Analytical toolbox studies

Introductory kinematicsby

Drs J. Whitty & B. Henderson

Learning Outcomes

• Use Distance-Time Graphs

•Use Velocity-Time Graphs

• Derivation of the Linear Equations of Kinematics

• Apply of the Equations of Kinematics

• Answer text book type questions on these topics

t

va

2

21 mc

mgh

FvP

Distance-Time Graphs

Consider an aeroplane travelling at 500km in the first hour of Flight and 750km in the second, plotting distance travelled against time

Renders:D

ista

nce

(km

)

time1 2

500

1000

The average speed is:

t

s652

2

750500

Velocity-Time Graphs

Consider a helicopter moves to velocity of 4m/s in seconds then increases its velocity to 8m/s in the next 3 seconds. Plotting the velocity against time

Renders:

Vel

ocity

(m

\s)

Time (s)1 5

2

10

The accelerations are:

t

v2

2

4

t

v322

3

8

Velocity-Time Graphs

Or more generally:

Vel

ocity

(m

\s)

Time (s)t

u

v

The average acceleration ist

uv

t

va

Where u is the initial velocity

and v is the final velocity

Slope

Area Under the v-t Graph

Consider a body with an initial velocity u at some time to which travels in a time t to a velocity of v

Vel

ocity

(m

\s)

Time (s)to t

u

v

tuA tu 1 tu 2 tu 3 tu 4 tu 5 tv

to sssssssA 54321

The Equations of Kinematics

Class Example: Derive the area of a trapezium

Now consider the aforementioned Velocity-Time Graph

Vel

ocity

(m

\s)

Time (s)to t

u

v

Area=The distance travelled in the time taken to change the velocity

The Equations of Kinematics

We have seen that:t

uva

And the area under the graph renders:

Hence atuv (1)

tvus 21 (2)

Substitution of (1) into (2) renders: tatuus 21

Therefore: 221 atuts (3)

Substitution of (2) into (1) for t renders:

uv

sauv

2Therefore: asuv 222 (4)

The Equations of Kinematics

Summary:

221 atuts (3)

asuv 222 (4)

atuv (1) The Gradient of the v-t Graph

tvus 21

(2) The Area Under the v-t Graph

t

sv The Slope of the s-t Graph

Class Problems

Class Problems

1. A train accelerates uniformly from rest to 60kph in 6min, after which the speed is constant, evaluate time taken to travel 6km.

2. If a train retards uniformly from 30m/s to 16m/s in 200m, evaluate the retardation and the total time of the retardation

3. A racing car engine is shut off. In the first 30s it covers 110m, and then comes to rest after a further 30s. Evaluate the initial speed and the total distance travelled.

Problem #1: solution

1. Here we have v=0, u=60kph, t=6min (implying eqn 2) as we need s.

km36062

0 6.360

3600

s

So the train has travelled half 3km, but is now at constant speedi.e. u=v=60/3.6=16.333m/s, we can use the same equation again here only need it to travel a further 3km, thus:

min954060

30006.3360

6.3

60

23000 6.3

606.3

60

tt

tt

Problem #1: alternative solution

We could obtain a direct solution from the velocity-time graph thus

5403000360

)360(30006000

)360(3606000

606.3

6.360

6.360

6.360

21

t

t

t

t-360s360s

This is the area of A triangle plus that of a rectangle

60/3.6

Problem #2: solution

2. Here v=16m/s, u=30m/s, s=200: t=?

s7.8

232

1630200

23200

t

tt

To evaluate the retardation we could simply just apply equation (1), however, we don’t have a time so in this case it is useful to use equation (2) first and then apply (1). i.e. answer the question backwards!

61.123/200

3016100161

a

a

t

uva

Now apply equation (1)

Problem #3: solution

3. Here we do our detective work and find that when t=30, s=210 s: u=? and v=?

60

230 45030

600

302

ua

aus

au

aus

1-

60450

ms89.45.730

110u :Hence

5.73030110

uuu u

m1473

44060

2

0 :now and 9

44

s

Note the ‘two times’ imply

the simultaneous

equations

Summary

Have we met our learning objectives: In particular are you able to: Use Distance-Time Graphs & Velocity-Time Graphs Derivation of the Linear Equations of Kinematics Apply of the Equations of Kinematics Apply Energy methods to solve engineering

problems Answer examination type questions on these topics