Linear Accelerated MotionPart 2

For the Higher Level Leaving Cert Course

©Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork

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Gravity

Signs are always how you start e.g. if you start throwing upwards, then up is +

Acceleration is g and always down. Sign depends on if you say up is + or down is +.

Watch for displacement e.g. throwing something off a cliff 30 m above ground, hence displacement is -30

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Example 3.1 [LC:1988 Q1 (b)]

A particle falls freely from rest from a point o, passing three points a, b and c, the distances ab and bc being equal. If the particle takes 3 s to pass from a to b and 2 s from b to c, calculate |ab|

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A

B

C

u=u

+

x

2x

Always go from A as initial velocity remains same

A to B

u=u, a=g, s=x, t=3

2.8862

)3(2

1)3(

2

1

2

2

ux

gux

atuts

A to C

u=u, a=g, s=2x, t=5

5.12252

)5(2

1)5(2

2

1

2

2

ux

gux

atuts

Solve to give x =147 m

Down is positivehence g is

positive here

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Example 3.2 [LC:2002 Q1(a)]

A stone is thrown vertically upwards under gravity with a speed of u m/s from a point 30 m above the horizontal ground. The stone hits the ground 5 s later.

i) Find the value of uii) Find the speed with which it hits the

ground.

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+

-30

u=+ua=-gs=-30t=5

Finding u

smu

u

atuts

/5.18

)5)(8.9(21)5(30

2

1

2

2

Finding v

smv

smv

v

atuv

/5.30||

/5.30

)5)(8.9(5,18

Displacement is -30 because up is

positive

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Example 3.3 [LC:1990 Q1(a)]

A particle is projected vertically upwards with velocity u m/s and is at a height h after t1 and t2 seconds respectively. Prove that:

g

htt

2. 21

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h

+

u=+ua=-gs=+ht=t (t1 on way up, t2 on way down)

Solving

022

22

2

21

2

2

2

2

hutgt

gtuth

tguth

atuts

g

htt

2. 21

Product of 2 rootsRemember α and β

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Example 3.4 [LC:1992 Q1(a)]

A balloon ascends vertically at uniform speed. 7.2 seconds after it leaves the ground, a particle is let fall from the balloon. The particle takes 9 seconds to reach the ground. Calculate the height from which the particle was dropped.

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u

h

+

Balloon

uh

hu

time

cedisspeed

2.72.7

tan

For the Particle

u= +ua= -9.8t= 9s= -h

2

2

)9)(8.9(2

1)9(

2

1

uh

atuts

smu

uu

/5.24

9.39692.7

muh 4.176)5.24(2.72.7

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Example 3.5 [LC:2007 Q1(a)]

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3rd second is between t=2 seconds and t=3 seconds

Let x = distance travelled in first 3 secondsLet y = distance travelled in first 2 seconds

smu

uyx

uy

ux

/4.5

9.295.24

)2)(8.9(2

1)2(

)3)(8.9(2

1)3(

2

2

From the Question

Equations always give zero to n time

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Finding Height of the tower

ms

s

atuts

100

)4)(8.9(2

1)4(4.5

2

1

2

2

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Now try some questions by yourself on the attached sheet