2) M1 Kinematics of a Particle Moving in a Straight Line

8/12/2019 2) M1 Kinematics of a Particle Moving in a Straight Line

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Introduction This chapter you will learn the SUVAT

equations

These are the foundations of many ofthe Mechanics topics

You will see how to use them to usemany types of problem involving motion


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Kinematics of a Particle moving in aStraight Line

You will begin by learning two of theSUVAT equations

s = Displacement (distance)u = Starting (initial) velocity

v = Final velocitya = Accelerationt = Time

2A

=

=

=

+ =

= +

Replace with theappropriate letters. Change in velocity =

final velocity initialvelocity

Multiply by t

Add u

This is theusual form!

( ) =

= +2

Replacewith the

appropriateletters


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You will begin by learning two of theSUVAT equations

s = Displacement (distance)u = Starting (initial) velocity

v = Final velocitya = Accelerationt = Time

2A

= +

= + 2

You need to consider using negative numbers insome cases

P Q

Positive direction

O4m 3m

2.5ms -1 6ms -1

If we are measuring displacements from O, and left to rightis the positive direction

For particle P: For particle Q: = 4

= 2.5 = 3

= 6 The particle is to the left of

the point O, which is thenegative side

The particle is moving at2.5ms -1 in the positive direction

The particle is tothe right of thepoint O, which isthe positive side

The particle is moving at 6ms -1 in the negative direction


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You will begin by learningtwo of the SUVAT

equations

s = Displacement (distance)

u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time

2A

= +

= +2

A particle is moving in a straight line from A to B with constantacceleration 3ms -2. Its speed at A is 2ms -1 and it takes 8 seconds tomove from A to B. Find:a) The speed of the particle at Bb) The distance from A to B

=? = 2 =? = 3 = 8 A B

2ms -1 Start with a

diagram

Write out suvat andfill in what you know

For part a) we needto calculate v, and we

know u, a and t = +

= 2 + (3 8)

= 26

Fill in thevalues you

know

Remember toinclude units!

You always need to set up the question in thisway. It makes it much easier to figure out whatequation you need to use (there will be more to

learn than just these two!)


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equations



2A

= +

= +2

A particle is moving in a straight line from A to B with constantacceleration 3ms -2. Its speed at A is 2ms -1 and it takes 8 seconds tomove from A to B. Find:a) The speed of the particle at B 26ms -1b) The distance from A to B

=? = 2 =? = 3 = 8 A B

2ms -1

= 26

For part b) we needto calculate s, and we

know u, v and t =

+ 2

= 2 + 262

8

= 14 8

= 112

Fill in thevalues you

knowShow

calculations

Rememberthe units!


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equations



2A

= +

= +2

A cyclist is travelling along a straight road. She accelerates at aconstant rate from a speed of 4ms -1 to a speed of 7.5ms -1 in 40seconds. Find:a) The distance travelled over this 40 secondsb) The acceleration over the 40 seconds

4ms -1 7.5ms -1

=? = 4 =? = 40 = 7.5

Draw a diagram(model the cyclist as

a particle)


= + 2

We are calculating s,and we already know

u, v and t

= 4 + 7.52

40

= 230

Sub in thevalues you

know

Rememberunits!


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equations



2A

= +

= +2

A cyclist is travelling along a straight road. She accelerates at aconstant rate from a speed of 4ms -1 to a speed of 7.5ms -1 in 40seconds. Find:a) The distance travelled over this 40 seconds 230mb) The acceleration over the 40 seconds

4ms -1 7.5ms -1

= 230 = 4 =? = 40 = 7.5

Draw a diagram(model the cyclist as

a particle)


For part b, we arecalculating a, and wealready know u, v and

t

Sub in thevalues you

know

Subtract 4

= +

7.5 = 4 + 40

7.5 = 40 Divide by

40 = 0.0875

=?


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equations



2A

= +

= +2

A particle moves in a straight line from a point A to B with constantdeceleration of 1.5ms -2. The speed of the particle at A is 8ms -1 and thespeed of the particle at B is 2ms -1. Find:a) The time taken for the particle to get from A to Bb) The distance from A to B

8ms -1 2ms -1

=? = 8 = 1.5 =? = 2

Draw a diagram


As the particle isdecelerating, a is

negative = +

2 = 8 1.5

6 = 1.5

4 =

Sub in thevalues you know

Subtract 8

Divide by -1.5

A B


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equations



2A

= +

= +2

A particle moves in a straight line from a point A to B with constantdeceleration of 1.5ms -2. The speed of the particle at A is 8ms -1 and thespeed of the particle at B is 2ms -1. Find:a) The time taken for the particle to get from A to B 4 secondsb) The distance from A to B

8ms -1 2ms -1

=? = 8 = 1.5 =? = 2

Draw a diagram


As the particle isdecelerating, a is

negativeSub in the

values you know

= 4

= + 2

= 8 + 22

4

= 20

Calculate theanswer!

A B


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equations



2A

= +

= +2

After reaching B the particle continues to move along the straight linewith the same deceleration. The particle is at point C, 6 seconds afterpassing through A. Find:a) The velocity of the particle at Cb) The distance from A to C

8ms -1 2ms -1

=? = 8 = 1.5 =? = 6 A B C

?Update the

diagram

Write outsuvat using

points A and C

= +

= 8 (1.5 6)

= 1

Sub in thevalues

Work itout!

As the velocity is negative, this means theparticle has now changed direction and isheading back towards A! (velocity has a

direction as well as a magnitude!)

The velocity is 1ms -1 in the direction C to A


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equations



2A

= +

= +2

After reaching B the particle continues to move along the straight linewith the same deceleration. The particle is at point C, 6 seconds afterpassing through A. Find:a) The velocity of the particle at C - -1ms-1b) The distance from A to C

8ms -1 2ms -1

=? = 8 = 1.5 =? = 6 A B C

?Update the

diagram

Write outsuvat using

points A and C

= 1

= + 2

= 8 12

6

= 21

Sub in thevalues

Work itout!

It is important to note that 21m is the distance from A to Conly

The particle was further away before it changeddirection, and has in total travelled further than 21m


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equations



2A

= +

= +2

A car moves from traffic lights along a straight road with constantacceleration. The car starts from rest at the traffic lights and 30seconds later passes a speed trap where it is travelling at 45 kmh -1. Find:a) The acceleration of the carb) The distance between the traffic lights and the speed-trap.

0ms -1 45kmh -1

Lights Trap

Standard units to use are metres and seconds, or kilometres and hoursIn this case, the time is in seconds and the speed is in kilometresper hourWe need to change the speed into metres per second first!

Draw a diagram

45

45,000

12.5

Multiply by 1000 (km to m)

Divide by 3600 (hours to seconds)


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equations



2A

= +

= +2


0ms -1 45kmh -1

Lights Trap

Draw a diagram

= 12.5ms-1

Write out suvat andfill in what you know =?

= 0 =? = 30 = 12.5

= +

12.5 = 0 + 30

512

=

Sub in thevalues

Divide by30

You can useexact answers!


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equations



2A

= +

= +2


0ms -1 45kmh -1

Lights Trap

Draw a diagram

= 12.5ms-1

Write out suvat andfill in what you know =?

= 0 =? = 30 = 12.5 =5

12

= + 2

=0 + 12.5

2 30

= 187.5

Sub in

values

Work itout!


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You can also use 3 more formulae linkingdifferent combination of SUVAT, for a

particle moving in a straight line withconstant acceleration

2B

= +

= + 2

= +

=

=

= + 2

= + 2

= 2

2 =

+ 2 =

= + 2

= + 2

Subtract u

Divide by a

Replace t with theexpression above

Multiply numerators anddenominators

Multiply by 2a

Add u2

This is the way it isusually written!


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2B

= +

= + 2

= + 2

= + 2

= + +

2

=2 +

2

= +12

= +12

Replace v with u + at

Group terms on thenumerator

Divide the numeratorby 2

Multiply out thebracket

= +1

2


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2B

= +

= + 2

= + 2

= +1

2

= +

=

= +12

= ( ) +12

= +12

= 12

= 12

Subtract at

Replace u with v- atfrom above

Multiply out thebracket

Group up the at 2 terms


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You can also use 3 more formulaelinking different combination of

SUVAT, for a particle moving ina straight line with constant

acceleration

2B

= + = +

2

= + 2 = +12

= 12

A particle is moving in a straight line from A to B with constantacceleration 5ms -2. The velocity of the particle at A is 3ms -1 in thedirection AB. The velocity at B is 18ms -1 in the same direction. Find thedistance from A to B.

3ms -1 18ms-1

A B

Draw a diagram

=? = 3 = 5 =? = 18 Write out suvat

with theinformation given

= + 2

18 = 3 + 2(5)

324 = 9 + 10

315 = 10

31.5 =

Replace v, u and a

Work out terms

Subtract 9

Divide by 10

We arecalculating s,

using v, u and a


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acceleration

2B

= + = +

2

= + 2 = +12

= 12

A car is travelling along a straight horizontal road with a constantacceleration of 0.75ms -2. The car is travelling at 8ms -1 as it passes apillar box. 12 seconds later the car passes a lamp post. Find:a) The distance between the pillar box and the lamp postb) The speed with which the car passes the lamp post

8ms -1

PillarBox

LampPost

=? = 8 = 0.75 = 12 =?

Draw a diagram

Write out suvat with the

information given


using u, a and t = +12

= (8 12) +12

(0.75 12 )

= 150

Replace u, aand t

Calculate


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acceleration

2B

= + = +

2

= + 2 = +12

= 12

A car is travelling along a straight horizontal road with a constantacceleration of 0.75ms -2. The car is travelling at 8ms -1 as it passes apillar box. 12 seconds later the car passes a lamp post. Find:a) The distance between the pillar box and the lamp post 150mb) The speed with which the car passes the lamp post

8ms -1

PillarBox

LampPost

=? = 8 = 0.75 = 12 =?

Draw a diagram


information given

We arecalculating v,

using u, a and tReplace u, a

and t

Calculate

= +

= 8 + (0.75 12)

= 17

Often you can use an answer you have calculated lateron in the same question. However, you must take care

to use exact values and not rounded answers!


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acceleration

2B

= + = +

2

= + 2 = +12

= 12

A particle is moving in a straight horizontal line with constantdeceleration 4ms -2. At time t = 0 the particle passes through a point Owith speed 13ms -1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through Ab) The total time the particle is beyond Ac) The time taken for the particle to return to O

13ms-1

O A = 20 = 13 = 4 =? =?

Draw a diagram


information given

We arecalculating t,

using s, u and a = +12

20 = (13) + 12

( 4)

20 = 13 2

2 13 + 20 = 0

(2 5)( 4) = 0

= 2.5 4

Replace s, u

and a

Simplify terms

Rearrange and set equal to 0

Factorise (or use the quadratic formula)

We have 2 answers. As theacceleration is negative, the

particle passes through A, thenchanges direction and passes

through it again!


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acceleration

2B

= + = +

2

= + 2 = +12

= 12

A particle is moving in a straight horizontal line with constantdeceleration 4ms -2. At time t = 0 the particle passes through a point Owith speed 13ms -1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through A 2.5 and 4 secondsb) The total time the particle is beyond Ac) The time taken for the particle to return to O

13ms-1

O A = 20 = 13 = 4 =? =?

Draw a diagram


information given

We arecalculating t,

using s, u and aThe particle passes through A at 2.5

seconds and 4 seconds, so it wasbeyond A for 1.5 seconds


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acceleration

2B

= + = +

2

= + 2 = +12

= 12

A particle is moving in a straight horizontal line with constantdeceleration 4ms -2. At time t = 0 the particle passes through a point Owith speed 13ms -1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through A 2.5 and 4 secondsb) The total time the particle is beyond A 1.5 secondsc) The time taken for the particle to return to O

13ms-1

O A = 20 = 13 = 4 =? =?

Draw a diagram


information given

The particlereturns to Owhen s = 0

= 0

= +12

0 = (13) + ( 2)

0 = 13 2

2 13 = 0

(2 13) = 0

Replace s, u and a

= 0 6.5

Simplify

Rearrange

FactoriseThe particle is at O when t = 0seconds (to begin with) and is

at O again when t = 6.5 seconds


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acceleration

2B

= + = +

2

= + 2 = +12

= 12

A particle is travelling along the x-axis with constant deceleration2.5ms -2. At time t = O, the particle passes through the origin, movingin the positive direction with speed 15ms -1. Calculate the distancetravelled by the particle by the time it returns to the origin.

15ms-1

O X

Draw a diagram

The total distancetravelled will be double the

distance the particlereaches from O (point X)

At X, the velocity is 0

=? = 15 = 2.5 =? = 0

= + 2

0 = 15 + 2( 2.5)

0 = 225 5

5 = 225

= 45

Replace v,u and a

Simplify

Add 5s

Divide by 5

= 90

45m is the distance from Oto X. Double it for the total

distance travelled


using u, v and a


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You can use the formulae for constant acceleration to model an object moving vertically in astraight line under the influence of gravity

Gravity causes objects to fall to the earth! (as you probably already know!)

The acceleration caused by gravity is constant (if you ignore air resistance)

This means the acceleration will be the same, regardless of the size of the object

On Earth, the acceleration due to gravity is 9.8ms -2 , correct to 2 significant figures.

When solving problems involving vertical motion you must carefully consider the direction. Asgravity acts in a downwards direction:

- An object thrown downwards will have an acceleration of 9.8ms -2- An object thrown upwards will have an acceleration of -9.8ms -2

The time of flight is the length of time an object spends in the air. The speed of projection isanother name for the objects initial speed (u)

2C


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You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the

influence of gravity

2C

A ball is projected vertically upwards from a point O with a speedof 12ms -1. Find:a) The greatest height reached by the ballb) The total time the ball is in the air

= + = + 2

= + 2 = +12

= 12

1 2 m s -

1

0 m s -

1

=?

= 12

= 9.8

=?

= 0

Draw a diagram

At its highest point, thevelocity of the ball is 0ms -1

As the ball has been projectedupwards, gravity is acting in theopposite direction and hence the

acceleration is negative

0 = 12 + 2( 9.8)

= + 2

0 = 144 19.6

19.6 = 144

= 7.4 (2 )

Replace v, u and a

Simplify

Add 19.6s

Divide and round to 2sf (sincegravity has been given to 2sf)

We are calculating s, using u, v and a


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2C

A ball is projected vertically upwards from a point O with a speedof 12ms -1. Find:a) The greatest height reached by the ball 7.4mb) The total time the ball is in the air

= + = + 2

= + 2 = +12

= 12

1 2 m s -

1

0 m s -

1

= 12

= 9.8

=?

Draw a diagram

For the total time the ball isin the air, the displacement

(s) will be 0

= 0

=?

Also, we will not know v (yet!)when the ball strikes the

ground

We are calculating t, using s, u and a

= + 12

0 = 12 4.9

0 = (12 4.9 )

= 0 2.4 (2 )

Replace s, u and a

Factorise

Choose theappropriate answer!

So the ball will be in the air for 2.4 seconds


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2C

A book falls off the top shelf of a bookcase. The shelf is 1.4above the ground. Find:a) The time it takes the book to reach the floorb) The speed with which the book strikes the floor

= + = + 2

= + 2 = +12

= 12

0 m s -1

= 0

= 9.8

=?

= 1.4

=?

Draw a diagram

1 . 4 m

The books initial speed willbe 0 as it has not been

projected to begin withAs the books initial movement is

downwards, we take theacceleration due to gravity as

positive

We are calculating t, usings, u and a = +

1

2

1.4 = (0) +12

(9.8)

1.4 = 4.9

1.44.9

=

= 0.53 (2 )

Replace s, u and a

Simplify

Divide by 4.9

Find the positivesquare root


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2C

A book falls off the top shelf of a bookcase. The shelf is 1.4above the ground. Find:a) The time it takes the book to reach the floor 0.53 secondsb) The speed with which the book strikes the floor

= + = + 2

= + 2 = +12

= 12

0 m s -1

= 0

= 9.8

=?

= 1.4

=?

Draw a diagram

1 . 4 m

The books initial speed willbe 0 as it has not been

projected to begin withAs the books initial movement is

downwards, we take theacceleration due to gravity as

positive

We are calculating v, usings, u and a

= + 2

= 0 + 2(9.8 1.4)

= 27.44

= 5.2 (2 )

Replace s, uand a

Calculate

Find the positivesquare root


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2C

A ball is projected upwards from a point X which is 7m above theground, with initial speed 21ms -1. Find the time of flight of theball.

= + = + 2

= + 2 = +12

= 12

2 1 m s -

1

7 m

= 21

= 9.8

=?

= 7

=?

Draw a diagram

The balls flight will last until ithits the ground

We want the ball to be 7mlower than it starts (in the

negative direction)Hence, s = -7

The ball is projected upwards, sothe acceleration due to gravity is

negative

= +12

7 = (21) +12 ( 9.8)

7 = 21 4.9

4.9 21 7 = 0

= 4.9 = 21 = 7

We are calculating t,using s, u and aReplace s, uand a

Simplify

Rearrange and set equal to 0

We will need the quadratic formulahere, so write down a, b and c


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2C

A ball is projected upwards from a point X which is 7m above theground, with initial speed 21ms -1. Find the time of flight of theball.

= + = + 2

= + 2 = +12

= 12

2 1 m s -

1

7 m

= 21

= 9.8

=?

= 7

=?

Draw a diagram

The balls flight will last until ithits the ground

We want the ball to be 7mlower than it starts (in the

negative direction)Hence, s = -7


negative = 4.9 = 21 = 7

= 4

2

=( 21) ( 21) (4 4.9 7)

(2 4.9)

= 4.6 0.3

Replace a, b and c(using brackets!)

Calculate and be careful withany negatives in the previous

step!)

= 4.6 (2 )


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2C

A particle is projected vertically upwards from a point O with initialspeed u ms-1. The greatest height reached by the particle is 62.5mabove the ground. Find:a) The speed of projectionb) The total time for which the ball is 50m or more above the ground

= + = + 2

= + 2 = +12

= 12

u m s -

1

62.5m Draw a diagram

The maximum height is 62.5m

At this point the balls velocityis 0ms -1


negative

=?

= 9.8

=?

= 62.5

= 0

We are calculating u,using s, v and a

= + 2

0 = + 2( 9.8 62.5)

0 = 1225

= 1225

= 35

Replace v, a and s

Simplify

Rewrite

Find the positive square root


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2C

A particle is projected vertically upwards from a point O with initialspeed u ms-1. The greatest height reached by the particle is 62.5mabove the ground. Find:a) The speed of projection 35ms -1b) The total time for which the ball is 50m or more above the ground

= + = + 2

= + 2 = +12

= 12

u m s -

1


The ball will pass the 50m marktwice we need to find these two

times!

= 9.8

=?

= 50

=?

= 35 50m

We are calculating t,using s, u and a

= +

1

2

50 = (35) +12

( 9.8)

50 = 35 4.9

4.9 35 + 50 = 0

= 4.9 = 35 = 50

Replace s, u and a

Simplify

Rearrange, and set equal to 0

We will need the quadratic

formula, and hence a, b and c


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2C

A particle is projected vertically upwards from a point O with initialspeed u ms-1. The greatest height reached by the particle is 62.5mabove the ground. Find:a) The speed of projection 35ms -1b) The total time for which the ball is 50m or more above the ground

= + = + 2

= + 2 = +12

= 12

u m s -

1


The ball will pass the 50m marktwice we need to find these two

times!

= 9.8

=?

= 50

=?

= 35 50m

We are calculating t,using s, u and a

= 4.9 = 35 = 50

Sub these into theQuadratic formula

= 4

2

=( 35) ( 35) (4 4.9 50)

4.9 2

= 5.1686 = 1.9742

We get the two times the ballpasses the 50m mark

Calculate the difference

between these times! = 3.2 (2 )


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2C

A ball, A, falls vertically from rest from the top of a tower 63m high.At the same time as A begins to fall, another ball, B, is projectedvertically upwards from the bottom of the tower with velocity 21ms -1.The balls collide. Find the height at which this happens.

= + = + 2

= + 2 = +12

= 12

6 3 m

s1

s221ms-1

Draw a diagram

In this case we need to considereach ball separately.

We can call the two distancess1 and s 2

The time will be the same forboth when they collide, so we

can just use tMake sure that acceleration

is positive for A as it istravelling downwards and

negative for B as it istravelling upwards

= 9.8 =

=

=?

= 0

= 9.8 =

=

=?

= 21

= + 12

= (0) +12

(9.8)

= 4.9

= + 12

= (21) +12

( 9.8)

= 21 4.9

Sub in s, u,a and t for

Ball B

Simplify

Sub in s, u,a and t for

Ball A

Simplify


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2C


= + = + 2

= + 2 = +12

= 12

6 3 m

s1

s221ms-1

Draw a diagram







= 9.8 =

=

=?

= 0

= 9.8 =

=

=?

= 21

= 4.9

= 21 4.9

1)

2) Add the two equations together(this cancels the 4.9t 2 terms)

+ = 21

63 = 21

3 =

s1 + s2 must be the height of thetower (63m)

Divide by 21

So the balls collide

after 3 seconds


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2C


= + = + 2

= + 2 = +12

= 12

6 3 m

s1

s221ms-1

Draw a diagram







= 9.8 =

=

=?

= 0

= 9.8 =

=

=?

= 21

= 21 4.9 2) Sub in t = 3 (we use thisequation since s

2 is the

height above the ground) = 21(3) 4.9(3)

= 18.9 (19 2 )


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You can represent the motion of anobject on a speed-time graph,

distance-time graph or an acceleration-time graph

2D

O

u

v

t

Initial velocity

Final velocity

Time taken

v - u

t

=

=

=

On a speed-time graph,the gradient of a section

is its acceleration!

= + 2

=

= +

2

v

u

t

= + 2

On a speed-time graph,the Area beneath it isthe distance covered!


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You can represent the motion of anobject on a speed-time graph,

distance-time graph or an acceleration-time graph

Gradient of a speed-time graph =Acceleration over that period

Area under a speed-time graph = distancetravelled during that period

2D

A car accelerates uniformly at 5ms -2 from rest for 20 seconds.It then travels at a constant speed for the next 40 seconds, thendecelerates uniformly for the final 20 seconds until it is at restagain.a) Draw an acceleration-time graph for this informationb) Draw a distance-time graph for this information

20 40 60 80

5

Acceleration(ms-2)

0

-5

For now, we assume therate of acceleration jumps between different

rates Time (s)

20 40 60 80Time (s)

As the speed increases thecurve gets steeper, but witha constant speed the curve is

straight. Finally the curvegets less steep as

deceleration takes place

Distance(m)


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You can represent the motionof an object on a speed-timegraph, distance-time graph or

an acceleration-time graph


Area under a speed-time graph =distance travelled during that

period

2D

The diagram below shows a speed-time graph for the motion of a cyclistmoving along a straight road for 12 seconds. For the first 8 seconds, shemoves at a constant speed of 6ms -1. She then decelerates at a constantrate, stopping after a further 4 seconds. Find:a) The distance travelled by the cyclistb) The rate of deceleration of the cyclist

v(ms-1)

t(s)0

6

8 12

= + 2

=8 + 12

2 6

8

12

6

= 60

60

Sub in the appropriate valuesfor the trapezium above

Calculate


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period

2D

The diagram below shows a speed-time graph for the motion of a cyclistmoving along a straight road for 12 seconds. For the first 8 seconds, shemoves at a constant speed of 6ms -1. She then decelerates at a constantrate, stopping after a further 4 seconds. Find:a) The distance travelled by the cyclist 60mb) The rate of deceleration of the cyclist

v(ms-1)

t(s)0

6

8 12

=

4

-6

Sub in the appropriate valuesfor the trapezium above

Calculate =

64

= 1.5

1.5


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period

2D

A particle moves along a straight line. It accelerates uniformly from restto a speed of 8ms -1 in T seconds. The particle then travels at a constantspeed for 5T seconds. It then decelerates to rest uniformly over thenext 40 seconds.a) Sketch a speed-time graph for this motionb) Given that the particle travels 600m, find the value of Tc) Sketch an acceleration-time graph for this motion

v(ms-1)

t(s)0

8

T 5T 40

= + 2

600 =5 + 6 + 40

2 8

5T

8

6T + 40

600 = 5.5 + 20 8

75 = 5.5 + 20

55 = 5.5

10 =

Sub invalues

Simplifyfraction

Divideby 8

Subtract 20

Divide by 5.5


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period

2D

A particle moves along a straight line. It accelerates uniformly from restto a speed of 8ms -1 in T seconds. The particle then travels at a constantspeed for 5T seconds. It then decelerates to rest uniformly over thenext 40 seconds.a) Sketch a speed-time graph for this motionb) Given that the particle travels 600m, find the value of T 10 secondsc) Sketch an acceleration-time graph for this motion

v(ms-1)

t(s)0

8

T 5T 405010

=

=8

10

= 0.8

=8

40

= 0.2

First section Last section

t(s)

a(ms-2)

20 40 60 80 100

0.8

-0.2


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period

2D

A car C is moving along a straight road with constant speed 17.5ms -1. Attime t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves thelay-by. Car D accelerates from rest to a speed of 20ms -1 in 15 seconds andthen maintains this speed. Car D passes car C at a road sign.a) Sketch a speed-time graph to show the motion of both carsb) Calculate the distance between the lay-by and the road sign

v(ms-1)

t(s)0

2017.5

15

CD

At the road sign, the cars have

covered the same distance in thesame time

We need to set up simultaneousequations using s and t

Let us call the time when theareas are equal T

T

17.5

T

=

= 17.5

= 17.5

= + 2

= + 15

2 20

T - 15

20

= 7.5 20

= 20 150

Sub invalues

Sub invalues

Simplifyfraction

Multiply

bracket


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period

2D

A car C is moving along a straight road with constant speed 17.5ms -1. Attime t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves thelay-by. Car D accelerates from rest to a speed of 20ms -1 in 15 seconds andthen maintains this speed. Car D passes car C at a road sign.a) Sketch a speed-time graph to show the motion of both carsb) Calculate the distance between the lay-by and the road sign

v(ms-1)

t(s)0

2017.5

15

CD

At the road sign, the cars have

covered the same distance in thesame time

We need to set up simultaneousequations using s and t

Let us call the time when theareas are equal T

T

= 17.5

= 20 150

17.5 = 20 150

0 = 2.5 150

150 = 2.5

60 =

= 17.5

= 17.5(60)

= 1050

Subtract17.5T

Add150Divideby 2.5

Subin T

Calculate!

Set theseequations equal to

each other!


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Summary This chapter we have seen how to solve problems

involving the motion of a particle in a straight line,with constant acceleration

We have extended the problems to vertical motioninvolving gravity

We have also seen how to solve problems involvingthe motion of two particles

We have also used graphs to solve some morecomplicated problems

2) M1 Kinematics of a Particle Moving in a Straight Line

Documents

Transcript of 2) M1 Kinematics of a Particle Moving in a Straight Line