an interval where v does not change sign, the object moves one way only, so the distance it covers is the size of its displacement, \left\lvert \int v(t)\,dt \right\rvert, which is \int \lvert v(t) \rvert\,dt because \lvert v \rvert is v or -v throughout. Cut [a, b] at every point where v changes sign and add the pieces:
To evaluate it, find the zeros of v and make a sign chart as in Position, velocity and acceleration, integrate v on each piece, and add the sizes of the results. Since \lvert v(t) \rvert is the speed, total distance is the integral of speed, and displacement the integral of velocity. The two agree exactly when v never changes sign on [a, b].
Velocity from acceleration. Acceleration is the derivative of velocity. Knowing a(t) and one velocity v(t_0) = v_0 is an initial value problem (Differential equations and their solutions):
Antidifferentiate (Antiderivatives and indefinite integrals) to get v(t) = A(t) + C, then use the condition to find C; equivalently v(t) = v_0 + \int_{t_0}^t a(u)\,du. The answer is the only one: two solutions have the same derivative, so their difference has derivative 0 and is a constant (by the mean value theorem), and that constant is 0 at t_0. A second initial value problem, s' = v with s(t_0) = s_0, then gives position.
Day 2: free fall, graphs, tables and averages
Free fall. Near the surface of the Earth, with air resistance ignored, a thrown or dropped object has constant acceleration -g (up taken as positive), with g \approx 9.8 meters per second squared, or 32 feet per second squared. This is a physical model, stated here without derivation. With v(0) = v_0 and s(0) = s_0, two initial value problems give
So s(t) = s_0 + v_0 t - 4.9t^2 in meters and s(t) = s_0 + v_0 t - 16t^2 in feet, the formulas Maximum, minimum and projectile problems took as given. The top of the flight is where v = 0; the landing is where s reaches ground level; the impact speed is \lvert v \rvert at the landing time.
From a velocity graph. The displacement is the area above the axis minus the area below it; the distance is the two added. When the graph is made of straight pieces, triangles and rectangles give the areas exactly.
From a velocity table. A Riemann or trapezoid sum of the readings estimates the displacement. A sum of the speeds \lvert v_i \rvert estimates the distance; the displacement estimate never does. Take care on an interval where the readings change sign: the object turned somewhere inside, at a time the table does not give. If v is assumed to change linearly from v_i to v_{i+1} over a step \Delta t, it crosses zero partway, and that step contributes two triangles,
rather than the trapezoid \displaystyle \frac{\Delta t}{2}\left(\lvert v_i \rvert + \lvert v_{i+1} \rvert\right), which treats the object as if it never slowed to a stop.
Average velocity and average speed. Over [a, b] the average value (Average value of a function) of v is the average velocity, displacement divided by elapsed time; the average value of \lvert v \rvert is the average speed, distance divided by elapsed time. Since distance is at least the size of the displacement, average speed is at least \lvert \text{average velocity} \rvert.
Worked examples
Day 1: a particle, net and total
A particle moves along a line with velocity v(t) = 2t^2 - 10t + 8 meters per second for 0 \le t \le 5, and s(0) = 2. Find its displacement, its position at t = 5, the total distance it travels, and its speed at t = 2.
An antiderivative is \displaystyle F(t) = \frac{2t^3}{3} - 5t^2 + 8t. The displacement is
so \displaystyle s(5) = 2 - \frac{5}{3} = \frac{1}{3}. For distance, v(t) = 2(t - 1)(t - 4) is positive on (0, 1), negative on (1, 4) and positive on (4, 5). With \displaystyle F(1) = \frac{11}{3} and \displaystyle F(4) = -\frac{16}{3}, the pieces are
The total distance is \displaystyle \frac{11}{3} + 9 + \frac{11}{3} = \frac{49}{3} \approx 16.33 meters. At t = 2, v(2) = 8 - 20 + 8 = -4, so the speed is 4 meters per second, moving in the negative direction.
Day 1: Kahurangi’s glass elevator
Kahurangi rides a glass elevator in a hotel atrium. Its velocity is \displaystyle v(t) = 3\sin\left(\frac{\pi t}{15}\right) meters per second for 0 \le t \le 20, positive upward, and she starts 12 meters above the lobby floor. Find her height at t = 20, the total distance she rides, and her greatest height.
SolutionBy the chain rule, \displaystyle -\frac{45}{\pi}\cos\left(\frac{\pi t}{15}\right) has derivative \displaystyle \frac{45}{\pi}\cdot\frac{\pi}{15}\sin\left(\frac{\pi t}{15}\right), which is v(t), so it is an antiderivative; at t = 20 the cosine is \displaystyle \cos\frac{4\pi}{3} = -\frac{1}{2}. The angle \displaystyle \frac{\pi t}{15} runs from 0 to \pi while t runs to 15, and from \pi to \displaystyle \frac{4\pi}{3} after, so v > 0 on (0, 15) and v < 0 on (15, 20). The pieces are
The displacement is \displaystyle \frac{90}{\pi} - \frac{45}{2\pi} = \frac{135}{2\pi} \approx 21.49 meters, so at t = 20 she is \displaystyle 12 + \frac{135}{2\pi} \approx 33.49 meters up. She rides \displaystyle \frac{90}{\pi} + \frac{45}{2\pi} = \frac{225}{2\pi} \approx 35.81 meters in all. She rises until t = 15 and falls after, so her greatest height is \displaystyle 12 + \frac{90}{\pi} \approx 40.65 meters.
Day 1: velocity and position from acceleration
A particle has acceleration a(t) = 10 - 6t meters per second squared, with v(0) = -3 and s(0) = 1. Find v(t) and s(t), check that s solves s'' = 10 - 6t, s(0) = 1, s'(0) = -3, and find the total distance traveled for 0 \le t \le 4.
SolutionAntidifferentiating, v(t) = 10t - 3t^2 + C_1, and v(0) = C_1 = -3. Again, s(t) = -3t + 5t^2 - t^3 + C_2, and s(0) = C_2 = 1:
Check: s'(t) = -3 + 10t - 3t^2 and s''(t) = 10 - 6t, with s(0) = 1 and s'(0) = -3. By the uniqueness argument above, this is the only such function.
Factor: v(t) = -(3t - 1)(t - 3), negative on \displaystyle \left(0, \frac{1}{3}\right), positive on \displaystyle \left(\frac{1}{3}, 3\right), negative on (3, 4). With \displaystyle s\left(\frac{1}{3}\right) = \frac{14}{27}, s(3) = 10 and s(4) = 5,
The total distance is \displaystyle \frac{13}{27} + \frac{256}{27} + 5 = \frac{404}{27} \approx 14.96 meters, against a displacement of s(4) - s(0) = 4 meters.
Day 2: Tavake’s ball from the roof deck
Tavake stands on a roof deck 24.5 meters above the street and tosses a ball straight up at 19.6 meters per second. It rises, falls past the deck and lands on the street. When is it highest, and how high? When does it land, and how fast? How far does it travel?
SolutionWith up positive and the street as height 0, a(t) = -9.8, v(0) = 19.6 and s(0) = 24.5, so
The top is where v = 0, at t = 2, and s(2) = 24.5 + 39.2 - 19.6 = 44.1 meters. It lands when s(t) = 0; dividing by -4.9 gives t^2 - 4t - 5 = (t - 5)(t + 1) = 0, and only t = 5 is in the motion. Then v(5) = 19.6 - 49 = -29.4, an impact speed of 29.4 meters per second.
Velocity changes sign at t = 2, so
while the displacement is s(5) - s(0) = -24.5 meters. Over the flight the average velocity is -4.9 meters per second and the average speed 12.74 meters per second.
Day 2: Kahurangi’s wrench down the shaft
Kahurangi drops a wrench from rest down an empty elevator shaft, and it hits the bottom at 96 feet per second. (a) How deep is the shaft? (b) Show that an object dropped from rest from height h lands after \displaystyle \sqrt{2h/g} seconds at speed \displaystyle \sqrt{2gh}. (c) How deep would a shaft need to be for twice the impact speed?
Solution(a) Measure height from the bottom, with depth h. From rest, v(t) = -32t and s(t) = h - 16t^2. A speed of 96 means 32t = 96, so t = 3; the wrench has then fallen to s = 0, so h = 16 \cdot 9 = 144 feet.
(b) In general \displaystyle s(t) = h - \frac{1}{2}gt^2, which is 0 when \displaystyle t = \sqrt{2h/g}. The speed then is
Check with (a): \displaystyle \sqrt{2 \cdot 32 \cdot 144} = \sqrt{9216} = 96.
(c) Impact speed grows like \displaystyle \sqrt{h}, so doubling it takes four times the height: 576 feet.
Day 2: Lagi reads an elevator’s velocity graph
Lagi logs an elevator ride: its velocity is the broken line in the figure, in meters per second with up positive, and it starts 10 meters above the ground. Find the displacement, the distance ridden, the final and greatest heights, and the average velocity and average speed over 0 \le t \le 12.
Above the axis are a triangle, a rectangle and a triangle: \displaystyle \frac{1}{2}(2)(3) + 3 \cdot 3 + \frac{1}{2}(2)(3) = 15. Below it, \displaystyle \frac{1}{2}(1)(2) + 3 \cdot 2 + \frac{1}{2}(1)(2) = 8. So
The ride ends at 10 + 7 = 17 meters. It rises until t = 7 and descends after, so the greatest height is 10 + 15 = 25 meters. The average velocity is \displaystyle \frac{7}{12} \approx 0.58 meters per second; the average speed is \displaystyle \frac{23}{12} \approx 1.92 meters per second.
Day 2: Kahurangi’s elevator from a table
A sensor in Kahurangi’s elevator logs its velocity every 2 seconds, up positive:
| t (s) | v(t) (m/s) |
|---|---|
| 0 | 0 |
| 2 | 1.8 |
| 4 | 2.4 |
| 6 | 1.0 |
| 8 | -0.6 |
| 10 | -1.5 |
| 12 | 0 |
Estimate the displacement and the distance with trapezoids, assuming v changes linearly between readings, and give the average velocity and average speed.
SolutionEach trapezoid has width 2, so the trapezoid sum for \int_0^{12} v(t)\,dt is
For distance, use speeds. On the five steps with no sign change the trapezoids of \lvert v \rvert are 1.8, 4.2, 3.4, 2.1 and 1.5, totaling 13.0. From t = 6 to t = 8 the reading goes from 1.0 to -0.6, so the elevator stopped and reversed inside the step. On a line from 1.0 to -0.6 the zero is \displaystyle \frac{1.0}{1.6} of the way along, at t = 7.25, leaving two triangles:
The distance estimate is 13.0 + 0.85 = 13.85 meters. The plain trapezoid \displaystyle \frac{2}{2}(1.0 + 0.6) = 1.6 on that step would give 14.6, counting the stretch around the stop as if the elevator kept moving. The average velocity is about \displaystyle \frac{6.2}{12} \approx 0.52 meters per second and the average speed about \displaystyle \frac{13.85}{12} \approx 1.15 meters per second.
Boundaries
- \int_a^b v(t)\,dt = s(b) - s(a) needs s to be an antiderivative of v on all of [a, b], as the second fundamental theorem requires. It gives a change, never a position: a position needs one known value of s.
- \int_a^b \lvert v(t) \rvert\,dt must be split at every point where v changes sign; a missed one gives a wrong distance. A zero where v touches 0 without changing sign may be split at or not, with the same result.
- The free-fall formulas assume constant acceleration: near the ground, no air resistance, and only while the object is in flight. They say nothing after it lands, and they do not describe a falling leaf or a parachute.
- A table only estimates. A turn between two readings of the same sign leaves no trace in the table, and the linear model across a sign change is an assumption, not a measurement.
Where this goes next
Position is an accumulation function of velocity, s(t) = s(a) + \int_a^t v(u)\,du, so the highest or farthest point of a motion is found as in Graphing an accumulation function: where velocity crosses from positive to negative. Free fall is the first of many motions found by solving an initial value problem; others, with acceleration that depends on position or velocity (a spring, a falling object with air resistance), need differential equations that cannot be solved by antidifferentiating alone. The same split between net and total, signed area against area, returns whenever a quantity can flow both ways.
Exercises
Day 1: displacement and distance from velocity
- A particle has velocity v(t) = 2t - 6 meters per second for 0 \le t \le 5. Find its displacement and the total distance it travels.
- A particle has velocity v(t) = t^3 - 4t for 0 \le t \le 3. Find its displacement and total distance.
- A particle has velocity v(t) = 2\cos t - 1 for 0 \le t \le \pi. Find its displacement and total distance.
- A particle has velocity v(t) = 4 - e^t for 0 \le t \le 2. Find its displacement and total distance, exactly and to three decimal places.
Day 1: position, speed and initial value problems
- A particle has velocity v(t) = 3t^2 + 4t and is at s(1) = 4. Find s(t) and s(3).
- Kahurangi’s elevator starts from rest 6 meters above the lobby with velocity v(t) = 0.12t(10 - t) meters per second for 0 \le t \le 10. How high does it end? What is its greatest speed?
- A particle has a(t) = 6t + 4, v(0) = -15 and s(0) = 0. Find v(t) and s(t), where it turns, and its displacement and total distance for 0 \le t \le 3.
- For t \ge 1 a particle has \displaystyle a(t) = \frac{6}{t^3} and v(1) = -2. Find v(t), check that it solves the initial value problem, and find the displacement and total distance for 1 \le t \le 4.
- A particle has v(t) = t^2 - 6t + 5 for 0 \le t \le 6. Find its speed at t = 2 and its greatest speed on the interval.
Day 1: reasoning
- Explain why \int_a^b \lvert v(t) \rvert\,dt \ge \left\lvert \int_a^b v(t)\,dt \right\rvert for a continuous v, and when the two are equal.
- Show that v'(t) = a(t), v(t_0) = v_0 has only one solution on an interval containing t_0.
Day 2: free fall
- A stone is dropped from rest from a bridge 122.5 meters above a river. When does it reach the water, and how fast is it going?
- Lagi throws a ball straight up at 48 feet per second from a ledge 64 feet above the ground. Find its greatest height, when it lands, its impact velocity, and the total distance it travels.
- How fast must a ball be thrown straight up to rise 20 meters, and how long does it take to reach the top?
- Tavake’s elevator is rising at 2 meters per second when a bolt comes loose from its floor 30 meters above the bottom of the shaft. The bolt starts with the elevator’s velocity. When does it reach the bottom, and how fast?
Day 2: graphs and tables
- A particle’s velocity is the broken line joining (0, 3), (2, 3), (5, -3), (8, -3) and (9, 0), and s(0) = 2. Find the displacement, the total distance, s(9), and the greatest position.
- A particle’s velocity readings are below. Estimate the displacement with trapezoids, and the distance assuming v changes linearly between readings.
| t | v(t) |
|---|---|
| 0 | 4 |
| 1 | 3 |
| 2 | 1 |
| 3 | -2 |
| 4 | -3 |
| 5 | -1 |
| 6 | 2 |
- Velocity readings of a particle are v = 2, 5, -1, -4 at t = 0, 3, 5, 9 seconds. Use right Riemann sums to estimate the displacement and the distance.
Day 2: average velocity and average speed
- For v(t) = t^2 - 2t - 3 meters per second on 0 \le t \le 4, find the average velocity and the average speed.
Day 2: reasoning
- A ball is thrown straight up at v_0 from height s_0. Show that it returns to height s_0 at twice the time it takes to reach the top, moving at speed v_0.
- Explain why average speed over an interval is never less than the size of the average velocity.
- A stone is dropped from rest, so its acceleration is -9.8 meters per second squared. Show that its speed increases as it falls, and explain why using the signs of v and a.
Answers
- Displacement -5 meters, distance 13 meters. (t^2 - 6t)\Big|_0^5 = -5; v changes sign at 3, and the pieces are -9 and 4.
- Displacement \displaystyle \frac{9}{4}, distance \displaystyle \frac{41}{4}. With \displaystyle F(t) = \frac{t^4}{4} - 2t^2: F(0) = 0, F(2) = -4, \displaystyle F(3) = \frac{9}{4}; pieces -4 and \displaystyle \frac{25}{4}.
- Displacement -\pi, distance \displaystyle 2\sqrt{3} + \frac{\pi}{3} \approx 4.511. v = 0 at \displaystyle \frac{\pi}{3}; with F(t) = 2\sin t - t, the pieces are \displaystyle \sqrt{3} - \frac{\pi}{3} and \displaystyle -\frac{2\pi}{3} - \sqrt{3}.
- Displacement 9 - e^2 \approx 1.611, distance 8\ln 4 + e^2 - 15 \approx 3.479. v = 0 at \ln 4; with F(t) = 4t - e^t the pieces are 4\ln 4 - 3 and 12 - e^2 - 4\ln 4, the second negative.
- s(t) = t^3 + 2t^2 + 1, s(3) = 46. s(3) = 4 + \left(t^3 + 2t^2\right)\Big|_1^3 = 4 + 45 - 3.
- 26 meters; 3 meters per second. v \ge 0, so it rises \displaystyle \int_0^{10} 0.12\left(10t - t^2\right)dt = 0.12 \cdot \frac{500}{3} = 20. The parabola v peaks at t = 5, where v = 3.
- v(t) = 3t^2 + 4t - 15 = (3t - 5)(t + 3), s(t) = t^3 + 2t^2 - 15t. It turns at \displaystyle t = \frac{5}{3}, where \displaystyle s = -\frac{400}{27}. Displacement s(3) - s(0) = 0, distance \displaystyle \frac{800}{27} \approx 29.63.
- \displaystyle v(t) = 1 - \frac{3}{t^2}: \displaystyle v'(t) = \frac{6}{t^3} and v(1) = -2. Displacement \displaystyle \frac{3}{4}, distance \displaystyle \frac{35}{4} - 4\sqrt{3} \approx 1.822. v = 0 at \displaystyle \sqrt{3}; with \displaystyle G(t) = t + \frac{3}{t}, G(1) = 4, \displaystyle G(\sqrt{3}) = 2\sqrt{3}, \displaystyle G(4) = \frac{19}{4}.
- Speed 3 at t = 2; greatest speed 5, at t = 0 and t = 6. v(2) = -3; v ranges from its minimum -4 at t = 3 up to 5 at the ends.
- Since -\lvert v \rvert \le v \le \lvert v \rvert, integrating gives -\int \lvert v \rvert \le \int v \le \int \lvert v \rvert. Equality holds exactly when v \ge 0 on all of [a, b] or v \le 0 on all of it, that is, when the motion never reverses.
- If v_1 and v_2 both solve it, (v_1 - v_2)' = a - a = 0 on the interval, so by the mean value theorem v_1 - v_2 is a constant, and at t_0 it is v_0 - v_0 = 0.
- After 5 seconds, at 49 meters per second. 4.9t^2 = 122.5 gives t^2 = 25; then \lvert v \rvert = 9.8 \cdot 5.
- Greatest height 100 feet at t = 1.5; lands at t = 4 with velocity -80 feet per second; distance 136 feet. s(t) = 64 + 48t - 16t^2, v(t) = 48 - 32t; s = 0 gives t^2 - 3t - 4 = 0; it rises 36 feet and falls 100.
- \displaystyle 14\sqrt{2} \approx 19.80 meters per second, reaching the top after \displaystyle \frac{10\sqrt{2}}{7} \approx 2.02 seconds. The top is at \displaystyle t = \frac{v_0}{9.8}, with rise \displaystyle \frac{v_0^2}{19.6} = 20, so v_0^2 = 392.
- At \displaystyle t = \frac{2 + 4\sqrt{37}}{9.8} \approx 2.687 seconds, at \displaystyle 4\sqrt{37} \approx 24.33 meters per second. s(t) = 30 + 2t - 4.9t^2 = 0, and \displaystyle v = 2 - 9.8t = -4\sqrt{37} there.
- Displacement -4.5, distance 21, s(9) = -2.5, greatest position 10.25 at t = 3.5. Areas: 6, then 2.25 and -2.25 on either side of t = 3.5, then -9 and -1.5.
- Displacement about 1, distance about \displaystyle \frac{35}{3} \approx 11.67. Trapezoids of v: 3.5 + 2 - 0.5 - 2.5 - 2 + 0.5 = 1. For distance, the steps [2, 3] and [5, 6] change sign and give \displaystyle \frac{1 + 4}{2 \cdot 3} = \frac{5}{6} each; the others give 3.5 + 2 + 2.5 + 2 = 10.
- Displacement about -3 meters, distance about 33 meters. 3(5) + 2(-1) + 4(-4) = -3 and 3(5) + 2(1) + 4(4) = 33. The particle turns somewhere in (3, 5), which the sum cannot locate.
- Average velocity \displaystyle -\frac{5}{3}, average speed \displaystyle \frac{17}{6} meters per second. With \displaystyle F(t) = \frac{t^3}{3} - t^2 - 3t: F(3) = -9, \displaystyle F(4) = -\frac{20}{3}; displacement \displaystyle -\frac{20}{3}, distance \displaystyle 9 + \frac{7}{3} = \frac{34}{3}, each divided by 4.
- \displaystyle s(t) - s_0 = v_0 t - \frac{1}{2}gt^2 = t\left(v_0 - \frac{1}{2}gt\right) is 0 again at \displaystyle t = \frac{2v_0}{g}, twice the time \displaystyle \frac{v_0}{g} at which v = 0. There v = v_0 - 2v_0 = -v_0, speed v_0.
- Average speed is \displaystyle \frac{1}{b - a}\int_a^b \lvert v \rvert\,dt and average velocity \displaystyle \frac{1}{b - a}\int_a^b v\,dt; by Exercise 10 the first is at least the size of the second.
- Dropped from rest, v(t) = -9.8t < 0 for t > 0. Velocity and acceleration have the same sign, so the stone speeds up: its speed 9.8t grows by 9.8 meters per second each second. A negative acceleration makes the velocity decrease, and a negative velocity that decreases grows in size, so the stone moves downward faster and faster.