1. Speed and velocity  the relationship between distance and time
Formula for speed,
calculations and interpreting distance  time graphs
Doc Brown's Physics Revision Notes
Suitable for GCSE/IGCSE Physics/Science courses or their
equivalent
This
page will help you answer questions like e.g.
What do we mean by displacement? How do you calculate speed or velocity? What is the difference between scalar and
vector quantities in motion? What is the difference and similarity
between speed and velocity? How do you interpret speed/velocity versus
time graphs?
Subindex for this page
(a)
A few technical terms
explained
(b)
Speed, distance and
time calculations  problem solving using the formula
(c)
Drawing and
interpreting distance  time graphs, calculations and problem solving
(a) A
few technical terms explained
A
scalar quantity only has a magnitude and no specified direction
e.g. distance, energy, mass,
speed (without change in direction), temperature, time etc.
A
vector quantity has both magnitude (size) and specified direction
e.g. acceleration, displacement,
force, momentum, velocity, weight etc.
When talking about vectors, you should
appreciate that their values can be positive and negative.
A car might have a speed of 10 m/s, but
its velocity might be 10 m/s in one direction, but if it changes direction
by 180^{o}, keeping the same speed, its velocity will be considered
to be 10 m/s.
It is possible to have an object moving
with a constant speed, but its velocity changes.
e.g. if an object is moving in a
circle at constant speed, the velocity is constantly changing, because
the direction of motion is constantly changing.
Distance: Distance is how far an object has moved in any direction,
the direction isn't specified, so its a
scalar measurement.
e.g. a car moves 30 km, a ball is tossed
in the air to a height of 3 m.
Displacement: Displacement is how far an object has been moved in a
straight line from specified starting point to a specified finishing
point, but in particular specified direction/directions, so its a
vector measurement.
It does not necessarily mean the
object has moved in a straight line  it may or may not.
Examples of displacements
(i) an object moves 1300 km north,
displacement =
1300 km
(ii) an object moves 300 m north and 400
m east, displacement is 500 m
northeast bearing 59^{o} from north)
(Draw this for your self and measure
the angle, or calculate it, Tan θ = (400/300) = 1.333, Tan^{1}θ
= 59^{o})
(iii) an object moves 3 m east and 3 m
west, displacement is 0 m, because object has returned to its starting point.
However, the object has moved a
distance of 6 m in
total, and you can consider it does two 3 m displacements, but take
in expressing the situation!
For a similar argument think of an
object that moves once in a circular path of circumference 6 m.
The total distance travelled is 6 m,
but the displacement is zero, because the object ends up where it
started.
Speed: Speed is how fast an object is moving but no direction is
specified, so its a scalar measurement.
So speed has magnitude but no
direction is indicated.
speed = distance travelled / time
taken
e.g. a car travelling at 30 m/s or a
train travelling at 200 km/hour, but real journeys obviously involve regularly changing
direction and speed  so strictly speaking, many changes in velocity (see
next).
In these examples, because the speed is variable,
can calculate the 'average speed'.
average speed = total distance
travelled / total time taken
To measure the constant speed of an
object need some means of measuring time (stopwatch) and distance (tape
measure). Speed/velocity calculations explained in the next section.
Velocity: Velocity is how fast an object is travelling in a
particular direction, so its a vector measurement.
So velocity has magnitude and a specified
direction.
e.g. a plane moving at a constant speed
of 600 km/hour at 50^{o} from due north.
Two cars, travelling at identical speeds,
passing in opposite directions have different velocities  different
directions.
You can having objects moving at a
constant speed but continually changing velocity.
e.g. any object moving in a circle at
constant speed is continuously changing velocity because it is continually
changing direction. Whirling an object around on the end of string is a simple
example.
Both speed and velocity tell you how
fast an object is moving.
(b)
Speed, distance and time calculations  problem solving
The speed
(or velocity) of a moving object is rarely constant.
When you walk, run or travel in a car or
train your/their speed
is constantly changing.
The speed that a person can walk, run or cycle
depends on many factors including: age, distance travelled, fitness and terrain.
Typical values may be taken as: (maths
reference to put speeds in perspective: 1 hour
≡ 3600 s, 1 mile = 1.61 km)
You are expected to know typical
values for commonly sited moving objects!
walking (briskly on the flat?) ̴ 1.5 m/s, ~5400
m/hour, ~5.4 km/hour, ~3.4 mph
You can do a simple experiment to
time a person walking a certain measured distance on the floor.
You multiply km/hr x 1.6 to
convert to mph.
km/hr into m/s: the factor is
x 1000 for m then divide by 3600 seconds/hr, so m/s = (km/hr) ÷ 3.6
running ̴ 3 m/s, ~11000 m/s,
~11 km/hour, ~6.7 mph
Again you can do a simple experiment
to time a running a certain measured distance on a running track.
light wind ~5 m/s, blowing gale ~20 m/s,
hurricane ~50 m/s
cycling ̴ 6 m/s, ~22000 m/s,
~22 km/hour, ~13 mph
A jet aircraft travelling at 600 km/hour
is moving at 168 m/s (from 600 x 1000/3600, 3 s.f.).
A
car travelling at 50.0 mph, speed = 50.0 x 1.61 = 80.5 km/hour, 1000 x 80.5
÷ 3600
= 22.4 m/s
Speed restricted in built up areas 30
mph, 47 km/hr, 13 m/s
Less restricted areas like motorways
with a higher speed limit of
70 mph, 112 km/hr, 31 m/s
A
train travelling at 200 km/hour (~ 124 mph) is moving at a speed
of ~56 m/s.
Sound travels at ~340 m/s (1224 km/hour,
~761 mph), but this varies
with air pressure (density) and temperature.
Sound travels much faster through and
liquids, ~1500 m/s in water, used in ultrasound depth measurement and surveying,
and even faster in solids >3000 m/s (e.g.
earthquake waves, at ~3 km/s you don't get much warning!)
The ultimate speed of anything is that of
'light' photons, all electromagnetic radiation travels at the 'speed of light' which is
greatest in vacuum and is ~3.0 x 10^{5} km/s (~3.0 x 10^{8} m/s,
186,000 miles/s), .
The formula for speed/velocity and example
calculations
 problem solving
speed (metres/second, m/s) = distance travelled
(metres, m) ÷ time (seconds, s)
distance (m) = average speed (m/s) x time (s)
time (s) = distance (m) ÷ average speed (m/s)
v = s
÷
t, so
s =
v x t
and t = s
÷
v
v = speed or velocity (m/s),
s =
distance (m), t = time (s)
sometimes distance is quoted as the
variable d as well as s.
Changes in distance might be called
Δd (or Δs) and changes in time Δt  the actual time taken for the
distance of actual moved in that time.
Make sure you are good at doing
equation rearrangements,
this one is easy, but some are not!
Beware! it is common in motion topics to
represent distance by the letter s, don't confuse with the time unit
second!
If the speed of an object is variable,
average speed is simply the total distance travelled divided by the total time
taken.
e.g. a sprinter completing a 200 m race in 25
seconds has an average speed of 200 / 25 = 8 m/s, but quite plainly the speed is
variable as the athlete starts from 0 m/s to perhaps a maximum speed of over10 m/s
since 100 m sprinters complete their distance in ~10 s giving an average speed
of ~10 m/s.
For v to be a velocity, then direction
of motion should be specified.
Example
of speed calculation questions
Q1.1 A train was timed to take 2.5 seconds when passing between two
posts 100 m apart.
(a) What is the speed of the train in
m/s?
v = s
÷
t = 100/2.5 =
40
m/s
(b) What is the speed of the train in
km/hour?
(I'm just deliberately asking a more
arithmetically awkward question of a sort you may have to deal with)
100 m = 100/1000 = 0.1 km
v = s
÷
t = 0.1/2.5 = 0.04 km/s
Since 1 hour equates to 60 x 60 =
3600 seconds
In 1 hour the train will travel
0.04 x 3600 km, so speed =
144 km/hour
Q1.2 How far will a car travel in 15 seconds at a speed of 20 m/s?
v = s ÷ t , so
s = vt = 20
x 15 = 300 m
Q1.3 A sprinter runs 400
m at an average speed of 8.4 m/s. To the nearest 0.1 s, how long did the sprint
run take?
v = s ÷ t, so
t = s
÷
v = 400/8.4 =
47.6 s
Q1.4
A car travels at a constant speed of 40.0 mph.
(a) If 1 mile = 1.61 km, calculate the
speed of the car in m/s.
speed = 40 x 1.61 = 64.4 km/hour
64.4 m = 66400 m, 1 hour = 60 x 60 =
3600 s
speed of car = 66400/3600 =
17.9 m/s
(3sf)
(b) How far will the car travel in 45.0
seconds?
speed = distance/time
distance = speed x time = 17.9 x 45 =
806 m (3sf)
(c) If the car speeds up and travels 5.2
km in 3 minutes, what is the car's speed?
3 minutes = 3 x 60 = 180 s, 5.2 km =
5200 m
speed = distance/time = 5200/180 =
29 m/s (2sf)
Q1.5
A car travels from motorway Junction 26 to Junction 27 with an average speed of
30 m/s (~67 mph).
It took 150 s to go from Junction 26 to
Junction 27.
Due to road works speed restrictions the
car moved on to Junction 28 at average speed of 20 m/s.
Junctions 27 and 28 are 15 km apart.
(a) Calculate how far apart Junctions
26 and 27 are.
v = Δd / Δt, Δd
= v x Δt = distance
Distance between Junction 26 and 27 =
30 x 150 = 4500 m
(4.5 km)
(b) Calculate how long it took to
travel between Junctions 27 and 28.
v = Δd / Δt, Δt = Δd / v
= time taken, don't forget to change from km to m (x 1000)
Time from Junction 27 to 28 = (15 x
1000) / 20 =
750 s.
(c) What is the average speed of the
car from Junction 26 to Junction 28?
The total distance is 4500 + 15000 =
19500 m
The total time = 150 + 750 = 900 s
Average speed = total distance /
total time
= 19500 / 900 =
22 m/s
(2 sf, 21.7 to 3 sf)
Q1.6
(c)
Drawing and interpreting
distance  time graph calculations  problem solving
The gradient (slope) at any point on the graph gives
you the speed at that point.
Since speed = distance ÷ time, then
for graphical work
speed = (change in vertical y axis
for distance) ÷ (change in horizontal x axis for time)
speed = ∆d / ∆t
The steeper the gradient the greater the
speed
(1) Distance  time graphs  acceleration  speeding up
Graph
curves upwards, showing
increasing speed/velocity with time (acceleration), for each incremental time
unit (e.g. minute or second) there is an ever increasing (larger) distance
(∆d) covered in the same time (∆t).
speed = ∆d/∆t = the gradient of graph
(∆d/∆t) is continuously increasing.
(2) Distance  time graphs  motionless  stationary
Graph
is
flat/horizontal, indicating
zero speed/velocity, there is no increase in distance with time, object has
stopped moving.
(3) Distance  time graphs  deceleration  slowing down
Graph
shows
the curve is levelling off, steadily decreasing speed/velocity with time (deceleration),
for each incremental time unit (e.g. minute or second) there is an ever
decreasing (smaller) distance covered in the same time.
(4) Distance  time graphs  constant speed
Graph
is
linear, showing
constant speed/velocity, the distance covered in any equal time increment is the
same.
The four graphs above illustrate of what you
might see at any point on a distancetime graph, BUT, in reality, any
distancetime graph will be more complicated than any of these specific graphs.
So, you find a graph with all four types of gradient in just one distancetime
graph  see the graph based questions below.
Examples of distancetime graph
calculation questions
Moving on from the stylised 'types of
graph' to full numerical graphs.
Graph (1) The graph lines for
objects A and B are linear, indicating constant speed.
Using a convenient triangle to
calculate the constant speeds of A and B
Speed of A = 500 / 10 = 50
m/s and speed of B = 300 / 25 =
12 m/s
Note the steeper the gradient, the
greater the speed.
Graph (2) The graph line is
horizontal, no change in distance, so object C is stationary at a distance of 300 m.
Graph (3) Lets call the object DE:
Its movement is first in a forward
direction, represented by line D at a constant speed of 400 / 5 = 80 m/s.
Its second movement is in the
opposite direction, line E, at a constant speed of 400 / 10 = 40 m/s.
(Strictly speaking, the line E
represents a speed of 40 m/s, because of the reversed direction.)
Note that object DE has travelled a
total distance of 800 m, BUT, its displacement is zero!
You get the speed from the gradient at any
given point on the graph, which is easy if the graph at that point is linear
(constant speed, like the three graphs above).
For example look at the linear portion
between the two purple 'blobs' on the righthand graph  that's a linear section
and easy to measure the gradient (speed).
If the graph is curved you will need to draw
a tangent to the curve at the point where you want to know the speed.
For example I've drawn a tangent at ~56
seconds and converted it to a purple triangle so you can measure the gradient 
see Q2.1 (c) below.
Q2.1 The
graph below shows part of a car journey.
Distance  time graph for a short car
journey
Interpretation question using speed =
distance / time
(a) Including calculating the average speed,
describe and explain the motion of the car between:
(i) 0 and 20 seconds?
speed increasing (acceleration)
because the gradient is steadily increasing
time interval = 20  0 = 20 s,
distance covered = 200  0 = 200 m
gradient = average speed = 200 / 20 =
10 m/s
(ii) 20 and 40 seconds?
gradient is constant, but NOT zero,
car moving at steady constant speed
time interval = 40  20 = 20 s,
distance covered = 600  200 = 400 m
gradient = constant speed = 400 / 20
= 20 m/s
(iii) 40 to 82 seconds?
gradient decreasing (deceleration),
gradient steadily decreasing
time interval = 82  40 = 42 s,
distance covered = 900  600 = 300 m
gradient = average speed = 300 / 42 =
7.14 m/s ( 1 dp, 3 sf)
(iv) 82 to 90 seconds?
At 82 seconds the gradient is zero,
graph line is flat, so car isn't moving  stationary from 82 to
90 s.
(b) What is the average speed of the car
while it is moving?
total time of journey (while moving) is
82 s.
Total distance covered until it stops =
900 m
average speed for journey = 900 / 82 =
10.8 m/s (1 dp, 3 sf)
(c) How can you find the specific speed of
the car at any point on the graph?
You can find the speed at any point on
the graph by drawing a tangent at that point on the graph.
No tangent is needed if the graph line is
linear, but you need to draw a tangent for any point on the curved sections
of a speed/velocitytime graph.
e.g. what is the speed of car at 56
seconds?
same
graph as above but tangent drawn on at 56 seconds.
You convert the purple tangent line into
a triangle so you read of the distance travelled (vertical y axis) and the
time taken (horizontal x axis) and then calculate the gradient from the two
dimensions of the triangle, which is the gradient at the point you are
interested in.
xaxis: time Δt = 80  16 = 64 s (easy),
yaxis: distance Δd = 1000  480 (watch the scale!) = 520 m
speed = distance / time taken = Δd/Δt = 520 /
64 = 8.1 m/s (to
2 s.f.)
Q2.2 The graph below summarises part of the journey of a train
(somewhat delayed at some point!).
Interpretation question using speed =
distance / time and giving your answers in km/hour and m/s
(a) What is the average speed between 1300
and 1400 hours?
time interval = 1 hour, distance
travelled = 100  0 = 100 km
average speed = 100/1 =
100 km/hour
1 km = 1000 m, 1 hour = 3600 s, average
speed = 100 x 1000 / 3600 = 27.8
m/s (1 dp, 3 sf)
(b) What is the average speed between 1530
and 1800 hours?
time interval = 2.5 hours, distance
travelled = 250  100 = 150 km
average speed = 150/2.5 =
60 km/hour
1 km = 1000 m, 1 hour = 3600 s, average
speed = 60 x 1000 / 3600 = 16.7
m/s (1 dp, 3 sf)
(c) How long was the train stopped for?
from 1400 to 1530 hours,
1.5 hours.
(d) What was the average speed for the whole
journey?
total time from 1300 to 1800 hours = 5
hours, total distance travelled = 250 km
average speed = 250/5 =
50 km/hour
1 km = 1000 m, 1 hour = 3600 s, average
speed = 50 x 1000 / 3600 = 13.9
m/s (1 dp, 3 sf)
Velocity  Time graphs and acceleration are dealt with in
detail in section 2
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