Uniform motion: speed, position and graphs
Why can the position of the Voyager 2 probe be calculated in advance for any day, even though it is more than 20 billion kilometres from the Sun?
Prior knowledge
- A straight line is fitted to measurement points, and its slope tells you how much $y$ changes when $x$ increases by onechapter Graphs and fitting a straight line
- In the equation of a straight line, the constant term is where the line crosses the $y$-axischapter Graphs and fitting a straight line
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 3.
Think about it
The Voyager 2 probe left the Earth in 1977. In November 2018 it crossed the heliopause, the boundary beyond which the Sun's solar wind no longer dominates space. According to NASA's JPL, the probe was then 119.7 AU from the Sun, and it is still moving away at a speed of $15.341\,\text{km/s}$. The astronomical unit AU is roughly the distance between the Earth and the Sun. You cannot hold a ruler next to the probe, but you can still give its distance for any day.
What information do you need to calculate the distance of the probe from the Sun, say a year from now, and what do you assume about its speed?
Optional Recap: the slope of a line
Predict
Make your predictions before investigating 2 questions
Prediction 1: the shape of the graph
Prediction 2: the time doubles
Investigate
A trolley moves along a straight track at constant speed. Change the initial position and the speed, start the motion or drag the moment, and see how the $(t, s)$ graph is drawn.
Simulation
How to use
Set the initial position $s_0$ and the speed $v$ with the sliders; the time returns to zero. Start plays the motion, Pause stops it and Reset returns to the beginning. You can also drag the Moment slider. On the track, the orange vertical lines are the positions of the trolley at whole seconds, the green flag is the initial position and the arrow shows the speed. A straight line is drawn in the graph up to the moment you have reached. When you have reached the moment $6\,\text{s}$, a slope triangle appears in the graph. Speed (replay) changes only the pace of the replay.
Assumptions of the model
The trolley moves along a straight track at constant speed: no friction and no acceleration, and the full speed is reached right at the start. The positions on the track and the values in the graph are values calculated by the model, not measurements. The trolley moves in one direction only, so the speed is positive.
How the calculation works
The position is $s = s_0 + vt$ and the distance travelled is $\Delta s = s - s_0 = vt$. The slope triangle in the graph goes from the moment $t = 2\,\text{s}$ to the moment $t = 6\,\text{s}$, so $\Delta t = 4.0\,\text{s}$, and the slope $\Delta s : \Delta t$ is the speed $v$.
Experiment 1: starting from the zero point
Test prediction 1. The initial position $s_0 = 0$ and the speed $v = 2.0\,\text{m/s}$ stay constant. Change only the moment $t$ and record the position at at least three moments $t > 0$, for example $2.0\,\text{s}$, $4.0\,\text{s}$ and $8.0\,\text{s}$. For each row, calculate the quotient $s : t$. Also look at how the graph is drawn.
Initial state: initial position = $0\,\text{m}$, speed = $2\,\text{m/s}$, moment = $0\,\text{s}$. Kept constant: initial position, speed. Varied: moment.
| moment | $s$ $(\text{m})$ |
|---|
Experiment 2: starting from 6.0 metres
Test prediction 2. Now the initial position is $s_0 = 6.0\,\text{m}$ and the speed is still $2.0\,\text{m/s}$. Change only the moment $t$ and record the position $s$ and the distance $\Delta s$ at at least three moments $t > 0$, for example $2.0\,\text{s}$, $4.0\,\text{s}$ and $8.0\,\text{s}$. For both columns, calculate the quotient (position or distance) divided by the time $t$.
Initial state: initial position = $6\,\text{m}$, speed = $2\,\text{m/s}$, moment = $0\,\text{s}$. Kept constant: initial position, speed. Varied: moment.
| moment | $s$ $(\text{m})$ | $\Delta s$ $(\text{m})$ |
|---|
Conclusion: when is the distance proportional to time?
Write, based on your tables: when is the position directly proportional to time, and when only the distance? What does the quotient $\Delta s : t$ tell you about the motion?
Model answer
In experiment 1 the initial position was zero, and the quotient of position and time $s : t$ was $2.0\,\text{m/s}$ in every row, which is the speed: the position was directly proportional to time. In experiment 2 the initial position was $6.0\,\text{m}$. The quotient of the position changed, for example $10 : 2.0 = 5.0\,\text{m/s}$ and $14 : 4.0 = 3.5\,\text{m/s}$, but the quotient of the distance $\Delta s : t$ was $2.0\,\text{m/s}$ in every row. So the distance travelled is always directly proportional to time, and the position only when the initial position is zero. The quotient $\Delta s : t$ is the speed. The values are values calculated by the model: the model assumes constant speed and no friction.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Uniform motion is motion along a straight line at constant speed: in equal intervals of time, the object travels equal distances. Speed tells you how far the object travels in one second, so its unit is metres per second (m/s). If a trolley moves two metres every second, its speed is $2\,\text{m/s}$.
Direct proportionality
Two quantities are directly proportional if their ratio is always the same. When uniform motion starts from the zero point, the position $s$, which here equals the distance travelled, and the time $t$ are directly proportional: double the time gives double the distance, and the ratio $s : t$ is the speed $v$. In a $(t, s)$ graph this shows as a straight line through the origin.
Linear model
Uniform motion is a model, not reality. A cyclist pedals unevenly, and Voyager 2 slows down slightly because of the Sun's gravity. The model is still good when the speed changes only a little during the time being considered. Its graph is a straight line, and a straight line is described by a linear model: the position changes by the same amount every second. Values calculated with the model are model values, which are compared with measurements.
Uniform motion
Suppose that at the start of the observation ($t = 0$) an object is at position $s_0$ and moves along a straight line at constant speed $v$. Its position at time $t$ is
$$s = s_0 + v t$$
where $s$ is the position (m) measured from the chosen zero point, $s_0$ the initial position (m), $v$ the speed (m/s) and $t$ the time from the start (s). The distance travelled by the object is $\Delta s = s - s_0 = vt$. In this chapter the object moves in one direction, so the speed is positive.
Speed from a graph
The $(t, s)$ graph is a straight line, and the slope of the line is the speed:
$$v = \frac{\Delta s}{\Delta t}$$
where $\Delta s$ is the change in position (m) and $\Delta t$ is its time interval (s). The line crosses the $s$-axis at the initial position $s_0$. If $s_0 = 0$, the position is directly proportional to time: $s = vt$. If $s_0 \neq 0$, the line is equally steep but shifted up or down. Then only the distance $\Delta s$, not the position $s$, is directly proportional to time.
Units
Always convert to compatible units (usually SI units) before substituting: substitute the value of the unit in its place. For example $1\,\text{km} = 1000\,\text{m}$ and $1\,\text{h} = 3600\,\text{s}$, so $72\,\text{km/h} = \dfrac{72\cdot1000\,\text{m}}{3600\,\text{s}} = 20\,\text{m/s}$. In the same way $1\,\text{min} = 60\,\text{s}$. In astronomy, long distances are given in astronomical units: $1\,\text{AU} = 149\,597\,870.7\,\text{km}$ is roughly the mean distance between the Earth and the Sun.
Common misconception
Many people think that in uniform motion the position doubles when the time doubles. This happens only if the motion starts from the zero point. When a trolley starts at $s_0 = 6\,\text{m}$ and moves at $2.0\,\text{m/s}$, it is at $10\,\text{m}$ at $2.0\,\text{s}$ and at $14\,\text{m}$ at $4.0\,\text{s}$, not at $20\,\text{m}$. The distance, however, does double: $4\,\text{m}$ becomes $8\,\text{m}$. Another mistake is to read the value of the position as the speed instead of the change in position per unit of time: the speed is $\Delta s : \Delta t$, not $s : t$, when $s_0 \neq 0$.
Summary
- In uniform motion the speed is constant, and the object travels equal distances in equal intervals of time: $s = s_0 + vt$.
- The distance travelled $\Delta s = vt$ is directly proportional to time. The position $s$ is proportional to time only when $s_0 = 0$.
- The $(t, s)$ graph is a straight line. Its slope $\Delta s : \Delta t$ is the speed, and the intercept on the $s$-axis is the initial position $s_0$.
- Convert units before calculating by substitution, for example $1\,\text{km} = 1000\,\text{m}$ and $1\,\text{h} = 3600\,\text{s}$.
- Uniform motion is a linear model. The values it gives are model values, which can be compared with measurements, such as the distance of Voyager 2.
Optional Back to your predictions
Compare your predictions with the experiments. Then return to the question at the start: what information is needed to calculate the distance of Voyager 2 for a future day, and what is assumed about its speed?
Would you change your prediction?
Apply
Worked example
Problem
Voyager 2 crossed the heliopause on 5 November 2018 at a distance of $s_0 = 119.7\,\text{AU}$ from the Sun (NASA/JPL). The speed of the probe relative to the Sun is $v = 15.341\,\text{km/s}$ (JPL Voyager mission status). Assume the speed is constant. According to the model, how far from the Sun is the probe on 5 November 2026?
Step 1/7
Given and find. $s_0 = 119.7\,\text{AU}$ at time $t = 0$ (5 November 2018), $v = 15.341\,\text{km/s}$ and $1\,\text{AU} = 149\,597\,870.7\,\text{km}$. The time interval from 5 November 2018 to 5 November 2026 is 8 years, including the leap days of 2020 and 2024: $8 \cdot 365 + 2 = 2922$ days. The quantity to find is the distance $s$ from the Sun.
Step 2/7
Principle. When the speed is assumed constant, the probe moves uniformly and its position follows the linear model. The result is a model value, not a measurement: the Sun's gravity slows the probe slightly.
Step 3/7
Rearranged formula. $$s = s_0 + v t$$
Step 4/7
Unit conversions. The time is already in days, $t = 2922\,\text{days}$, so the speed is converted to AU/day:
and dividing by $149\,597\,870.7\,\text{km/AU}$ gives $v = 0.0088602\,\tfrac{\text{AU}}{\text{day}}$.
Step 5/7
Substitution. $$\begin{aligned} s &= s_0 + v t \\ &= 119.7\,\text{AU} \\ &\quad + 0.0088602\,\tfrac{\text{AU}}{\text{day}} \cdot 2922\,\text{days} \\ &= 119.7\,\text{AU} + 25.89\,\text{AU} \\ &= 145.59\,\text{AU} \end{aligned}$$
Step 6/7
Result. According to the model, the probe is $s \approx 145.6\,\text{AU}$ from the Sun, about $2.18 \cdot 10^{10}\,\text{km}$. The initial distance is given to the nearest tenth, so the result is rounded to the nearest tenth.
Step 7/7
Reasonableness. A speed of $15\,\text{km/s}$ covers in a year about
so in 8 years the probe advances about 26 AU and $119.7 + 26 \approx 146\,\text{AU}$. For 15 February 2026 the same model gives $143.3\,\text{AU}$, and JPL reports $143.05\,\text{AU}$ for February 2026. The difference is about $0.2\,\text{AU}$, less than $0.2\,\%$ of the distance, so the constant-speed model works well here. The difference is partly because the Sun's gravity slows the probe very slightly and the initial distance is given only to the nearest tenth. So the model is accurate to a fraction of a percent.
Find the error
Apply to a new situation
At time $0\,\text{s}$ a cyclist is already $120\,\text{m}$ from the starting line and rides away from it uniformly at $21.6\,\text{km/h}$.
How far from the starting line is the cyclist $45\,\text{s}$ later? Give the answer in metres.
Hint 1: a guiding question
What is the initial position, and what is the speed? Are the units compatible?
Hint 2: what is the connection
Use the formula $s = s_0 + vt$. First convert the speed to m/s by substituting $1\,\text{km} = 1000\,\text{m}$ and $1\,\text{h} = 3600\,\text{s}$.
Hint 3: an intermediate step
and $s = 120\,\text{m} + 6.0\,\tfrac{\text{m}}{\text{s}} \cdot 45\,\text{s}$.
Show the solution
Given
initial position $s_0 = 120\,\text{m}$, speed $v = 21.6\,\text{km/h}$, time $t = 45\,\text{s}$
Find
position $s$ from the starting line in metres
Principle
The speed of the cyclist is constant, so the motion is uniform, and the position follows the linear model. The initial position is not zero, so it must be added to the distance travelled.
Formula
$s = s_0 + vt$
Conversions
Substitute $1\,\text{km} = 1000\,\text{m}$ and $1\,\text{h} = 3600\,\text{s}$:
The time and the initial position are already in SI units.
Substitution
Answer
$390\,\text{m}$
Two significant figures as in the time $45\,\text{s}$.
Reasonableness
The speed $6\,\text{m/s}$ is about $22\,\text{km/h}$, and $45\,\text{s}$ is three quarters of a minute, so the distance travelled is about $6 \cdot 45 \approx 270\,\text{m}$. The initial position $120\,\text{m}$ is added to this, and the position is greater than the initial position, as it should be when the cyclist moves away.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A train travels uniformly at $90\,\text{km/h}$. How far does it travel in $20\,\text{min}$?
Answer
The time in hours is
$t$$= 20\,\text{min}$$= \tfrac{1}{3}\,\text{h}$so
$s$$= vt$$= 90\,\tfrac{\text{km}}{\text{h}} \cdot \tfrac{1}{3}\,\text{h}$$= 30\,\text{km}$The units fit together when the time is in hours.
1. Concept
2. Interpreting a graph
3. Reading a graph
The line of a $(t, s)$ graph of uniform motion passes through the points $(1.0\,\text{s}$; $7.0\,\text{m})$ and $(5.0\,\text{s}$; $19\,\text{m})$.
What is the initial position $s_0$ of the motion? Give the answer in metres.
Hint 1: a guiding question
The initial position is the position at time $t = 0$, that is, the point where the line crosses the $s$-axis. What is the slope of the line?
Hint 2: what is the connection
The slope is $v = \dfrac{s_2 - s_1}{t_2 - t_1}$. You get the initial position by going back in time from the point $(t_1, s_1)$ to the start: $s_0 = s_1 - v t_1$.
Hint 3: an intermediate step
and $s_0 = 7.0\,\text{m} - 3.0\,\text{m/s} \cdot 1.0\,\text{s}$.
Show the solution
Given
points $(t_1, s_1) = (1.0\,\text{s}$; $7.0\,\text{m})$ and $(t_2, s_2) = (5.0\,\text{s}$; $19\,\text{m})$ on the same line
Find
initial position $s_0$, that is, the position at time $t = 0$, in metres
Principle
The motion is uniform, so the line of the $(t, s)$ graph follows the equation $s = s_0 + vt$. The slope of the line is the speed, and the point where the line crosses the $s$-axis is the initial position.
Reading the graph
The two given points are enough. The change in position between them is divided by the time interval, which gives the slope. Then we go back from the first point to the moment $t = 0$: in one second the position decreases by $v \cdot 1.0\,\text{s}$.
Formula
$v = \dfrac{s_2 - s_1}{t_2 - t_1}$ and $s_1 = s_0 + v t_1$, so $s_0 = s_1 - v t_1$
Conversions
None needed: the positions are in metres and the times in seconds.
Substitution
Answer
$4.0\,\text{m}$
Two significant figures as in the positions $7.0\,\text{m}$ and $19\,\text{m}$.
Reasonableness
Check with the second point:
which is the given value. The initial position is smaller than $7.0\,\text{m}$, as it should be, because the object moves forwards.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 4. Explain and evaluate the model
Explain why the distance of the Voyager 2 probe from the Sun can be calculated with a linear model, and why the calculated value is a model value and not a measurement.
Model answer
The speed of the probe is almost constant, so its distance from the Sun grows by the same amount every second: $s = s_0 + vt$, where $s_0 = 119.7\,\text{AU}$ at the moment of the heliopause crossing on 5 November 2018 and $v = 15.341\,\text{km/s}$. With this model the distance on 5 November 2026 is about $145.6\,\text{AU}$. The value is a model value, because the speed is not exactly constant: the Sun's gravity slows the probe slightly. For 15 February 2026 the model gives $143.3\,\text{AU}$, while JPL reports $143.05\,\text{AU}$ for February 2026. The difference is about $0.2\,\text{AU}$. The difference is partly because the Sun's gravity slows the probe slightly and the initial distance is given only to the nearest tenth. The model is accurate to a fraction of a percent, so it is good for studying a few years.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Direct proportionality
- When one quantity doubles, so does the other, and the graph passes through the origin.
- Position
- Where an object is in a chosen coordinate system at a given instant.
- Uniform motion
- The position changes by equal amounts in equal time intervals.
- Velocity
- The change in position per unit time, together with the direction of motion.
- Average velocity
- The whole displacement divided by the time it took.
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 5-1Core
A cyclist rides uniformly at $5.5\,\text{m/s}$. How far does the cyclist travel in 20 seconds?
Answer
$110\,\text{m}$
Show the solution
Given
speed $v = 5.5\,\text{m/s}$, time $t = 20\,\text{s}$; the motion starts from the zero point, so $s_0 = 0$
Find
distance travelled $s$ in metres
Principle
The speed is constant, so the motion is uniform, and the distance is directly proportional to time.
Formula
$s = s_0 + vt$, and since $s_0 = 0$, $s = vt$
Conversions
None needed: the speed is in m/s and the time in seconds, so the product is in metres.
Substitution
$s$$= 5.5\,\tfrac{\text{m}}{\text{s}} \cdot 20\,\text{s}$$= 110\,\text{m}$Answer
$110\,\text{m}$
Two significant figures as in the speed $5.5\,\text{m/s}$.
Reasonableness
About $5\,\text{m}$ every second for twenty seconds is about $5 \cdot 20 = 100\,\text{m}$, and the unit $\tfrac{\text{m}}{\text{s}} \cdot \text{s} = \text{m}$ is the unit of distance.
- Exercise 5-2Core
A runner covers $360\,\text{m}$ in $45\,\text{s}$ at a steady pace. What is the runner's speed?
Answer
$8.0\,\text{m/s}$
Show the solution
Given
distance $s = 360\,\text{m}$, time $t = 45\,\text{s}$; the motion is uniform and starts from the zero point
Find
speed $v$ in m/s
Principle
In uniform motion, the speed is the distance travelled divided by the time it took.
Formula
$s = vt$, so $v = \dfrac{s}{t}$
Conversions
None needed: the distance is in metres and the time in seconds.
Substitution
$v$$= \dfrac{360\,\text{m}}{45\,\text{s}}$$= 8.0\,\tfrac{\text{m}}{\text{s}}$Answer
$8.0\,\text{m/s}$
Two significant figures as in the time $45\,\text{s}$.
Reasonableness
Check backwards: $8.0\,\text{m/s} \cdot 45\,\text{s} = 360\,\text{m}$. The speed corresponds to about $29\,\text{km/h}$, which matches the pace of a good runner in a 400-metre race.
- Exercise 5-3
The graph shows the position of a walker as a function of time. The walker keeps moving uniformly in the same direction also beyond the graph.
How many seconds pass from the start until the walker has travelled $150\,\text{m}$?
Answer
$60\,\text{s}$
Show the solution
Given
a graph in which the position $s$ of a walker as a function of time $t$ is a straight line from the origin; the motion continues uniformly beyond the graph; distance $s = 150\,\text{m}$
Find
time $t$ in seconds when the distance is $150\,\text{m}$
Principle
A straight line from the origin means uniform motion, and the slope of the line is the speed. When the speed is known, the time beyond the graph is obtained by dividing the distance by the speed.
Reading the graph
The line passes through the grid crossing where $t_1 = 0\,\text{s}$ and $s_1 = 0\,\text{m}$ (the origin), and through the crossing where $t_2 = 8\,\text{s}$ and $s_2 = 20\,\text{m}$. Other crossings are for example $(4\,\text{s}$; $10\,\text{m})$ and $(10\,\text{s}$; $25\,\text{m})$. The reading is exact from the grid, so an answer that deviates by at most $3\,\%$ is accepted.
Formula
$v = \dfrac{\Delta s}{\Delta t}$ and $s = vt$, so $t = \dfrac{s}{v}$
Conversions
None needed: the distance is in metres and the time in seconds.
Substitution
$v$$= \dfrac{20\,\text{m} - 0\,\text{m}}{8\,\text{s} - 0\,\text{s}}$$= 2.5\,\dfrac{\text{m}}{\text{s}}$$t$$= \dfrac{150\,\text{m}}{2.5\,\frac{\text{m}}{\text{s}}}$$= 60\,\text{s}$Answer
$60\,\text{s}$
That is one minute; the speed was read with two significant figures, so the answer is given with the same precision.
Reasonableness
The speed $2.5\,\text{m/s}$ is about $9\,\text{km/h}$, a brisk walk or a slow jog. In the graph $10\,\text{s}$ takes $25\,\text{m}$, so $150\,\text{m}$, a distance six times as long, takes six times as long, $60\,\text{s}$.
- Exercise 5-4
An object moves uniformly from its starting point and travels $18\,\text{m}$ in $4.0\,\text{s}$. How far does it travel in $10\,\text{s}$?
Answer
$45\,\text{m}$
Show the solution
Given
$s_1 = 18\,\text{m}$ in time $t_1 = 4.0\,\text{s}$; the time asked about is $t_2 = 10\,\text{s}$; the motion starts from the starting point, so $s_0 = 0$
Find
distance $s_2$ in time $t_2$ in metres
Principle
The motion is uniform and starts from the zero point, so the distance is directly proportional to time: the ratio $s : t$ is the same for both times.
Formula
$\dfrac{s_2}{t_2} = \dfrac{s_1}{t_1}$, so $s_2 = s_1\,\dfrac{t_2}{t_1}$
Conversions
None needed: the times are in seconds, and the seconds cancel.
Substitution
$s_2$$= 18\,\text{m} \cdot \dfrac{10\,\text{s}}{4.0\,\text{s}}$$= 45\,\text{m}$Answer
$45\,\text{m}$
Two significant figures as in the distance $18\,\text{m}$.
Reasonableness
The time $10\,\text{s}$ is $2.5$ times the time $4.0\,\text{s}$, so the distance must also be $2.5$ times as large: $2.5 \cdot 18\,\text{m} = 45\,\text{m}$. The speed is $18\,\text{m} : 4.0\,\text{s} = 4.5\,\text{m/s}$, and $4.5\,\text{m/s} \cdot 10\,\text{s} = 45\,\text{m}$.
- Exercise 5-5
The position of a uniformly moving trolley, measured from the starting point, is $7.0\,\text{m}$ at $2.0\,\text{s}$ and $21\,\text{m}$ at $6.0\,\text{s}$. Calculate the speed of the trolley using the change in position and the time interval.
Answer
$3.5\,\text{m/s}$
Show the solution
Given
at time $t_1 = 2.0\,\text{s}$ the position is $s_1 = 7.0\,\text{m}$, and at time $t_2 = 6.0\,\text{s}$ the position is $s_2 = 21\,\text{m}$; the motion is uniform
Find
speed $v$ in m/s
Principle
In uniform motion the speed is constant, so it equals the change in position divided by the time interval. The initial position does not matter, because it cancels in the difference.
Formula
$v$$= \dfrac{\Delta s}{\Delta t}$$= \dfrac{s_2 - s_1}{t_2 - t_1}$Conversions
None needed: the positions are in metres and the times in seconds.
Substitution
$v$$= \dfrac{21\,\text{m} - 7.0\,\text{m}}{6.0\,\text{s} - 2.0\,\text{s}}$$= \dfrac{14\,\text{m}}{4.0\,\text{s}}$$= 3.5\,\tfrac{\text{m}}{\text{s}}$Answer
$3.5\,\text{m/s}$
Two significant figures as in the positions $7.0\,\text{m}$ and the times $2.0\,\text{s}$.
Reasonableness
The quotients $7.0\,\text{m} : 2.0\,\text{s}$ and $21\,\text{m} : 6.0\,\text{s}$ are both $3.5\,\text{m/s}$, so the position is proportional to time and the initial position is zero. The speed $3.5\,\text{m/s}$ corresponds to slow cycling, about $13\,\text{km/h}$.
- Exercise 5-6
The position $s$ of a walker has been measured as a function of time $t$, and the measurement points lie on a straight line.
What does the slope $\Delta s : \Delta t$ of the line tell?
Answer
The walker's speed.
Show the solution
Find
what the slope of the line in the $(t, s)$ graph tells about uniform motion.
Principle
In uniform motion, the $(t, s)$ graph is a straight line. The slope is the change in position divided by the time interval, $\Delta s : \Delta t$, and it is the speed $v$ of the object (unit m/s). The initial position $s_0$ is where the line crosses the $s$-axis, not the slope.
Model answer
The slope tells the speed of the walker. The steeper the line, the greater the speed. It is not the initial position (the intercept) or the total distance (just $\Delta s$), and the quotient $s : t$ is the speed only when the line passes through the origin.
- Exercise 5-7
In the $(t, v)$ graph, the speed of a uniformly moving trolley is a horizontal line at $v = 0.80\,\text{m/s}$ during the interval $0\,\text{s} - 12\,\text{s}$. The area of the rectangle between the line and the $t$-axis corresponds to the distance travelled.
How far does the trolley travel during this time?
Answer
$9.6\,\text{m}$
Show the solution
Given
speed $v = 0.80\,\text{m/s}$ constant during the interval $0\,\text{s} - 12\,\text{s}$, so the interval is $t = 12\,\text{s}$
Find
distance travelled $\Delta s$ in metres
Principle
In a $(t, v)$ graph, a horizontal line is the top edge of a rectangle. The height of the rectangle is the speed $v$ and the width is the time $t$, so its area $vt$ is the distance travelled in uniform motion.
Reading the graph
The height is $v = 0.80\,\text{m/s}$ and the width is
$t$$= 12\,\text{s} - 0\,\text{s}$$= 12\,\text{s}$Formula
$\Delta s = vt$ (area of the rectangle = height $\cdot$ width)
Conversions
None needed: the speed is in m/s and the time in seconds, and $\tfrac{\text{m}}{\text{s}} \cdot \text{s} = \text{m}$.
Substitution
$\Delta s$$= 0.80\,\tfrac{\text{m}}{\text{s}} \cdot 12\,\text{s}$$= 9.6\,\text{m}$Answer
$9.6\,\text{m}$
Two significant figures as in the speed $0.80\,\text{m/s}$ and the time $12\,\text{s}$.
Reasonableness
The speed $0.80\,\text{m/s}$ is less than a metre per second, so in $12\,\text{s}$ the distance is less than $12\,\text{m}$. The same result is obtained with the formula $s = vt$, which is the basic equation of uniform motion.
- Exercise 5-8
A cyclist travels uniformly $2.4\,\text{km}$ in $16\,\text{min}$. What is the cyclist's speed? Give the answer in m/s.
Answer
$2.5\,\text{m/s}$
Show the solution
Given
distance $s = 2.4\,\text{km}$, time $t = 16\,\text{min}$; the motion is uniform and starts from the zero point
Find
speed $v$ in m/s
Principle
In uniform motion, the speed is the distance travelled divided by the time it took. Kilometres and minutes are converted to metres and seconds before dividing.
Formula
$s = vt$, so $v = \dfrac{s}{t}$
Conversions
substituting $1\,\text{km} = 1000\,\text{m}$ and $1\,\text{min} = 60\,\text{s}$:
$s$$= 2.4 \cdot 1000\,\text{m}$$= 2400\,\text{m}$$t$$= 16 \cdot 60\,\text{s}$$= 960\,\text{s}$Substitution
$v$$= \dfrac{2400\,\text{m}}{960\,\text{s}}$$= 2.5\,\tfrac{\text{m}}{\text{s}}$Answer
$2.5\,\text{m/s}$
Two significant figures as in the data $2.4\,\text{km}$ and $16\,\text{min}$.
Reasonableness
Check backwards: $2.5\,\text{m/s} \cdot 960\,\text{s} = 2400\,\text{m}$. The speed is $9.0\,\text{km/h}$, which suits leisurely cycling.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 5-9Core
A car drives at a constant speed of $54\,\text{km/h}$. How far does it travel in 14 seconds? Give the answer in metres.
Answer
$210\,\text{m}$
Show the solution
Given
speed $v = 54\,\text{km/h}$, time $t = 14\,\text{s}$; the motion starts from the zero point
Find
distance $s$ in metres
Principle
The car drives at constant speed, so the motion is uniform and $s = vt$. The speed and the time must be in compatible units.
Formula
$s = vt$
Conversions
substitute $1\,\text{km} = 1000\,\text{m}$ and $1\,\text{h} = 3600\,\text{s}$:
$v$$= 54\,\text{km/h}$$= \dfrac{54\cdot1000\,\text{m}}{3600\,\text{s}}$$= 15\,\text{m/s}$Substitution
$s$$= 15\,\tfrac{\text{m}}{\text{s}} \cdot 14\,\text{s}$$= 210\,\text{m}$Answer
$210\,\text{m}$
Two significant figures as in the speed $54\,\text{km/h}$ and the time $14\,\text{s}$.
Reasonableness
The speed $54\,\text{km/h}$ is $0.9\,\text{km}$ per minute, and $14\,\text{s}$ is about $0.23\,\text{min}$, so the distance is about
$0.9 \cdot 0.23\,\text{km}$$\approx 0.21\,\text{km}$$= 210\,\text{m}$ - Exercise 5-10Core
A bus drives uniformly at $36\,\text{km/h}$. How many seconds does a distance of $4.5\,\text{km}$ take?
Answer
$450\,\text{s}$
Show the solution
Given
speed $v = 36\,\text{km/h}$, distance $s = 4.5\,\text{km}$; the motion is uniform and starts from the zero point
Find
time $t$ in seconds
Principle
The bus drives at constant speed, so the distance is proportional to time and the time is the distance divided by the speed.
Formula
$s = vt$, so $t = \dfrac{s}{v}$
Conversions
$v$$= 36\,\text{km/h}$$= \dfrac{36\cdot1000\,\text{m}}{3600\,\text{s}}$$= 10\,\text{m/s}$$s = 4.5\,\text{km} = 4500\,\text{m}$.
Substitution
$t$$= \dfrac{4500\,\text{m}}{10\,\tfrac{\text{m}}{\text{s}}}$$= 450\,\text{s}$Answer
$450\,\text{s}$
Two significant figures as in the data $36\,\text{km/h}$ and $4.5\,\text{km}$.
Reasonableness
In kilometres and hours:
$4.5\,\text{km} : 36\,\text{km/h}$$= 0.125\,\text{h}$$= 7.5\,\text{min}$$= 450\,\text{s}$which is the same result.
- Exercise 5-11
A hiker is at a point $1.5\,\text{km}$ from a cabin. The hiker walks along a straight path away from the cabin, uniformly at $1.5\,\text{m/s}$.
How far from the cabin is the hiker after 10 minutes? Give the answer in kilometres.
Answer
$2.4\,\text{km}$
Show the solution
Given
initial position $s_0 = 1.5\,\text{km}$ from the cabin, speed $v = 1.5\,\text{m/s}$ away from the cabin, time $t = 10\,\text{min}$
Find
distance from the cabin $s$ in kilometres
Principle
The hiker walks at constant speed, so the motion is uniform. The initial position is not zero, so the position is the initial position plus the distance travelled.
Formula
$s = s_0 + vt$
Conversions
$s_0 = 1.5\,\text{km} = 1500\,\text{m}$ and $t = 10\,\text{min} = 600\,\text{s}$. Finally $1\,\text{km} = 1000\,\text{m}$.
Substitution
$s$$= 1500\,\text{m} + 1.5\,\tfrac{\text{m}}{\text{s}} \cdot 600\,\text{s}$$= 1500\,\text{m} + 900\,\text{m}$$= 2400\,\text{m}$$= 2.4\,\text{km}$Answer
$2.4\,\text{km}$
Two significant figures as in the data $1.5\,\text{km}$ and $1.5\,\text{m/s}$.
Reasonableness
The speed $1.5\,\text{m/s}$ is about $5.4\,\text{km/h}$, so in ten minutes, which is one sixth of an hour, the hiker advances about $0.9\,\text{km}$. Added to the initial position, $1.5\,\text{km} + 0.9\,\text{km} = 2.4\,\text{km}$, and the result is greater than the initial position, as it should be.
- Exercise 5-12
After a flash of lightning, the rumble of thunder is heard $5.0\,\text{s}$ later. Sound travels in air uniformly at $343\,\text{m/s}$, and the travel time of light is so small that it is ignored.
How far away did the lightning strike? Give the answer in kilometres.
Answer
$1.7\,\text{km}$
Show the solution
Given
time $t = 5.0\,\text{s}$, speed of sound $v = 343\,\text{m/s}$; the travel time of light is ignored
Find
distance of the lightning $s$ in kilometres
Principle
Sound travels at constant speed, so the distance it travels is $s = vt$. Light arrives practically at once, so the delay of the thunder is the travel time of sound over the whole distance.
Formula
$s = vt$
Conversions
The distance is calculated in metres, and finally $1\,\text{km} = 1000\,\text{m}$.
Substitution
$s$$= 343\,\tfrac{\text{m}}{\text{s}} \cdot 5.0\,\text{s}$$= 1715\,\text{m}$$= 1.7\,\text{km}$Answer
$1.7\,\text{km}$
Two significant figures as in the time $5.0\,\text{s}$.
Reasonableness
Rule of thumb: three seconds corresponds to about a kilometre, so five seconds corresponds to about $1.7\,\text{km}$. Light would travel this distance in $1715\,\text{m} : (3.0 \cdot 10^{8}\,\text{m/s}) \approx 6\,\mu\text{s}$, which is negligible compared with the delay $5.0\,\text{s}$.
- Exercise 5-13
The motion of a walker was filmed, and the position $s$ of the walker from the starting mark was measured by video analysis as a function of time $t$. The walker moves uniformly, and the points deviate from a straight line because of measurement uncertainty.
Draw a graph of the points and fit a straight line to them. Determine the speed of the walker from the slope of the line and use it to predict the position of the walker at time $12\,\text{s}$. How many metres from the starting mark is the walker then? Justify why the prediction is a model value and not a measurement.
Position of the walker as a function of time $t\,(\text{s})$ $s\,(\text{m})$ 0.0 1.2 1.0 2.7 2.0 3.9 3.0 5.5 4.0 6.7 5.0 8.3 Answer
$18\,\text{m}$
Show the solution
Given
measurement points of the table $(t$; $s)$: $(0.0\,\text{s}$; $1.2\,\text{m})$, $(1.0\,\text{s}$; $2.7\,\text{m})$, $(2.0\,\text{s}$; $3.9\,\text{m})$, $(3.0\,\text{s}$; $5.5\,\text{m})$, $(4.0\,\text{s}$; $6.7\,\text{m})$ and $(5.0\,\text{s}$; $8.3\,\text{m})$; the walker moves uniformly
Find
position $s$ of the walker at time $t = 12\,\text{s}$ in metres
Principle
In uniform motion, the position is a linear function of time, $s = s_0 + vt$. In real measurement series, the measurement points usually do not lie exactly on the same straight line because of measurement uncertainty, so the line is fitted through the middle of the set of points and the speed is read along it.
Reading the graph
The points lie on a rising line with points on both sides of it. The fitted line passes approximately through the points $(0\,\text{s}$; $1.2\,\text{m})$ and $(5.0\,\text{s}$; $8.2\,\text{m})$. They are read from the line and chosen far apart. The initial position is $s_0 = 1.2\,\text{m}$, because the line crosses the $s$-axis there.
Formula
$v = \dfrac{\Delta s}{\Delta t}$ and $s = s_0 + vt$
Conversions
None needed: the positions are in metres and the times in seconds.
Substitution
$v$$= \dfrac{8.2\,\text{m} - 1.2\,\text{m}}{5.0\,\text{s} - 0\,\text{s}}$$= 1.4\,\tfrac{\text{m}}{\text{s}}$$s$$= 1.2\,\text{m} + 1.4\,\tfrac{\text{m}}{\text{s}} \cdot 12\,\text{s}$$= 1.2\,\text{m} + 16.8\,\text{m}$$= 18\,\text{m}$Answer
about $18\,\text{m}$, two significant figures. A fit by eye varies, so about $17.3\text{–}18.7\,\text{m}$ is a good answer. The prediction is a model value, not a measurement: the moment $12\,\text{s}$ is outside the measured range $0\,\text{s} - 5\,\text{s}$, and it is correct only if the walker continues at the same speed. This is an assumption that the measurement has not checked.
Reasonableness
The speed $1.4\,\text{m/s}$ is about $5\,\text{km/h}$, an ordinary walking pace. The prediction clearly exceeds the last measurement $8.3\,\text{m}$, as it should, because $12\,\text{s}$ is more than twice as long as the measured range.
- Exercise 5-14
A trolley moves uniformly on a track. It is initially at $s_0 = 8.0\,\text{m}$, and at time $2.0\,\text{s}$ it is at $11\,\text{m}$. A student reasons: “When the time triples to $6.0\,\text{s}$, the position triples, so the trolley is then at $33\,\text{m}$.” Find the student's error and calculate the correct position of the trolley at time $6.0\,\text{s}$.
Answer
$17\,\text{m}$
Show the solution
Given
initial position $s_0 = 8.0\,\text{m}$, position $s_1 = 11\,\text{m}$ at time $t_1 = 2.0\,\text{s}$; the student's conclusion is $s = 33\,\text{m}$ at time $t_2 = 6.0\,\text{s}$
Find
the student's error and the correct position $s$ at time $6.0\,\text{s}$ in metres
Principle
In uniform motion, only the distance travelled $\Delta s = vt$ is directly proportional to time. The position $s = s_0 + vt$ is proportional to time only when $s_0 = 0$. The student mixed up position and distance: the student tripled the position, although the distance should have been tripled.
Formula
$v = \dfrac{\Delta s}{\Delta t}$ and $s = s_0 + vt$
Conversions
None needed: the positions are in metres and the times in seconds.
Substitution
$v$$= \dfrac{11\,\text{m} - 8.0\,\text{m}}{2.0\,\text{s}}$$= 1.5\,\tfrac{\text{m}}{\text{s}}$$s$$= 8.0\,\text{m} + 1.5\,\tfrac{\text{m}}{\text{s}} \cdot 6.0\,\text{s}$$= 8.0\,\text{m} + 9.0\,\text{m}$$= 17\,\text{m}$Answer
The trolley is at $17\,\text{m}$, not at $33\,\text{m}$. The error is that the position was tripled: the initial position $s_0 = 8.0\,\text{m}$ is not zero, so only the distance triples ($3.0\,\text{m} \to 9.0\,\text{m}$), and the initial position must be added to it.
Reasonableness
The distance tripled: $3.0\,\text{m} \cdot 3 = 9.0\,\text{m}$. The student's answer $33\,\text{m}$ would require the trolley to travel $25\,\text{m}$ in six seconds, that is, a speed of $4.2\,\text{m/s}$, which is a different speed from the $1.5\,\text{m/s}$ obtained from the data.
- Exercise 5-15
A toy car drives uniformly at about $0.80\,\text{m/s}$. The travel time of the car is measured by hand, and the uncertainty of the timing is about $0.2\,\text{s}$ regardless of the length of the time measured.
You are planning a measurement of the speed in which the relative uncertainty of the time must be at most $2.0\,\%$. Assume that the distance is measured so accurately that its uncertainty does not affect the result.
What is the minimum distance the car must travel during the measurement? Explain why a long distance improves the measurement.
Answer
$8.0\,\text{m}$
Show the solution
Given
speed about $v = 0.80\,\text{m/s}$, uncertainty of the timing $u_t = 0.2\,\text{s}$ (the same for all times), limit of the relative uncertainty $2.0\,\%$
Find
minimum distance $s$ in metres
Principle
The relative uncertainty of the time is $u_t : t$. Since $u_t$ is constant, the relative uncertainty decreases as the measured time grows. The limit gives the minimum time, and in uniform motion the distance is speed times time.
Formula
$\dfrac{u_t}{t} \le 0.020$, so $t \ge \dfrac{u_t}{0.020}$, and $s = vt$
Conversions
None needed: the times are in seconds and the speed in m/s. The percentage is $2.0\,\% = 0.020$.
Substitution
$t$$\ge \dfrac{0.2\,\text{s}}{0.020}$$= 10\,\text{s}$$s$$= 0.80\,\tfrac{\text{m}}{\text{s}} \cdot 10\,\text{s}$$= 8.0\,\text{m}$Answer
The car must travel at least $8.0\,\text{m}$. A long distance improves the measurement, because it takes longer and the timing uncertainty $0.2\,\text{s}$ is then a smaller part of the whole time. On a short distance, for example $1.0\,\text{m}$ ($t = 1.25\,\text{s}$), the relative uncertainty would be $16\,\%$.
Reasonableness
Check: at the time $10\,\text{s}$ the relative uncertainty is
$0.2\,\text{s} : 10\,\text{s}$$= 0.020$$= 2.0\,\%$so the limit is met exactly. A distance of $8.0\,\text{m}$ fits in a school corridor.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 5-16
The Voyager 2 probe crossed the heliopause on 5 November 2018 at a distance of $119.7\,\text{AU}$ from the Sun (NASA/JPL), and its speed relative to the Sun is $15.341\,\text{km/s}$ (JPL). Assume the speed is constant.
How many days after the heliopause crossing is the probe, according to the model, $150\,\text{AU}$ from the Sun? Use the values $1\,\text{AU} = 149\,597\,870.7\,\text{km}$ and $1\,\text{day} = 86\,400\,\text{s}$. The result is a model value.
Answer
$3420\,\text{days}$
Show the solution
Given
$s_0 = 119.7\,\text{AU}$ at time $t = 0$ (5 November 2018), $v = 15.341\,\text{km/s}$, $s = 150\,\text{AU}$, $1\,\text{AU} = 149\,597\,870.7\,\text{km}$, $1\,\text{day} = 86\,400\,\text{s}$
Find
time $t$ in days when the probe is at a distance of $150\,\text{AU}$
Principle
The speed is assumed constant, so the uniform motion model $s = s_0 + vt$ applies. The result is a model value, because the Sun's gravity slows the probe slightly and the real time is therefore a little longer.
Formula
$s = s_0 + vt$, so $t = \dfrac{s - s_0}{v}$
Conversions
The speed in astronomical units per day:
$v$$= 15.341\,\tfrac{\text{km}}{\text{s}} \cdot 86\,400\,\tfrac{\text{s}}{\text{day}}$$= 1\,325\,462\,\tfrac{\text{km}}{\text{day}}$and
$\dfrac{1\,325\,462\,\text{km/day}}{149\,597\,870.7\,\text{km/AU}}$$= 0.0088602\,\tfrac{\text{AU}}{\text{day}}$Substitution
$t$$= \dfrac{150\,\text{AU} - 119.7\,\text{AU}}{0.0088602\,\text{AU/day}}$$= \dfrac{30.3\,\text{AU}}{0.0088602\,\text{AU/day}}$$= 3419.8\,\text{days}$Answer
$3420\,\text{days}$
Three significant figures, when $150\,\text{AU}$ is interpreted as accurate to three figures and the difference is $30.3\,\text{AU}$.
Reasonableness
The rate is about $3.2\,\text{AU}$ per year, so $30.3\,\text{AU}$ takes about $30.3 : 3.2 \approx 9.5$ years. The result $3420\,\text{days} \approx 9.4$ years fits this and falls in March 2028. The model differs from the JPL value by about 0.2 AU, which corresponds to an error of about 20 days in the date.
- Exercise 5-17
According to the model, Voyager 2 is at a distance of $150\,\text{AU}$ from the Earth. A radio signal travels in vacuum uniformly at $299\,792\,458\,\text{m/s}$.
How many hours pass from the moment a command leaves the Earth to the moment the reply of the probe reaches the Earth? The motion of the probe and its processing time for the reply are ignored. Use the value $1\,\text{AU} = 149\,597\,870.7\,\text{km}$. Justify why the motion can be ignored.
Answer
$41.6\,\text{h}$
Show the solution
Given
distance $d = 150\,\text{AU}$, speed of the signal $c = 299\,792\,458\,\text{m/s}$, $1\,\text{AU} = 149\,597\,870.7\,\text{km}$
Find
round-trip travel time $t$ in hours
Principle
A radio signal travels in vacuum at constant speed, so the distance it travels is $s = ct$. The command and the reply each travel the distance $d$, so the total distance is $2d$. During this time the probe moves about $15\,\text{km/s} \cdot 1.5 \cdot 10^{5}\,\text{s} \approx 2.3 \cdot 10^{6}\,\text{km} \approx 0.015\,\text{AU}$, which is only $0.01\,\%$ of the distance $150\,\text{AU}$. In the same time the Earth moves about $30\,\text{km/s} \cdot 1.5 \cdot 10^{5}\,\text{s} \approx 0.03\,\text{AU}$, only about $0.02\,\%$, so neither motion needs to be taken into account.
Formula
$2d = ct$, so $t = \dfrac{2d}{c}$
Conversions
$d$$= 150 \cdot 149\,597\,870.7\,\text{km}$$= 2.2440 \cdot 10^{10}\,\text{km}$$= 2.2440 \cdot 10^{13}\,\text{m}$Finally $1\,\text{h} = 3600\,\text{s}$.
Substitution
$t$$= \dfrac{2 \cdot 2.2440 \cdot 10^{13}\,\text{m}}{299\,792\,458\,\text{m/s}}$$= 1.497 \cdot 10^{5}\,\text{s}$$= \dfrac{1.497 \cdot 10^{5}\,\text{s}}{3600\,\text{s/h}}$$= 41.58\,\text{h}$Answer
$41.6\,\text{h}$
Three significant figures, when the distance $150\,\text{AU}$ is interpreted as accurate to three figures.
Reasonableness
Light travels one AU in about $499\,\text{s} \approx 8.3\,\text{min}$, so $150\,\text{AU}$ takes about
$150 \cdot 8.3\,\text{min}$$\approx 1250\,\text{min}$$\approx 20.8\,\text{h}$and there and back $41.6\,\text{h}$. A delay of more than a day is reasonable, because the probe is tens of billions of kilometres away.
- Exercise 5-18
Two friends set off from the school door along the same straight street in the same direction. The first walks uniformly at $1.5\,\text{m/s}$. The second leaves the door $40\,\text{s}$ later on a bicycle and rides uniformly at $4.5\,\text{m/s}$.
How far from the door does the cyclist catch up with the walker?
Answer
$90\,\text{m}$
Show the solution
Given
speed of the walker $v_1 = 1.5\,\text{m/s}$, speed of the cyclist $v_2 = 4.5\,\text{m/s}$, delay of the cyclist $t_0 = 40\,\text{s}$; both start from the door, so $s_0 = 0$
Find
distance from the door $s$ at the point where the cyclist catches up with the walker
Principle
Both move uniformly. They are at the same position when their distances from the door are equal. The time $t$ is measured from the start of the walker, so the cyclist has been moving for the time $t - t_0$.
Formula
$s_1 = v_1 t$ and $s_2 = v_2 (t - t_0)$. From the condition $s_1 = s_2$ we get $v_1 t = v_2 (t - t_0)$, so $t = \dfrac{v_2 t_0}{v_2 - v_1}$ and $s = v_1 t$.
Conversions
None needed: the speeds are in m/s and the delay in seconds.
Substitution
$t$$= \dfrac{4.5\,\tfrac{\text{m}}{\text{s}} \cdot 40\,\text{s}}{4.5\,\tfrac{\text{m}}{\text{s}} - 1.5\,\tfrac{\text{m}}{\text{s}}}$$= 60\,\text{s}$$s$$= 1.5\,\tfrac{\text{m}}{\text{s}} \cdot 60\,\text{s}$$= 90\,\text{m}$Answer
$90\,\text{m}$
Two significant figures as in the speeds.
Reasonableness
The cyclist rides for $60\,\text{s} - 40\,\text{s} = 20\,\text{s}$, and $4.5\,\text{m/s} \cdot 20\,\text{s} = 90\,\text{m}$, the same result. Another check: the walker's lead after $40\,\text{s}$ is $1.5 \cdot 40 = 60\,\text{m}$, and the cyclist closes it at the speed difference $3.0\,\text{m/s}$ in $60 : 3.0 = 20\,\text{s}$.
- Exercise 5-19
The position $s$ of a cyclist from the start of the track was measured by video analysis as a function of time $t$. Draw a graph of the points.
The cyclist's speed changes partway, so the motion cannot be modelled with a single straight line. Find in the table the moment when the speed changes, and model the motion as two uniform parts.
Use the model to calculate the cyclist's average speed in the interval $5.0\,\text{s} - 35\,\text{s}$, where the average speed is the distance travelled in the interval divided by the length of the interval. The moment $35\,\text{s}$ is not in the table. Why is the average speed not the mean of the speeds of the parts?
Position of the cyclist as a function of time $t\,(\text{s})$ $s\,(\text{m})$ 0.0 0.0 5.0 15.0 10.0 30.0 15.0 35.0 20.0 40.0 30.0 50.0 40.0 60.0 Answer
$1.3\,\text{m/s}$
Show the solution
Given
measurement points of the table $(t$; $s)$: $(0\,\text{s}$; $0\,\text{m})$, $(5.0\,\text{s}$; $15\,\text{m})$, $(10\,\text{s}$; $30\,\text{m})$, $(15\,\text{s}$; $35\,\text{m})$, $(20\,\text{s}$; $40\,\text{m})$, $(30\,\text{s}$; $50\,\text{m})$ and $(40\,\text{s}$; $60\,\text{m})$
Find
average speed $v_\text{avg}$ in the interval $5.0\,\text{s} - 35\,\text{s}$ and the reason why it is not the mean of the speeds of the parts
Principle
In the graph, the points lie on two straight parts with different slopes, so a single line does not fit the whole motion. The motion is modelled piecewise: each part has its own uniform speed, and the position at the moment $35\,\text{s}$ is calculated from the model of the latter part. The average speed is the distance travelled divided by the time.
Reading the graph
The points $0\,\text{s} - 10\,\text{s}$ are on the same steep line. From the moment $10\,\text{s}$ on, the line is gentler, and the points $(10\,\text{s}$; $30\,\text{m})$, $(20\,\text{s}$; $40\,\text{m})$ and $(40\,\text{s}$; $60\,\text{m})$ are on the same gentle line. So the speed changes at the moment $t = 10\,\text{s}$.
Formula
speed of a part $v = \dfrac{\Delta s}{\Delta t}$, position on the gentle part $s = s_{10} + v_2 (t - 10\,\text{s})$ and average speed
$v_\text{avg}$$= \dfrac{s(35\,\text{s}) - s(5.0\,\text{s})}{35\,\text{s} - 5.0\,\text{s}}$Conversions
None needed: the positions are in metres and the times in seconds.
Substitution
Part 1:
$v_1$$= \dfrac{30\,\text{m} - 0\,\text{m}}{10\,\text{s} - 0\,\text{s}}$$= 3.0\,\tfrac{\text{m}}{\text{s}}$Part 2:
$v_2$$= \dfrac{60\,\text{m} - 30\,\text{m}}{40\,\text{s} - 10\,\text{s}}$$= 1.0\,\tfrac{\text{m}}{\text{s}}$Position at $35\,\text{s}$:
$s$$= 30\,\text{m} + 1.0\,\tfrac{\text{m}}{\text{s}} \cdot 25\,\text{s}$$= 55\,\text{m}$Average speed:
$v_\text{avg}$$= \dfrac{55\,\text{m} - 15\,\text{m}}{35\,\text{s} - 5.0\,\text{s}}$$= \dfrac{40\,\text{m}}{30\,\text{s}}$$= 1.33\,\tfrac{\text{m}}{\text{s}}$Answer
$1.3\,\text{m/s}$
Two significant figures as in the data. The mean of the speeds of the parts would be $(3.0 + 1.0)\,\text{m/s} : 2 = 2.0\,\text{m/s}$, but that would be wrong, because the parts last for different times: the speed $3.0\,\text{m/s}$ is in effect for only $5.0\,\text{s}$ and the speed $1.0\,\text{m/s}$ for as long as $25\,\text{s}$.
Reasonableness
The partial distances are $3.0\,\text{m/s} \cdot 5.0\,\text{s} = 15\,\text{m}$ and $1.0\,\text{m/s} \cdot 25\,\text{s} = 25\,\text{m}$, together $40\,\text{m}$, the same as above. The average speed $1.3\,\text{m/s}$ is closer to the slower speed $1.0\,\text{m/s}$ than to the faster speed $3.0\,\text{m/s}$, as it should be, because most of the time is spent riding slowly.
- Exercise 5-20
A fast bullet is filmed with a camera that takes a picture every $0.010\,\text{s}$. The position $s$ of the bullet from the starting point, measured from the pictures, is in the table. The speed of the bullet is constant during this time. Calculate the speed of the bullet.
A $(t, s)$ graph is drawn in which $1.0\,\text{s}$ on the horizontal axis is $10\,\text{cm}$ and $1.0\,\text{m}$ on the vertical axis is $1.0\,\text{cm}$.
How many centimetres does the line rise during one imaging interval, and how long is the imaging interval on the horizontal axis? Use these to justify why the line looks almost vertical but cannot be exactly vertical.
Position of the bullet as a function of time $t\,(\text{s})$ $s\,(\text{m})$ 0.000 2.0 0.010 5.0 0.020 8.0 0.030 11.0 Answer
$300\,\text{m/s}$
Show the solution
Given
imaging interval $\Delta t = 0.010\,\text{s}$, positions of the bullet $s = 2.0\,\text{m}$, $5.0\,\text{m}$, $8.0\,\text{m}$ and $11.0\,\text{m}$ at the moments $0.000\,\text{s}$, $0.010\,\text{s}$, $0.020\,\text{s}$ and $0.030\,\text{s}$; scale of the drawing: $1.0\,\text{s}$ is $10\,\text{cm}$ and $1.0\,\text{m}$ is $1.0\,\text{cm}$
Find
speed $v$ of the bullet in m/s, and the rise of the graph and the length of the imaging interval in centimetres
Principle
The position grows by the same amount in every imaging interval ($3.0\,\text{m}$), so the motion is uniform and the speed is the slope of the graph, $\Delta s : \Delta t$. An object has only one position at every moment, so every value of time corresponds to one value of position. A vertical line would have $\Delta t = 0$ and $\Delta s \neq 0$, and then the quotient would not be any number.
Reading the graph
The points of the table lie on the same straight line, and between two consecutive points the position changes by
$\Delta s$$= 5.0\,\text{m} - 2.0\,\text{m}$$= 3.0\,\text{m}$in the time $\Delta t = 0.010\,\text{s}$.
Formula
$v = \dfrac{\Delta s}{\Delta t}$; in the drawing, the length is the value times the scale
Conversions
Not needed for calculating the speed. In the drawing,
$\Delta s$$= 3.0 \cdot 1.0\,\text{cm}$$= 3.0\,\text{cm}$$\Delta t$$= 0.010 \cdot 10\,\text{cm}$$= 0.10\,\text{cm}$Substitution
$v$$= \dfrac{3.0\,\text{m}}{0.010\,\text{s}}$$= 300\,\tfrac{\text{m}}{\text{s}}$In one imaging interval the line rises $3.0\,\text{cm}$ and advances horizontally only $0.10\,\text{cm}$, that is, $1.0\,\text{mm}$.
Answer
$300\,\text{m/s}$
Two significant figures as in the difference of positions $3.0\,\text{m}$. The line rises $3.0\,\text{cm}$ and advances $0.10\,\text{cm}$, so it is about $30$ times as steep as it is wide and looks almost vertical. It is still not vertical: the imaging interval always takes a time $\Delta t > 0$, and the bullet is at exactly one position at every moment. A vertical line would require the bullet to be at two positions at the same moment, that is, the speed would be infinite.
Reasonableness
The speed $300\,\text{m/s}$ is of the same order of magnitude as the speed of sound $343\,\text{m/s}$ and suits a fast bullet. Check: $300\,\text{m/s} \cdot 0.010\,\text{s} = 3.0\,\text{m}$, the same as in the table. A stopped object would be drawn as a horizontal line, because $\Delta s = 0$.
Photographs
- Long exposure of highway traffic light trails at night. Tao Yuan. source, Unsplash License