Chapter 3

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Chapter 3

3.1 Distance & Displacement

3.2 Speed & Velocity

3.3 Displacement-Time Graph

3.4 Acceleration

3.5 Velocity-Time Graph

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At the end of this chapter you should be able to:

• state what is meant by speed, velocity and acceleration

• recognise motion for which the acceleration is constant and calculate the acceleration

• recognise motion for which the acceleration is not constant • plot and interpret a speed-time graph

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At the end of this chapter you should be able to:

• calculate the area under a speed-time graph to determine the

distance travelled for motion with constant speed or constant acceleration

Chapter 3

 recognise from the shape of a speed-time graph when a body is a) at rest

b) moving with constant speed c) moving with constant acceleration

d) moving with an acceleration that is not constant

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Unit 3.1

Displacement

Definition:

Displacement is defined as the distance moved in a

specified direction.

Example:

Q. A car travels 10 km due East and makes a

U-turn back and travels a further 7 km due West.

What is the distance covered by the car and

what is its displacement at the end of the journey?

A. Distance

covered

= (10 + 7) km = 17 km

Displacement

= (10 - 7) km = 3 km

10 km

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Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Do the following physical quantities have any direction associated with them? i.e. are they scalars or vectors?

Physical Quantities Scalars or Vectors

Length

Mass

Weight

Electric Current

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Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Length

Mass

Weight

Electric Current

Density

(7)

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Length

Mass

Weight

Electric Current

Density

Scalar

(8)

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Length

Mass

Weight

Electric Current

Density

Scalar

(9)

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Length

Mass

Weight

Electric Current

Density

Scalar

Vector

(10)

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Length

Mass

Weight

Electric Current

Density

Scalar

Vector

Scalar

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Chapter 3

3.1 Distance & Displacement

(Completed)

3.2 Speed & Velocity

3.3 Displacement-Time Graph

3.4 Acceleration

3.5 Velocity-Time Graph

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Unit 3.2

Speed

Definition:

Speed

is defined as the rate of change of distance with time.

Equation: v = s / t where v = speed; s = distance; t = time interval

Note:

 This equation is for instantaneous speed (actual speed).

 <v> is the average speed.

 1 km = 1000 m and 1 hour = 60 × 60 s (= 3600 s)

Thus,

1 km/h = 1000/3600 m/s 1 km/h = 5/18 m/s

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Unit 3.2

Example:

A car covers a distance of 15 km in 30 minutes.

Q. What is the car’s average speed?

Q. Does the above answer tell us anything about the maximum or minimum

speed of the car?

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Unit 3.2

Example:

A car covers a distance of 15 km in 30 minutes.

Q. What is the car’s average speed?

Q. Does the above answer tell us anything about the maximum or minimum

speed of the car?

Q. The above answer seems slow for a car. Explain why it is probably correct.

<v> = s / t = 15 km / (30/60) h = 15 km / 0.5 h = 30 km h

-1

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Unit 3.2

Example:

A car covers a distance of 15 km in 30 minutes.

Q. What is the car’s average speed?

Q. Does the above answer tell us anything about the maximum or minimum

speed of the car?

Q. The above answer seems slow for a car. Explain why it is probably correct.

<v> = s / t = 15 km / (30/60) h = 15 km / 0.5 h = 30 km h

-1

No, it does not.

Since the answer is just the average speed, it just shows that it is

possible for the car to actually have larger or smaller speeds.

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Unit 3.2

Question 1:

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Unit 3.2

Question 1:

An aeroplane travelling at a constant speed covers a distance of 2750 km in a

period of 2 hours 30 minutes. What is the speed of the aeroplane?

s = 2750 km

t = 2 h 30 min = 2.5 h

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Unit 3.2

Question 2:

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Unit 3.2

Question 2:

You run a race in 25 seconds. If your average speed for the race was 8 m/s, what

distance was the race?

t = 25 s

<v> = 8 m/s

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Unit 3.2

Question 3:

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Unit 3.2

Question 3:

A man walks a distance of 2 km in 40 minutes. Calculate his average speed in

(i) km/h & (ii) m/s.

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Unit 3.2

Question 3:

A man walks a distance of 2 km in 40 minutes. Calculate his average speed in

(i) km/h & (ii) m/s.

(i) <v> = s / t = 2km / (40/60)h = 3 km/h

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Unit 3.2

Question 3:

A man walks a distance of 2 km in 40 minutes. Calculate his average speed in

(i) km/h & (ii) m/s.

(i) <v> = s / t = 2km / (40/60)h = 3 km/h

(ii) <v> = s / t = (2 x 1000)m / (40 x 60)s = 0.833 m/s (to 3 sig fig)

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Unit 3.2

Question 4:

Convert the following speeds between m/s and km/h.

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Unit 3.2

Question 4:

Convert the following speeds between m/s and km/h.

i) 10 m/s & ii) 90 km/h (the speed limit on the Singapore expressways.)

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Unit 3.2

Question 4:

Convert the following speeds between m/s and km/h.

i) 10 m/s & ii) 90 km/h (the speed limit on the Singapore expressways.)

(i) v = 10 m/s = (10/1000)km / (1/3600)h = 36 km/h

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Unit 3.2

Question 5:

You walk a distance of 2 km at an average speed of 6 km/h. You then run for a

further 10 minutes with an average speed of 12 km/h.

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Unit 3.2

Question 5:

You walk a distance of 2 km at an average speed of 6 km/h. You then run for a

further 10 minutes with an average speed of 12 km/h.

What is your average velocity for the whole journey?

s₁ = 2 km = 2000 m v₁ = 6 km/h

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Unit 3.2

Question 5:

You walk a distance of 2 km at an average speed of 6 km/h. You then run for a

further 10 minutes with an average speed of 12 km/h.

What is your average velocity for the whole journey?

s₁ = 2 km = 2000 m v₁ = 6 km/h

t₁ = s₁/ v₁ = 2/6 h = 1/3 h = 20 min = (20 x 60) s = 1200 s

v₂ = 12 km/h

t₂ = 10 min = 1/6 h = (10 x 60) s = 600 s

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Unit 3.2

Question 5:

You walk a distance of 2 km at an average speed of 6 km/h. You then run for a

further 10 minutes with an average speed of 12 km/h.

What is your average velocity for the whole journey?

s₁ = 2 km = 2000 m v₁ = 6 km/h

t₁ = s₁/ v₁ = 2/6 h = 1/3 h = 20 min = (20 x 60) s = 1200 s

v₂ = 12 km/h

t₂ = 10 min = 1/6 h = (10 x 60) s = 600 s

s₂ = v₂x t₂ = 12 km/h x 1/6 h = 2 km = 2000 m

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Unit 3.2

Velocity

Velocity is a vector form of speed.

Definition:

Velocity is defined as the rate of change of distance in a specified direction

with time.

or

Velocity is defined as the rate of change of displacement with time.

or

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Unit 3.2

Example:

A car drives 20 km in an eastward direction and then 10 km in a westward

direction. It takes 20 minutes to complete its journey. Calculate:

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Unit 3.2

Example:

A car drives 20 km in an eastward direction and then 10 km in a westward

direction. It takes 20 minutes to complete its journey. Calculate:

a) average speed for the journey ; b) the average velocity for the journey.

a) Distance = (20 + 10) km = 30 km

Time = 20 min = 1/3 h

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Unit 3.2

Example:

A car drives 20 km in an eastward direction and then 10 km in a westward

direction. It takes 20 minutes to complete its journey. Calculate:

a) average speed for the journey ; b) the average velocity for the journey.

a) Distance = (20 + 10) km = 30 km

Time = 20 min = 1/3 h

Average speed = distance / time = 30 km / (1/3)h = 90 km/h

b) Displacement = (20 –10) km = 10 km

Time = 20 min = 1/3 h

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Chapter 3

3.1 Distance & Displacement

(Completed)

3.2 Speed & Velocity

(Completed)

3.3 Displacement-Time Graph

3.4 Acceleration

3.5 Velocity-Time Graph

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Unit 3.3

A graph can be a very useful way to show information about how an object moves.

One type of graph we can use is the Displacement-Time Graph.

s

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Unit 3.3

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Unit 3.3

We can measure the distance of the balls from the start point and plot this information on a graph.

Time (s) Displacement (cm)

1

2

3

4

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Unit 3.3

1

2

3

4

5

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Unit 3.3

1

2

3

4

5

2

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Unit 3.3

1

2

3

4

5

2

4

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Unit 3.3

1

2

3

4

5

2

4

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Unit 3.3

1

2

3

4

5

2

4

6 8

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Unit 3.3

Use a suitable scale to copy the information onto the displacement-time graph below.

Displacement cm Time s 0 2 4 6 8 10

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Unit 3.3

You should now have a straight-line graph.

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Unit 3.3

0 1 2 3 4 5 6 7

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Unit 3.3

0 1 2 3 4 5 6 7

Q.What type of motion will produce this type of graph?

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Unit 3.3

Look at the displacement-time graphs below to determine how the objects are moving.

Type of motion:

Graph 1.

S

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Unit 3.3

Type of motion:

Object is not moving at all

Graph 1.

S

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Unit 3.3

Type of motion:

Graph 2.

S

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Unit 3.3

Type of motion:

Constant negative velocity

(moving object is approaching observer)

Graph 2.

S

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Unit 3.3

Type of motion:

Graph 3.

S

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Unit 3.3

Type of motion:

Object is not moving.

It is also at the same spot as the observer.

Graph 3.

S

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Unit 3.3

Type of motion:

Graph 4.

S

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Unit 3.3

Type of motion:

Increasing velocity

Graph 4.

S

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Unit 3.3

Type of motion:

Graph 5.

S

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Unit 3.3

Type of motion:

Increasing Negative Velocity (object is speeding towards observer)

Graph 5.

S

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Unit 3.3

Gradient of a Displacement-Time Graph

In general the instantaneous speed (instantaneous velocity) can be calculated from the gradient of the graph at a point.

Example:

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Unit 3.3

Gradient of a Displacement-Time Graph

In general the instantaneous speed (instantaneous velocity) can be calculated from the gradient of the graph at a point.

Example:

What is the speed at t=6 s?

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Unit 3.3

Questions:

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Unit 3.3

Questions:

Look at the displacement-time graph below and answer the questions.

Question 1 :

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Unit 3.3

Questions:

Question 1 :

At which position(s) is the velocity zero?

Solution:

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Unit 3.3

Questions:

Question 2 :

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Unit 3.3

Questions:

Question 2 :

At which position(s) is the velocity the greatest?

Solution:

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Unit 3.3

Questions:

Question 3 :

At which position is it farthest from the reference point?

Solution:

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Chapter 3

3.1 Distance & Displacement

(Completed)

3.2 Speed & Velocity

(Completed)

3.3 Displacement-Time Graph

(Completed)

3.4 Acceleration

3.5 Velocity-Time Graph

(67)

Unit 3.4

Definition:

Acceleration is defined as the rate of change of velocity with respect to time.

Equation:

(v – u)

a = t

where a = acceleration in m s-2_;

v = final velocity in m s-1 _;

u = initial velocity in m s-1_;

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Unit 3.4

Example 1:

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Unit 3.4

Example 1:

A car starting from rest is accelerated to a velocity of 48 m/s in a time of 10 seconds. Calculate the acceleration of the car.

Solution:

Since the car starts from rest, u = 0 m s-1_.

Final velocity = v = 48 m s-1_.

Time interval = t = 10 s.

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Unit 3.4

Example 2:

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Unit 3.4

Example 2:

A sports car is travelling at a constant velocity of 10 m/s before accelerating to a velocity of 55 m/s in 3 seconds. What is the acceleration of the car?

Solution:

u = 10 m s-1_.

v = 55 m s-1_.

t = 3 s.

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Unit 3.4

Example 3:

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Unit 3.4

Example 3:

A car initially has a velocity of 58 m/s. After braking for 10 seconds it has a velocity of 33 m/s. Calculate the acceleration of the car.

Solution:

u = 58 m s-1_.

v = 33 m s-1_.

t = 10 s.

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Unit 3.4

Deceleration

When an object changes its velocity (or speed) to become slower, we can state this in several way.

We can say that the object

(a) is experiencing deceleration ;

(b) is experiencing retardation ;

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Unit 3.4

Example 4:

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Unit 3.4

Example 4:

A train slows down from 60 m/s to rest in one minute. Calculate the retardation of the train.

Solution: u = 60 m/s ; v = 0 m/s ; t = 60 s ;

a = (v-u) / t = [(0 – 60) / 60] m/s2_{= - 1 m/s}2

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Unit 3.4

Uniform Acceleration

Acceleration is said to be uniform if it remains constant over a period of time.

Example 5:

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Unit 3.4

Example 5:

The velocity of a car increases from rest to 9 m/s in the first 3 s and then it increases from 9 - 27 m/s over the next 6 seconds.

Is the acceleration of the car uniform?

Solution:

During the first 3 s, a = (v-u) / t = [(9 – 0) / 3] m/s2_{= 3 m/s}2_.

During the next 6 s, a = (v-u) / t = [(27 – 9) / 6] m/s2_{= 3 m/s}2_.

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Unit 3.4

Example 6:

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Unit 3.4

Example 6:

A car starts travelling from rest and undergoes constant acceleration for 10 seconds. If it is travelling at 6 m/s after 3 seconds what will be its final velocity?

Solution:

For the first 3 s, t = 3 s, u = 0 m/s, v = 6 m/s; a = (v - u) / t = [(6 – 0) / 3] m/s2_{= 2 m/s}2_.

For the next 7 s, a = (v – u) / t 2 = (v – 6) / 7

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Chapter 3

3.1 Distance & Displacement

(Completed)

3.2 Speed & Velocity

(Completed)

3.3 Displacement-Time Graph

(Completed)

3.4 Acceleration

(Completed)

3.5 Velocity-Time Graph

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Unit 3.5

In a similar manner to the Displacement-Time (s-t) graph, the Velocity-Time (v-t) graph can be used

to find out a lot about the motion of an object.

Note:

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Unit 3.5

You sit in a car and record the velocity from the speedometer as the car pulls away from some traffic lights. You obtain the data shown below.

Time (s) Velocity (m/s)

0 0

2 15

4 30

6 40

8 45

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Unit 3.5

Using the data, a v-t graph is plotted. Draw a smooth line through the points and we have a velocity-time graph of the motion of the car.

Time (s) Velocity (m/s)

0 0

2 15

4 30

6 40

8 45

10 45

2 4 6 8 10 12 14 16

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Common Shapes of Velocity-Time Graphs

The following 8 v-t graphs show the basic shapes that you will encounter.

Unit 3.5

1. At Rest (not moving) 2. Constant Velocity

v v

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Unit 3.5

3. Constant Acceleration 4. Constant Deceleration

v v

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Unit 3.5

5. Decreasing Acceleration 6. Increasing Acceleration

v v

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Unit 3.5

7. Decreasing Deceleration 8. Increasing Deceleration

v v

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Examples:

Look at the following velocity-time graph showing the velocity of an MRT train travelling between two stations.

Unit 3.5

0 5 10 15 20 25 30 35

0 10 20 30 40 50 60 70 80 90 100 110 120 130

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Unit 3.5

0 5 10 15 20 25 30 35

0 10 20 30 40 50 60 70 80 90 100 110 120 130

Time (s) V el oc it y (m /s ) B A C D

AB: a = (v – u) / t = (30 – 0) / (20 – 0) = 1.5 m/s2 _{(constant acceleration)} BC: a = (v – u) / t = (30 – 30) / (90 - 20) = 0 m/s2 _{(constant velocity)}

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Examples:

The following v-t graph shows the velocity of a car over a period of 12 seconds. Describe the motion of the car.

Unit 3.5

0 5 10 15 20 25 30 35 40 45

0 1 2 3 4 5 6 7 8 9 10 11 12 13

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Unit 3.5

0 5 10 15 20 25 30 35 40 45

0 1 2 3 4 5 6 7 8 9 10 11 12 13

Time (s) V el oc it y (m /s )

First 2 s:

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Unit 3.5

0 5 10 15 20 25 30 35 40 45

0 1 2 3 4 5 6 7 8 9 10 11 12 13

Next 2 s:

a = (v – u) / t = (40 – 10) / (4 –2) = 30 / 2 m/s2_{= 15 m/s}2_.

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Unit 3.5

0 5 10 15 20 25 30 35 40 45

0 1 2 3 4 5 6 7 8 9 10 11 12 13

t = 4 s to t = 6 s

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Unit 3.5

0 5 10 15 20 25 30 35 40 45

0 1 2 3 4 5 6 7 8 9 10 11 12 13

t = 6 s to t = 8 s

a = (v – u) / t = (20 – 40) / (8 – 6) = - 20 / 2 = - 10 m/s2_.

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Unit 3.5

0 5 10 15 20 25 30 35 40 45

0 1 2 3 4 5 6 7 8 9 10 11 12 13

t = 8 s to t = 12 s

a = (v – u) / t = (0 – 20) / (12 – 8) = - 20 / 4 = - 5 m/s2_.

Constant deceleration of – 5 m/s2_{for the last 4 s.}

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Instantaneous Velocity from the V-T Graph

Unit 3.5

Look at the graph on the left.

We can easily determine the velocity at any instant.

E.g.

at t = 2 s, v = 40 m/s

at t = 0.5 s, v = 20 m/s

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0

0 1 2 3 4 5 6

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Acceleration from the V-T Graph

Unit 3.5

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5

0 1 2 3 4 5

From the shape of the graph we know that the object is undergoing constant acceleration.

What is the initial velocity?

u = 0 m/s

What is the velocity after 4 s?

v = 40 m/s

Acceleration can be calculated using

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Distance from the V-T Graph

Unit 3.5

If a car moves at 20 m/s for 5 seconds how far has it travelled?

Distance = Velocity  Time (because car moves at constant velocity)

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Unit 3.5

Alternatively we could have seen this from the v-t graph.

1. Shade the area under the graph.

0 5 1 0 1 5 2 0 2 5

0 1 2 3 4 5 6

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Unit 3.5

Alternatively we could have seen this from the v-t graph.

1. Shade the area under the graph.

2. Calculate the area you have shaded.

Area = 5 x 20 = 100 m

0 5 1 0 1 5 2 0 2 5

0 1 2 3 4 5 6

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Unit 3.5

Notes:

You should see the area under the graph is equal to the distance covered.

This is actually true for any v-t graph.

Can the equation "Distance = Speed  Time" be used on all occasions?

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Summary

The velocity-time graph can be used to find:

* Instantaneous velocity

* Acceleration - from the gradient of the line

* Distance travelled - from the area underneath the graph

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Example 1:

What distance does the car cover as it moves according to the v-t graph shown below?

Unit 3.5

0 5 10 15 20 25

0 5 10 15 20 25 30 35

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Solution:

Distance = [(0.5 x 5 x 20) + (10 x 20) + (0.5 x 5 x (10 + 20)) + (0.5 x 10 x 10)]m = [ 50 + 200 + 75 + 50 ]m

= 375 m

Unit 3.5

0 5 10 15 20 25

0 5 10 15 20 25 30 35

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Example 2:

The graph below shows the motion of an MRT train as it passes from Yio Chu Kang to Ang Mo Kio to Bishan.

Unit 3.5

0 5 1 0 1 5 2 0 2 5 3 0 3 5

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0

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Solution:

Distance = [(0.5 x 20 x (40+70)) + (0.5 x 20 x 20) + (0.5 x 20 x (20+30)) + (0.5 x 30 x (20+40))

= [ 1100 + 20 + 500 + 900 ] m = 2520 m

Unit 3.5

0 5 1 0 1 5 2 0 2 5 3 0 3 5

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0

Time (s) V e lo c it y ( m /s )

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Solution:

Distance = [(0.5 x 20 x 20) + (0.5 x 20 x (20+30)) + (0.5 x 30 x (20+40)) = [ 20 + 500 + 900 ] m = 1420 m

Unit 3.5

0 5 1 0 1 5 2 0 2 5 3 0 3 5

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0

Time (s) V e lo c it y ( m /s )

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Chapter 3

3.1 Distance & Displacement

(Completed)

3.2 Speed & Velocity

(Completed)

3.3 Displacement-Time Graph

(Completed)

3.4 Acceleration

(Completed)

3.5 Velocity-Time Graph

(Completed)

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Unit 3.6

Have you ever wondered why objects fall?

According to Sir Isaac Newton,

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Unit 3.6

Free-Fall

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Unit 3.6

Free-Fall

Drop a small object and a large object together.

Which one will reach the floor the fastest?

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Unit 3.6

Free-Fall

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Unit 3.6

Free-Fall

A feather and a hammer falling together will follow the same pattern.

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Unit 3.6

Free-Fall

Using the information on the previous diagram, the following v-t graph for an object in free-fall near to the Earth is plotted.

0 10 20 30 40

0 1 2 3 4

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Unit 3.6

Free-Fall

What does the shape of your graph tell you about the motion of the objects?

Objects move with constant acceleration during free fall.

0 10 20 30 40

0 1 2 3 4

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Unit 3.6

Free-Fall

In what ways, if any, would the same experiment be different if it had been carried out on the surface of the moon?

The moon’s gravity is smaller than that of the earth.

As such, the acceleration due to gravity will be lesser on the moon. 0 10 20 30 40

0 1 2 3 4

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Unit 3.6

Free-Fall

What is the true value for the

acceleration due to gravity near Earth?

The true value for the

acceleration due to gravity near Earth is 9.8 m/s2_.

In most cases, you may assume g = 10 m/s2_.

0 10 20 30 40

0 1 2 3 4

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Unit 3.6

Free-Fall

But what if you had dropped a feather would it really fall at the same rate as a coin?

Probably not.

vacuum

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Unit 3.6

Free-Fall

But what if you had dropped a feather would it really fall at the same rate as a coin?

Probably not.

Problem with this fact

:

Why does the paper behave like this and

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Unit 3.6

Air-Resistance

Air resistance is the answer…

Walking through water is hard because it offers resistance to our motion.

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Unit 3.6

Air-Resistance

Air resistance depends on several factors:

1. Velocity of the object moving through air.

2. Density of the atmosphere.

3. Surface area exposed to air resistance .

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Unit 3.6

Terminal Velocity

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Unit 3.6

Example 1:

The following v-t graph shows the motion of a man as he parachutes out of an aeroplane.

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Unit 3.6

Example 1:

At which point on the graph does his parachute open?

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Unit 3.6

Example 1:

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Unit 3.6

Example 1:

What is his terminal velocity without a parachute?

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Unit 3.6

Example 1:

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Unit 3.6

Example 1:

What is his terminal velocity with a parachute?

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Unit 3.6

Example 1:

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Unit 3.6

Example 1:

A heavier man also parachutes. Will his terminal velocity be identical to the lighter man?

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Unit 3.6

Example 1:

A heavier man also parachutes. Will his terminal velocity be identical to the lighter man?

No, his terminal velocity is greater than the lighter man.

This is because his weight is larger than the lighter man;

as a result the downward force is larger. By the time the air resistance manages to balance the downward force (i.e.

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Unit 3.6

Example 2:

Draw the velocity-time graph for a stone thrown vertically into the air with a velocity of 20 m/s. Assume that the acceleration due to gravity is 10 m/s².

-20 -10 0 10 20

0 1 2 3 4

(134)

Unit 3.6

Example 2:

-20 -10 0 10 20

0 1 2 3 4

(135)

Unit 3.6

Example 2:

-20 -10 0 10 20

0 1 2 3 4

(136)

Unit 3.6

Example 2:

-20 -10 0 10 20

0 1 2 3 4

(137)

Unit 3.6

Example 2:

-20 -10 0 10 20

0 1 2 3 4

(138)

Unit 3.6

Example 2:

The corresponding speed-time graph is shown on the right..

-20 -10 0 10 20

0 1 2 3 4

Time (s) V e loc it y ( m /s ) -20 -10 0 10 20

0 1 2 3 4

(139)

Chapter 3

3.1 Distance & Displacement

(Completed)

3.2 Speed & Velocity

(Completed)

3.3 Displacement-Time Graph

(Completed)

3.4 Acceleration

(Completed)

3.5 Velocity-Time Graph

(Completed)

(140)

Chapter 3

Chapter 3

Contents

3.1

Distance & Displacement

3.2

Speed & Velocity

3.3

Displacement-Time Graph

3.4

Acceleration

3.5

Velocity-Time Graph

Chapter 3

Unit 3.1

Displacement

Definition:

Displacement is defined as the distance moved in a

specified direction.

Example:

Q.

A car travels 10 km due East and makes a

U-turn back and travels a further 7 km due West.

What is the distance covered by the car and

what is its displacement at the end of the journey?

A. Distance

covered

= (10 + 7) km = 17 km

Displacement

= (10 - 7) km = 3 km

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and

direction

.

Unit 3.1

We say that :

Distance is a

scalar

quantity — it has

magnitude only.

Displacement is a

vector

quantity — it has

magnitude

and