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

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 js the difference and similarity between speed and velocity?

How do you interpret speed/velocity versus time graphs?

A few technical terms explained

A scalar quantity only has a magnitude and no specified direction

e.g. distance, energy, mass, speed, 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 180o, keeping the same speed, its velocity might be considered to be -10 m/s.

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. e.g.

(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 north-east bearing 59o from north)

(Draw this for your self and measure the angle, or calculate it, Tan θ = (400/300) = 1.333, Tan-1θ = 59o)

(iii) an object moves 3 m east and 3 m west, displacement is 0 m, object returned to starting point.

However, the object has moved 6 m in total, and you can consider it does two 3 m displacements!

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.

e.g. a car travelling at 30 m/s or a train travelling at 200 km/hour, both obviously regularly changing direction.

In these examples the speed is variable so you will come across the idea of 'average speed'.

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 50o 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.

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 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 105 km/s (~3.0 x 108 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.

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 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

Drawing and interpreting distance - time graphs

The gradient (slope) at any point on the graph gives you the speed at that point.

Since speed = distance ÷ time, then speed = (change in vertical y axis) ÷ (change in horizontal x axis)

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 covered in the same time.

(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 distance-time graph, BUT, in reality, any distance-time 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 distance-time graph - see the graph based questions below.

Examples of distance-time graph questions

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).

For example look at the linear portion between the two purple 'blobs' on the right-hand graph.

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?

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?

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/velocity-time 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.

time = 80 - 16 = 64 s (easy), distance = 1000 - 480 (watch the scale!) = 520 m

speed = distance / time taken = 520 / 64 = 8.1 m/s (to 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

Motion and associated forces notes index (including Newton's Laws of Motion)

gcse physics

gcse physics revision notes

3. Acceleration, friction, drag effects and terminal velocity experiments gcse physics revision notes

gcse physics revision notes

gcse physics revision notes

IGCSE revision notes Speed velocity formula calculations analysing distance-time graphs KS4 physics Science notes on Speed velocity formula calculations analysing distance-time graphs GCSE physics guide notes on Speed velocity formula calculations analysing distance-time graphs for schools colleges academies science course tutors images pictures diagrams for Speed velocity formula calculations analysing distance-time graphs science revision notes on Speed velocity formula calculations analysing distance-time graphs for revising physics modules physics topics notes to help on understanding of Speed velocity formula calculations analysing distance-time graphs university courses in physics careers in science physics jobs in the engineering industry technical laboratory assistant apprenticeships engineer internships in physics USA US grade 8 grade 9 grade10 AQA GCSE 9-1 physics science revision notes on Speed velocity formula calculations analysing distance-time graphs GCSE notes on Speed velocity formula calculations analysing distance-time graphs Edexcel GCSE 9-1 physics science revision notes on Speed velocity formula calculations analysing distance-time graphs for OCR GCSE 9-1 21st century physics science notes on Speed velocity formula calculations analysing distance-time graphs OCR GCSE 9-1 Gateway  physics science revision notes on Speed velocity formula calculations analysing distance-time graphs WJEC gcse science CCEA/CEA gcse science

 KS3 SCIENCE QUIZZES ALPHABETICAL INDEX GCSE grade 9-1 & IGCSE CHEMISTRY Doc Brown's Travel Pictures & Notes ADVANCED LEVEL CHEMISTRY [SEARCH BOX] - see below All website content © Dr Phil Brown 2000 onwards. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to science course specifications are unofficial. Email doc b: chem55555@hotmail.com

Doc Brown's Physics

*