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 js the difference and similarity between speed and velocity?

 How do you interpret speed/velocity versus time graphs?



A few technical terms explained

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 falls 1.7 m when dropped.

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.

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 simple example.


Speed, distance and time calculations - problem solving

The speed 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, terrain, fitness and distance travelled.

Typical values may be taken as: walking  ̴ 1.5 m/s, running  ̴ 3 m/s, cycling  ̴ 6 m/s.

A jet aircraft travelling at 600 km/hour is moving at 168 m/s (from 600,000/3600, 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

A train travelling at 200 km/hour (from 200/1.61 = 124 mph) is moving at a speed of 55.6 m/s (from 200000/3600)

(1 mile = 1.61 km, 1 km = 1000 m)

Sound travels at ~340 m/s but this varies with air pressure (density) and temperature.

Sound travels much faster through and liquids (~1500 m/s in water, 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 299792 km/s (~3.0 x 108 m/s), .

 

The formula for speed/velocity and example calculations - problem solving

Speed (metres/second) = distance travelled (metres) time (seconds)

v = s t,  so  s = vt  and  t = s v 

v = speed/velocity (m/s), s = distance (m), t = time (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 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 ~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 ?


Drawing and interpreting distance - time graphs

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

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.

Graph is flat/horizontal, indicating zero speed/velocity, there is no increase in distance with time, object has stopped moving.

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.

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

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/velocity-time graph.

(need to do example)

 

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.(under development)


INDEX

 


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