Lesson 5 of 5Kinematics

Unit 1 · Lesson 5 of 5

Analysing Motion Data from an Experiment

Turning a column of raw position and time readings into a number you can defend — interval velocities, a line of best fit, a slope with its units, and an honest account of the error.

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Key things to remember

1An interval velocity is $Δ d /Δ t$ across one timing interval, and it is plotted at the midpoint time of that interval.
2A line of best fit is a single straight line drawn through the scatter — never a dot-to-dot, and never forced through the origin.
3Take slope points from the drawn line, far apart, and divide the units as well as the numbers.
4Slope of $d$ against $t$ is velocity (m/s); slope of $v$ against $t$ is acceleration ( $m/s^{2}$ ).
5For an object starting from rest, plotting $d$ against $t^{2}$ straightens the curve and gives a slope of $\frac{1}{2} a$ .
6Random errors scatter both ways and shrink when you average; systematic errors push one way and averaging never helps.
7 $% error = \frac{∣ measured - accepted ∣}{accepted} \times 100%,$ with the accepted value always on the bottom.

Every number in this unit so far arrived already measured. A real investigation is not like that: it hands you a strip of paper with dots on it, or a spreadsheet of positions, and nothing else. The physics is no longer in the equations — it is in knowing which quantity to calculate, which graph to draw, and how much of your answer you are entitled to believe. This topic is the analysing-and-communicating half of a scientific investigation, applied to the kinematics you already know.

Getting the data

Method	How it records position	Time between readings	Watch out for
Ticker tape	A vibrating needle prints a dot on a paper tape towed behind the cart	Set by the mains frequency — a $60 Hz$ timer prints a dot every $\frac{1}{60} s$	The tape dragging through the timer slightly slows the cart
Motion sensor	Ultrasound pulses reflect off the cart and the return time gives the distance	Whatever sampling rate you choose, often 20–50 readings per second	It reports the nearest object, so a stray hand ruins the run
Video analysis	You step through the frames and mark the object against a metre stick placed in shot	The frame rate, commonly 30 or 60 frames per second	The camera must be square to the motion, or the distances are distorted

Fig 1A strip of ticker tape, the table it becomes, and the graph drawn from that table.

Whichever method you use, the raw result is the same pair of columns: a time and a position. Counting off every twelfth dot on a $60 Hz$ tape makes each interval $0.200 s$ long, which is short enough to follow the motion and long enough that the gaps can be measured with a ruler. The interval velocity is then just the average velocity across one interval, $Δ d /Δ t$ , and calculating one for every interval turns a position table into a velocity table.

$t$ (s)	Position $d$ (m)	$Δ d$ (m)	Interval velocity (m/s)	Midpoint time (s)
0.000	0.000	—	—	—
0.200	0.030	0.030	0.150	0.100
0.400	0.120	0.090	0.450	0.300
0.600	0.270	0.150	0.750	0.500
0.800	0.480	0.210	1.050	0.700
1.000	0.750	0.270	1.350	0.900

A cart released from rest at the top of a ramp, timed at 0.200 s intervals.

An interval velocity belongs to the middle of its interval

The value

0.750 m/s

is the average across the whole stretch from

0.400 s

0.600 s

, so the moment it best describes is the midpoint,

0.500 s

. For uniformly accelerated motion the average velocity over an interval is exactly the instantaneous velocity at its midpoint, which is why that column exists. Plotting these values against the end of the interval instead shifts the whole line sideways.

Plotting, and the line of best fit

Independent variable on the horizontal axis. You chose the times, so time goes across; the measured quantity goes up.
Label both axes with the quantity and its unit, because the units on the axes are what give the slope its units later.
Choose a scale that uses most of the paper. A graph squeezed into one corner magnifies every reading error when you take the slope.
Plot the points, then draw one straight line through them with a ruler — a line of best fit, not a dot-to-dot. Aim for roughly as many points above the line as below, and as close to each as you can manage.
Do not force the line through the origin unless the data genuinely go there. A non-zero intercept is a result, not a mistake to be tidied away.
Circle any point that sits far off the line and say so in the report rather than quietly deleting it.

The slope, and what its units tell you

slope = \frac{rise}{run} = \frac{1.350 m/s - 0.150 m/s}{0.900 s - 0.100 s} = \frac{1.200 m/s}{0.800 s} = 1.50 m/s^{2}

Take the two points off the drawn line, not off the plotted data, and keep them far apart.

The units of a slope name the quantity

Divide the units as well as the numbers. Metres divided by seconds gives m/s, so the slope of a position-time graph is a velocity. Metres per second divided by seconds gives

m/s^{2}

, so the slope of a velocity-time graph is an acceleration. If you carry the units through, you never have to remember which slope is which — the unit tells you, and it also catches the case where you plotted the wrong pair of columns.

Linearizing a curved graph

The position-time graph for that cart is a curve, and a curve has no single slope. Rather than give up, change what you plot. For an object released from rest, $Δ d = v_{i} Δ t + \frac{1}{2} a Δ t^{2}$ loses its first term and becomes $d = \frac{1}{2} a t^{2}$ . Compare that with the equation of a straight line through the origin, $y = m x$ : if you put $d$ up the vertical axis and $t^{2}$ along the horizontal one, the points fall on a straight line whose slope is $\frac{1}{2} a$ . Linearizing is this trick in general — square, invert or otherwise transform one axis until the relationship becomes a line, because a line is the only shape whose slope you can read reliably by hand.

d = (\frac{1}{2} a) t^{2} ⟹ a = 2 \times slope of the d vs t^{2} graph

The horizontal axis is now in seconds squared, so the slope comes out in metres per second squared.

Fig 2Slope and area are the two things a graph is for. Everything in this analysis is one or the other.

Random error and systematic error

ARandom error

Scatters readings both above and below the true value
Comes from reaction time, judging a ruler mark, a wobbling cart
Shows up as points scattered either side of the line of best fit
Reduced by repeating the measurement and averaging
Never removed completely — it sets the precision

BSystematic error

Pushes every reading the same way, by roughly the same amount
Comes from a mis-set zero, a stretched tape, friction in the timer, air resistance
Shows up as points that fit the line beautifully but give the wrong slope or intercept
Repeating and averaging does not help at all
Removed only by finding the cause and fixing or correcting for it

This is why a result can be very precise and completely wrong. Five runs that agree with each other to within $1%$ but sit $4%$ below the accepted value are telling you about a systematic problem, not a random one. Finish the report by comparing your value with the accepted one using $% error = \frac{∣ measured - accepted ∣}{accepted} \times 100%,$ quoting your result to the number of significant figures your instruments actually justified, and naming the most likely source of the discrepancy along with something specific you would change in the next run.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Using the ticker-tape table above, find the interval velocity between

t = 0.400 s

and

t = 0.600 s

, and state the time it should be plotted at.

Example 2

The five interval velocities are plotted against their midpoint times and a line of best fit is drawn. It passes through

(0.100 s, 0.150 m/s)

and

(0.900 s, 1.350 m/s)

. Find the slope, give its units, and say what physical quantity it is.

Example 3

The same run is analysed a second way. The position-time graph is a curve, so

d

is plotted against

t^{2}

instead. The result is a straight line through the origin that passes through

(1.000 s^{2}, 0.750 m)

. Find the acceleration.

Example 4

A group uses video analysis of a dropped ball to measure

g

and obtains

9.42 m/s^{2}

. Find the percent error and identify the kind of error the result points to.

Where marks get lost

Plotting an interval velocity at the end of its interval instead of at the midpoint, which shifts the whole velocity-time line sideways and biases the intercept.
Joining the plotted points dot-to-dot. A line of best fit is one straight line through the scatter; a zig-zag has no slope worth taking.
Taking the slope from two plotted data points rather than from two points on the drawn line — the line is the averaged result, the individual points are not.
Reporting a slope as a bare number. Divide the units as well: metres per second divided by seconds is $m/s^{2}$ , which is what identifies the slope as an acceleration.
Calling a consistently low result a random error. Random errors scatter both ways; an error that always pushes the same direction is systematic and averaging will never remove it.

Flashcards 9

Interval velocity

Line of best fit

How do you take a slope from a graph?

Slope of a

d

t^{2}

graph

+ 5 more

Practice

A cart is at

0.480 m

t = 0.800 s

and at

0.750 m

t = 1.000 s

. What is the interval velocity, in m/s?

m/s

At which time should that interval velocity be plotted on the velocity-time graph?

A line of best fit on a velocity-time graph runs from

(0.20 s, 0.40 m/s)

(1.20 s, 2.90 m/s)

. What is its slope, in

m/s^{2}

m/s^2

A graph of position against time squared is plotted for a cart released from rest, and the slope comes out as

0.60 m/s^{2}

. What is the cart's acceleration?

What is the name for the single straight line drawn through a set of scattered data points?

A stopwatch is started a fraction of a second late on every single trial. What kind of error is this?

A group measures the acceleration of a free-falling mass as

10.20 m/s^{2}

. What is the percent error, to two significant figures?

Five trials give results that agree with each other to within

0.5%

, but all sit

6%

below the accepted value. What does this indicate?

Put the stages of analysing a set of motion data in the order they are carried out.

1Report the value with correct significant figures and a percent error
2Take the slope from two widely separated points on the line, with its units
3Plot velocity against midpoint time and draw a line of best fit
4Tabulate the raw times and positions with their units
5Calculate an interval velocity for each interval and record its midpoint time

Match each graph to the quantity its slope gives.

A.Acceleration, in $m/s^{2}$
B.Half the acceleration
C.Rate of change of acceleration — not used in this course
D.Velocity, in m/s

Position against time

Velocity against time

Position against time squared, from rest

Acceleration against time

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