Linear Accelerated Motion Part 1 For the Higher Level Leaving Cert Course ©Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork.

Linear Accelerated MotionPart 1

For the Higher Level Leaving Cert Course

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

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Ordinary LevelBefore you start this course, it is highly recommended you complete the basics in the ordinary level course.This will give you the basics in the different sections

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Types of Questions

Velocity Time GraphPeriod of Constant SpeedNo Period of Constant Speed

Gravity/Vertical Motion2 Bodies in Motion/OvertakingPassing successive Points/2 points

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Velocity Time GraphArea under Curve = distance travelledSlope (y over x) is accelerationUse equations of motion for constant

acceleration period only.

a

d

t

t

2

1avelo

city

time

d

t1 t2

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Example 1.1 [LC:1997 Q1(a)]

A particle, moving in a straight line, accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance travelled while acceleration is 6 m. The total distance travelled is 30 m and the total time taken is 6 s.

Draw a speed time graph and hence or otherwise, find the value of v.

Calculate the distance travelled at v m/s

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Vel

ocity

Time2t t

v

6-3t

Area = 6

Area of whole shape is 30

Area of acceleration = 6

6

6)2(2

1

tv

vt

Relate Times together

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Total Area = 30 m

smv

v

tv

tvtvv

tvvt

/5.6

30)6(2

1)6(366

)6(

302

1366

302

1)36(6

13

12

5.6

66

vt

sm

vtArea

/21

5.6))13

12(36(

)36(

Finding area of triangles and rectangle

Finding Distance (area of rectangle)

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Example 1.2 [LC: 1994 Q1(a)]

A lift in a continuous descent, had uniform acceleration of 0.6 m/s2 for the first part of its descent and a retardation of 0.8 m/s2 for the remainder. The time, from rest to rest was 14 seconds.

Draw a velocity time graph and hence, or otherwise, find the distance descended.

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Vel

ocity

Timet 14-t

0.60.8

Fact: Ratio of accelerations

83

4

6.0

8.0

14

tt

t

Ratios are easier as fractions

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sm

tv

atuv

/8.4)8(6.0

6.0

m

Area

6.33

)8.4)(6(2

1)8.4)(8(

2

1

When Accelerating

Total Distance = Total Area

Area of left triangle and area of right

triangle

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Example 1.3 [LC: 2006 Q1(a)]

A lift starts from rest. For the first part of its descent, it travels with uniform acceleration f. It then travels with uniform retardation 3f and comes to rest. The total distance travelled is d and the total time taken is t.

i. Draw a speed time graph for the motionii. Find d in terms of f and t.

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33

2

1 f

f

t

t

4

3t

4

t

4

343

ftv

ftv

f3f

Vel

ocity

Timet1

t2

Ratios of Times

Total Area

Told t in the question is total time

Use info on acceleration

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8

3)

4

3(22

))(4(2

1))(

4

3(2

1

2ftfttvt

vt

vt

Area

Area of left

Triangle

Area of right triangle

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Now try some questions by yourself on the attached sheet

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Section 2Passing a number of

points

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Passing number of points

Always start from same point to use same initial velocity

i.e. if question is a to b, b to c and c to d; you treat it as

a to ba to ca to d

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Example 2.1 [LC:2003]

The points p, q, and r all lie in a straight line.A train passes point p with speed u m/s. The

train is travelling with uniform retardation f m/s2.The train takes 10 seconds to travel from p to q and 15 seconds to travel from q to r, where |pq|=|qr|=125 metres.

i) Show that f=1/3ii) The train comes to rest s metres after

passing r. Find s, giving your answer correct to the nearest metre.

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125 m 125 m

p q r

u = ua = -f

p to q

2

2

1atuts

fu

fu

5010125

)10(2

110125 2

fu

fu

2

62525250

)25(2

125250 2

p to rStart from same point

each time i.e. p

w

s

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17.14;3

1 uf

ms

mx

xu

fxu

51250301

301

)3

1(20

20

2

22

Solving

p to w

Get total distance from p to w and

then subtract distance p to r

from it

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Example 2.2 [LC:1988 Q1 (a)]

A particle moving in a straight line with uniform acceleration describes 23 m in the fifth second of its motion and 31 m in the seventh second. Calculate its initial velocity

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Fifth Second is between t=4 and t=5 Initial Velocity is u and acc=a

t=4, u=u, s=s, t=4, a=a

aus

aus

84

)4(2

1)4(

4

24

t=5, u=u, s=s, t=5, a=a

aus

aus

5.125

)5(2

1)5(

5

25

235.445 auss From Question

t=7, u=u, s=s, t=7, a=at=6, u=u, s=s, t=6, a=a

aus

aus

186

)6(2

1)6(

6

6

2

aus

aus

5.247

)7(2

1)7(

7

27

315.667 auss

smu /5

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Now try some questions by yourself on the attached sheet