Motion in Physics

At the end of this unit you should:

1. Be able to state the relationship between speed,

velocity and acceleration.

2. Be able to draw and interpret distance–time graphs.

3. Be able to draw and interpret velocity–time graphs.

4. Be able to calculate the speed, velocity and

acceleration of objects.

acceleration

distance

distance–time graph

rate of change

speed

time

velocity

velocity–time graph

If a car is travelling at a constant

speed and turns a corner (we assume

it has not lost speed as it turns), both

its velocity and acceleration change.

Since the direction of speed has

changed (say from east to north to

west as in illustration) the velocity has

changed. Since the velocity has

changed, so has the acceleration.

Speed:

The distance an object in one unit of time. It can also be

described as the rate of change of distance with respect to time.

Unit: metre per second (m/s).

Distance:

The length of space between two points. Unit: metre (m).

Time:

A universal unit of measurement through which we measure

events in the past, present and future. Unit: second (s)

Equipment: 

Three wind-up toys, a long, smooth surface and a

stopwatch.

Instructions: 

Calculate the average speed of three wind-up toys.

Investigation 10.02.01: To calculate the average speed of three

wind-up toys

1. How would you make sure this was a fair test?

All cars receive the same amount of wind-up. They

compete over the same terrain and length of track.

2. Write a report on your experiment.

Your report writing should contain the following:

Statement of the problem/investigation being carried

out; list of equipment; list of variables (independent and

dependent); brief description of the procedure carried

out; data collected (description and visually); limitations

of study (error reporting etc.); conclusions.

3. Which toy was the fastest? Can you explain why?

Consider examining the cars to determine what makes

the fastest one so fast; for example, is it wheel size, or the

length of the car, or the weight?

(a)

What is the formula for calculating distance?

Speed = Distance/Time.

(b) In what units are speed, distance and time measured?

Speed is measured in metres per second (m/s). Distance is

measured in metres (m). Time is measured in seconds (s).

(c) Would it be a fair test to run 10 m to calculate your average

speed for 100 m? Why/why not?

No, it would not be a fair test as you are accelerating in the first

10–20 m of any sprint.

(d) Calculate the speed of a person walking 100 m in 60 s.

1.67 m/s.

(e) Calculate the speed of a person running 200 m in 24 s.

8.3 m/s.

(f) How long does it take a car to travel 7 km at a speed of 14 m/s?

Don’t forget to convert 7 km into metres. 500 s or 8.3 min.

(g) How far will a person walk in 1 hour if they walk at a speed of

1.5 m/s?

Convert 1 hour into seconds (60 min x 60 s) = 3600 s x 1.5=  5400

m or 5.4 km.

(h) Calculate the speed of a car travelling 2 km in 45 seconds.

Don’t forget to convert to SI units. 44.44 m/s. If you want to

convert to km/h, multiply by 3 600 (converts seconds to hours)

and divide that answer by 1 000 (converts metres to km) = 159.98

km/h.

(i) Calculate the speed of a car travelling 10 km in 15 minutes.

Don’t forget to convert to S.I. 11.11 m/s.

(j) How long does it take a truck to travel 15 km if it has a speed of

80 km/h?

Don’t forget to convert to SI units. 675.1 s or 11.25 min.

(k) How far can a person get if they run at a speed of 8 m/s for 1 h

and 10 min?

Convert to SI units. 33 600 m or 33.6 km.

(l) Compare the speed of the cars in (h) and (i).

Car H is travelling approximately four times faster than car I.

Velocity:

Speed in a given direction. Unit: metre per second (m/s).

(a)

A person walks north for 190 m. They walk this distance in

24 seconds. What is their velocity?

7.92 m/s north.

(b) Calculate the velocity of a rollercoaster if it covers 40 m in

1.5 s.

26.67 m/s. (We do not have a direction to supply as the

information is not given.)

(c) Which is faster: this rollercoaster in part (b) or a car covering 1

km in 35 s? What is the speed of this car in km/h?

The rollercoaster travels at 26.67 m/s (as per previous

questions). The car travels at 28.57 m/s. So the car travels

faster.

28.57 m/s x 3600 = 102 857.14 m/h.

102 857.14 m/h/1000 = 

102.86 km/h.

(d) A car travels from Dublin to Galway – a distance of 210 km –

in 2 h and 5 min. What is the velocity of the car?

Convert to SI units. 28 m/s west.

Yes, as velocity changes with direction because it is a vector

quantity (it has a size and direction).

Acceleration:

The change in velocity over time, or the rate of change of

velocity with respect to time.

Speed is not the same as

velocity. Since speed does not

have a direction, we cannot

equate speed and

acceleration. We can only

equate velocity and

acceleration.

(a) Which of these three examples involves acceleration?

(i)

A car changing its velocity from 10 m/s to 14 m/s.

Car is accelerating as there is a change in velocity.

(ii) A car travelling at a constant velocity.

No acceleration, as no change in direction or velocity.

(iii) A car changing its velocity from 14 m/s to 10 m/s.

Change in velocity (slowing down), so there is a change in

acceleration.

(b) In 

Table 10.02.02

you will see a list of

animals and their top

speeds. These are not

the speeds of these

animals in nature.

These are the speeds of

the animals if they

were the size of a

human. Use the table

to answer these

questions.

(b) (i) Which

animal is the

fastest?

White-

throated

needletail.

(b) (ii) What

animal is the

slowest?

African bush

elephant.

(b) (iii) Explain, in your own words, why you think the mouse is as

fast as the cheetah according to the table?

The table lists the animals’ speed if they were all human-sized. So,

if they were the same size, a mouse would be as fast as a cheetah.

(b) (iv) Suggest one possible method for calculating the speed of

the animals in their natural habitat. State how you would ensure it

is a fair test.

An observer would have to monitor the animals in their habitat,

measure a set distance and time the animals when they run that

distance. For it to be a fair test, all animals must run the same

distance.

(c) A car starting from rest accelerates to 15 m/s in 4 s. What is the

acceleration of the car?

3.75 m/s

2

.

(d) A car is moving at 27 m/s when it slams on the breaks and

comes to a stop 1.2 s later. What is the acceleration of the car?

-22.5 m/s2. Note the minus symbol indicates negative

acceleration (deceleration).

Table 10.02.04 

shows the distance covered by a car starting

from rest to 20 s later.

(a) Represent the information in Table 10.02.04 in a graph.

(i) How far had the car travelled after 2 min?

120 m.

(ii) At what time did the car reach 60 m?

1 s.

(iii) What was the speed of the car at 8 s and 10 s? Explain your

answer.

At both 8 s and 10 s the speed of the car is 60 m/s. The car is

travelling at a constant velocity.

(iv) Give the time taken for the car to reach 800 m.

800 m/60 m/s = 13.33 s.

(v) Is the car accelerating? Justify your answer.

Since the car is not changing its velocity (speed) and there is

now mention of direction, we can say there is no acceleration.

Tracking a jogger around a field, a graph of their movement can

be seen in Fig. 10.02.07. Using this graph, answer these questions.

(i)  What is the velocity of the runner at 1 s?

4 m/s

(ii) What is the velocity of the runner at 11 s?

6 m/s

(iii) In your own words, describe the motion of the runner

between 0–10 s.

The runner is accelerating from rest to 8 m/s within 2 s. After

these two seconds the runner maintains this velocity for 8 s.

(iv) What do we call the motion of the runner between 2–10 s?

Constant velocity.

(v) Calculate the acceleration of the runner between 0 and 2 s.

4m/s

2

(vi) Calculate the acceleration of the runner between 10 and 12

s.

-2 m/s

2

. Note the minus symbol indicates negative acceleration

(deceleration).

(vii) What is the acceleration of the runner between 18 and 20 s?

-5 m/s

2

.

Analysing Distance–Time Graphs

Graph showing an object moving with constant velocity

Graph showing an object moving with constant velocity

towards its starting point

Analysing Distance–Time Graphs

Graph showing an object that is not moving (stationary)

Analysing Distance–Time Graphs

Some graphs, like the one in 

Fig. 10.02.11

, include all three

conditions. Tell the story of this graph.

From 0 to 4 s the object is moving at a constant speed.

Between 4 and 6 s the object does not move; it stops at 40

metres. From 6–10 s the object is moving at a constant speed

back towards the 0 metre mark (start line) (where the object

started).

Analysing Velocity–Time Graphs

Graph showing an object moving with constant acceleration

Graph showing an object moving with constant deceleration

Analysing Velocity–Time Graphs

Analysing Velocity–Time Graphs

You may see a graph with all of these three conditions

involved, like the one shown in 

Fig. 10.02.15

. Can you explain

the motion of the object?

From 0–4 s the object is moving at constant acceleration, i.e. the

object has accelerated for those four seconds. Between 4–6 s the

object is travelling at a constant velocity so it is not accelerating.

Between 6–10 s the object is slowing down (decelerating) at a

constant acceleration.

Copy and Complete

In this topic I learned about 

distance

, speed and time. I also

learned the difference between speed, 

velocity

and 

acceleration

.

Speed just gives us a value of how fast a car is going, whereas

velocity gives us the speed and the 

direction

. The unit of speed and

velocity is 

metre per second (m/s)

.  

Acceleration is the 

rate

 of

change

 of velocity. The unit of acceleration is the 

metre per second

squared (m/s

2

). 

An alternative word for negative acceleration is

deceleration

. A distance-time graph allows me to see how far an

object has travelled in a certain amount of 

time

.

A 

velocity-time

 graph tells me how 

fast

 an object is moving at a

certain point in time.

1. Define speed. What is the unit of speed?

The speed of an object is the distance it travels in a certain

amount of time, or the rate of change of distance with

respect to time. Unit: metre per second (m/s).

2. Define velocity. What is the unit of velocity?

Velocity is speed in a given direction.

Unit: metre per second (m/s).

3. Define acceleration. What is the unit of acceleration?

Acceleration is the change in velocity over time, or the rate

of change of velocity with respect to time. The unit of

acceleration is the metre per second squared m/s

2

(also represented as m/s/s).

4. A ball was dropped from a balcony in a sports hall. The

ball’s approximate velocity was measured each second as

it fell. The data collected during this experiment is given in

the graph (

Fig 10.02.16

).

4. (i) What was the velocity of the ball after 3 s?

12 m/s

4. (iii) What was the acceleration of the ball?

4 m/s

2

Time (s)

(ii) What is the distance travelled after 4 s?

8 metres.

(iii) How long did it take to travel 12 m?

6 seconds.

(iv) What was the walker’s speed at 5 s?

2 m/s.

(v) Comment on the shape of the graph.

Straight line graph through the origin.

A proportional relationship.

(vi) What does the shape of the line tell you about the speed of

the walker?

The walker was moving at a constant speed.

Time (s)

(ii) What is the cyclist’s velocity after 10 seconds?

5m/s

(iii) At what times was the cyclist’s velocity 10 m/s?

15 s and 60 s.

(iv) What was the cyclist’s acceleration after 20 seconds?

Acceleration = change in velocity/time taken

Acceleration = 12 – 0/20  Acceleration= 12/20 = 0.6 m/s

2

(v) Comment on the general shape of the graph, referring to specific

phases.

Acceleration between 0 and 20 s. Constant velocity between 20 and

50 s. The bike comes to an instant stop at 40 s until 45 s.

Acceleration again between 50 and 60 s.

(vi) Give a possible explanation for the sudden drop in velocity at 40 s

and 45 s.

Any reasonable explanation; e.g. red light, car pulling out, cyclist

tiring.

7. A stone was

dropped from

the top of a cliff

and the distance

that it fell was

measured at the

intervals of time

as given in 

Table

10.02.08.

(ii) Use the graph to find how far the stone had fallen in 3.5 s.

60 m.

(iii) Calculate the average speed of the falling stone between the

second and the fourth second. Give the unit with your answer.

30 m/s.

(iv) In this experiment, is distance fallen directly proportional to

time? Justify your answer.

No. The graph is not a straight line.