UNIT 1: 1-D KINEMATICS Lesson 4: Graphical Analysis of Accelerating Motion CENTRE HIGH: PHYSICS 20

UNIT 1: 1-D KINEMATICS

Lesson 4:

Graphical Analysis of Accelerating Motion

CENTRE HIGH: PHYSICS 20

Reading Segment #1:

Average and Instantaneous Velocity on Position-Time Graphs

To prepare for this section, please read:

Unit 1: p.14

D1. Position-Time Graphs (Accelerated Motion)

Recall, when the motion of an object is uniform,

- the velocity is constant

- since velocity is the slope of a position-time graph,

the graph is a straight line (constant slope).

But now, the velocity is going to change.

As a result, the line is no longer going to be straight.

Consider the following position-time graph

for an accelerating object: (Forward is positive)

d (m)

t (s)

Can you describe the motion?

d (m)

Zero slope

t (s)

At the start, the slope is zero.

Thus, the object starts from rest.

d (m)

Slopes are increasing

t (s)

But, over time, the slope increases.

Thus, the object's velocity increases, which in this case

means that the object is speeding up.

d (m)

Slopes are increasing

t (s)

Since the velocity is changing,

the object is accelerating.

For non-linear position-time graphs, there are two types

of velocities (i.e. slopes) that could be calculated:

1. Average velocity

- from one time to another

2. Instantaneous velocity

- at one point in time

1. Average velocity

To find the average velocity from t = t1 to t = t2,

Find the initial position (at t1) and

the final position (at t2) on the graph

Draw a straight line from the two positions

(called a secant line)

Find the slope of this secant line

e.g. How would we find the average velocity from t = 3.0 s

to t = 5.0 s on the graph below?

d (m)

3.0 5.0 t (s)

d (m)

3.0 5.0 t (s)

First, find the initial and final position on the line.

d (m)

secant line

3.0 5.0 t (s)

Draw a straight line between the two points.

This is called the secant line.

d (m)

slope = vavg

3.0 5.0 t (s)

Find the slope of the secant line.

The slope is the average velocity between 3.0 and 5.0 s.

2. Instantaneous velocity

To find the instantaneous velocity at t = t1 :

Find the position (at t1) on the graph

Draw a tangent line at this point

Choose 2 points on the tangent line and find its slope

e.g. How would we find the instantaneous velocity at t = 4.0 s ?

d (m)

4.0 t (s)

d (m)

4.0 t (s)

First, find the position at t = 4.0 s on the line.

d (m)

tangent line

4.0 t (s)

Next, draw a tangent line at this point.

A tangent line touches the curve only at this point,

and if done correctly, it should show the slope at that point.

d (m)

slope of tangent

= velocity

4.0 t (s)

Finally, locate two points on the tangent line

and find its slope.

The slope is the instantaneous velocity.

Ex. 1 For the graph below, find: (North is positive)

a) the average velocity from 0.25 s to 5.0 s

b) the instantaneous velocity at 1.0 s

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

a) Locate the positions at t = 0.25 s and t = 5.0 s

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

secant line

a) Draw a straight line between the two points.

This the secant line.

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

02468101214161820222426283032343638404244464850

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

(5.0 s, 50.0 m)

secant line

(0.25 s, 11.0 m)

a) Choose two points on the secant line and find the slope.

a) Average velocity between 0.25 s and 5.0 s:

vavg = slope of secant line

= y2 - y1 = (50.0 m) - (11.0 m)

x2 - x1 (5.0 s) - (0.25 s)

= 8.2 m/s North

Be sure to include the direction for velocity.

b) Locate the position at t = 1.0 s

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

tangent

line

b) Draw the tangent line at this point.

It should touch the curve only at this point.

02468101214161820222426283032343638404244464850

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

(3.25 s, 48.0 m)

tangent

line

(1.0 s, 22.0 m)

b) Find two points on the tangent line. Calculate the slope.

02468101214161820222426283032343638404244464850

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (s)

Po

siti

on

(m

)

b) Instantaneous velocity between 1.0 s:

v1.0 = slope of tangent line

= y2 - y1 = (48.0 m) - (22.0 m)

x2 - x1 (3.25 s) - (1.0 s)

= 12 m/s North

Be sure to include the direction for velocity.

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 1 p. 16 #1

Reading Segment #2:

Velocity-Time Graphs

To prepare for this section, please read:

Unit 1: p.15

D2. Velocity-Time Graphs (Accelerated Motion)

There are two major calculations we find for velocity-time

graphs:

1. Acceleration

- this will be the slope of a v-t graph

2. Displacement

- this will be the area "under" the v-t graph

1. Acceleration on a velocity-time graph

The slope of a v-t graph is the acceleration

Thus, if the object has uniform acceleration:

- the acceleration is constant

- the slope is constant on the v-t graph

- this produces a straight line

e.g. Ref: Forward is positive

Backward is negative

If the graph has a positive slope (i.e. a positive acceleration),

it means the object's velocity is increasing.

v v

t

t

Ref: Forward is positive

Backward is negative

If the graph has a negative slope (i.e. a negative acceleration),

it means the object's velocity is decreasing.

v v

t

t

Ref: Forward is positive

Backward is negative

If the graph has a zero slope (i.e. a zero acceleration),

it means the object has a constant velocity.

v v

t

t

Ex. 2 For the graph below, find the acceleration at t = 3.0 s.

(East is positive)

Velocity 8.0

( 103 m/s)

2.0

4.0 t (s)

Velocity 8.0

( 103 m/s)

2.0

4.0 t (s)

Acceleration is the slope of a v-t graph.

Since this is a straight line, you can choose any 2 points

on the line to calculate the slope.

Velocity 8.0 (0 s, 8.0 103 m/s)

( 103 m/s)

(4.0 s, 2.0 103 m/s)

2.0

4.0 t (s)

Choose 2 points on the line.

Acceleration at 3.0 s:

a3.0 = slope of line

= y2 - y1 = (2.0 103 m/s) - (8.0 103 m/s)

x2 - x1 (3.0 s) - (0 s)

= -1.5 103 m/s2

= 1.5 103 m/s2 West

Be sure to include the direction for acceleration.

Ex. 3 Describe the motion showed by the following

velocity-time graph: (Forward is positive)

v (m/s)

10

t (s)

-6

v (m/s) Forward is positive

10 A

t (s)

-6

At the very start of the motion (A), the object is moving

forward at 10 m/s.

v (m/s) Forward is positive

10 A

B t (s)

-6

Along AB, the object is slowing down at a constant acceleration (constant slope), from 10 m/s to 0.

It is still moving forward, since the velocity is positive.

At B, the object is at rest.

v (m/s) Forward is positive

10 A

B t (s)

-6 C

Along BC, the object is speeding up at a constant acceleration (constant slope), from 0 to 6 m/s.

It is now moving backward, since the velocity is negative.

At C, the object is at moving backwards at 6 m/s.

v (m/s) Forward is positive

10 A

B t (s)

-6 C D

Along CD, the object is at a constant velocity of -6.0 m/s.

Acceleration (slope) is zero.

At D, the object is at moving backwards at 6 m/s.

v (m/s) Forward is positive

10 A

B E t (s)

-6 C D

Along DE, the object is slowing down at a constant

acceleration (slope), from 6 m/s to zero.

It is moving backwards, since the velocities are negative.

At E, the object has come to rest.

2. Displacement on a velocity-time graph

The area between the line and the t-axis on a

v-t graph is the displacement

Ref: Forward is positive

Backward is negative

e.g. When the line is above the t-axis,

displacement (area) is positive.

i.e. A forward displacement

v v

t t

Ref: Forward is positive

Backward is negative

e.g. When the line is below the t-axis,

displacement (area) is negative.

i.e. A backward displacement

v v

t t

Ex. 4 Calculate the displacement for the first 8.0 seconds.

(North is positive)

v (m/s)

33

14

8.0 t (s)

v (m/s)

33

14

8.0 t (s)

Displacement is the area between the line

and the t-axis

v (m/s)

33

1

14

2

8.0 t (s)

To calculate the area, find the area of the triangle (1)

and the rectangle (2).

v (m/s)

33

1 19 m/s

14

2 14 m/s

8.0 s 8.0 t (s)

Show the dimensions on the diagram.

Displacement for the first 8.0 seconds:

A1 = Area of triangle

= 0.5 b h

= 0.5 (8.0 s) (19 m/s) = 76 m

A2 = Area of rectangle

= L w

= (8.0 s) (14 m/s) = 112 m

Thus, d = A1 + A2

= 1.9 102 m North

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 1 pp. 16 - 17 #2 - 5