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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:

time1 2

500

1000

The average speed is:

t

s

6522

750500

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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:

Time (s)1 5

2

10

The accelerations are:

t

v2

2

4

t

v3

22

3

8

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Area Under the v-t Graph

Consider a body with an initial velocity u at some time

to

which travels in a time tto a velocity ofv

Time (s)to t

u

v

tuA tu 1

tu 2

tu 3

tu 4

tu 5 tv

to

sssssssA 54321

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Class Problems

1. A train accelerates uniformly from rest to60kph in 6min, after which the speed isconstant, evaluate time taken to travel 6km.

2. If a train retards uniformly from 30m/s to16m/s in 200m, evaluate the retardation andthe total time of the retardation

3. A racing car engine is shut off. In the first

30s it covers 110m, and then comes to restafter a further 30s. Evaluate the initial speedand the total distance travelled.

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Problem #1:solution

1. Here we have v=0, u=60kph, t=6min (implying eqn

2) as we need s.

km36062

06.360

3600

s

So the train has travelled half 3km, but is now at constant speed

i.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

30006.3

360

6.3

60

23000 6.3

606.360

tt

tt

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Problem #1: alternative solution

We could obtain a direct solution from the

velocity-time graph thus

5403000360

)360(30006000

)360(3606000

60

6.3

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

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Problem #2:solution

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

s7.8

2321630200

23200

t

tt

To evaluate the retardation we could simply just

apply equation (1), however, we dont 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!

61.1

23/200

3016100161

a

a

t

uva

Now apply

equation (1)

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Problem #3:solution

3. Here we do our detective work and find that

when t=30, s=210 s: u=?and v=?

60

2

30 45030

60030

2

ua

aus

auaus

1-

60450

ms89.45.730

110u:Hence

5.73030110

uuu u

m1473

440

602

0

:nowand

944

s

Note the two

times imply

thesimultaneous

equations

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