Kinematics

KinematicsKinematics

DefinitionsDefinitions

DisplacementDisplacement Distance moved in a particular directionDistance moved in a particular direction

ScalarScalar Quantity that only has magnitude (size)Quantity that only has magnitude (size)

eg speed, distance, temperature, pressureeg speed, distance, temperature, pressure

VectorVector Quantity that has magnitude (size) and Quantity that has magnitude (size) and

directiondirection eg velocity, displacement, acceleration, forceeg velocity, displacement, acceleration, force

DefinitionsDefinitions

ExamplesExamplesThe aircraft flew due north at 300 msThe aircraft flew due north at 300 ms-1-1 - -

velocityvelocityThe aircraft flew 600km due north – The aircraft flew 600km due north –

displacementdisplacementThe ship sailed SW for 200 miles - ?The ship sailed SW for 200 miles - ?I averaged 7mph during the marathon - ?I averaged 7mph during the marathon - ?The snail crawled at 2 mmsThe snail crawled at 2 mms-1-1 along the bench along the bench

- ?- ?The sales rep’s round trip was 200km - ?The sales rep’s round trip was 200km - ?

displacementdisplacement

speedspeed

velocityvelocity

distancedistance

Symbols and UnitsSymbols and Units

QuantityQuantity Symbol for Symbol for quantityquantity

Symbol for Symbol for unitunit

displacemedisplacementnt

s (or x)s (or x) mm

timetime tt ss

velocityvelocity vv msms-1-1

These symbols may be different to what you have been used to at GCSE – beware!

Vector QuestionsVector Questions

1.1. A spider runs along two sides of a table. What is A spider runs along two sides of a table. What is its final displacement?its final displacement?

O

A B1.2 m

0.8 m

θ

Vector QuestionsVector Questions

2.2. You walk 3km due north, then 4km due east.You walk 3km due north, then 4km due east.a)a) What is the total distance you have travelled?What is the total distance you have travelled?b)b) Make a scale drawing of your walk, and use it to find Make a scale drawing of your walk, and use it to find

your final displacement. Remember to give both your final displacement. Remember to give both distance and direction.distance and direction.

c)c) Check your answer to (b) by calculating your Check your answer to (b) by calculating your displacementdisplacement

3.3. A boat leaves harbour and travels due north for A boat leaves harbour and travels due north for a distance of 3km and then due west for a a distance of 3km and then due west for a distance of 8km. What is the final displacement distance of 8km. What is the final displacement of the boat with respect to the harbour? The of the boat with respect to the harbour? The boat then travels a further distance of 1km due boat then travels a further distance of 1km due south. What is the new displacement with south. What is the new displacement with respect to the harbour?respect to the harbour?

Using the components of a vectorUsing the components of a vector

A vector quantity is a quantity with both magnitude (size) and A vector quantity is a quantity with both magnitude (size) and direction. Sometimes it can be helpful to find the direction. Sometimes it can be helpful to find the component (or effect) of a vector in a particular direction. In component (or effect) of a vector in a particular direction. In this activity, you will gain further practice in calculating this activity, you will gain further practice in calculating vector components, using trigonometry, and with scale vector components, using trigonometry, and with scale drawing.drawing.

Part 1: Working out the component of a vector in a Part 1: Working out the component of a vector in a particular directionparticular direction

You will look at two ways of doing this: by drawing and by You will look at two ways of doing this: by drawing and by calculation (using trigonometry).calculation (using trigonometry).

Using the components of a vectorUsing the components of a vectorFinding components by drawingFinding components by drawing

You are walking at 4.0 m sYou are walking at 4.0 m s–1–1 in a direction 30° N of E. What is the in a direction 30° N of E. What is the component of your velocity in an easterly direction?component of your velocity in an easterly direction?

Make a scale drawing of this on graph paper. The x-axis points Make a scale drawing of this on graph paper. The x-axis points east, the y-axis points north. You need to choose a suitable east, the y-axis points north. You need to choose a suitable scale. scale.

You need to find the component of velocity in the easterly You need to find the component of velocity in the easterly direction. From the end of the velocity arrow, draw a line direction. From the end of the velocity arrow, draw a line straight line down to the x-axis, point A.straight line down to the x-axis, point A.

Measure from the origin O to A. Convert this distance using the Measure from the origin O to A. Convert this distance using the scale into msscale into ms–1–1::

This method is limited by the precision with which you can draw This method is limited by the precision with which you can draw and measure. How precisely can you measure the angle? How and measure. How precisely can you measure the angle? How precisely can you draw the velocity arrow? How accurately can precisely can you draw the velocity arrow? How accurately can you measure the length OA? All of these factors affect the you measure the length OA? All of these factors affect the precision of your final answer precision of your final answer


Finding components by calculationFinding components by calculation

You have a right-angled triangle OAB. You know the length of the You have a right-angled triangle OAB. You know the length of the long side (the hypotenuse) and you need to find the length of long side (the hypotenuse) and you need to find the length of one of the other sides. They are related by the cosine of the one of the other sides. They are related by the cosine of the angle AOB. Now:angle AOB. Now:

Cos = adjacentCos = adjacent

hypotenusehypotenuse

so you have:so you have:

cos 30cos 30° = OA = OA° = OA = OA

OB 4.0 msOB 4.0 ms-1-1


Re-arranging gives:Re-arranging gives:

OA = 4.0 m sOA = 4.0 m s–1–1 × cos 30° × cos 30°

and calculation gives:and calculation gives:

OA = 3.46 m sOA = 3.46 m s–1–1

This is the same answer as you found by scale drawing, but This is the same answer as you found by scale drawing, but without the uncertainties introduced by drawing. without the uncertainties introduced by drawing.

Take care not to be misled by the apparent accuracy of this Take care not to be misled by the apparent accuracy of this answer. answer.

(The calculator gave 3.464 1016... m s–1.) If your speed is given (The calculator gave 3.464 1016... m s–1.) If your speed is given as 4.0 m s–1, you can only give the value of the component to as 4.0 m s–1, you can only give the value of the component to two significant figures, i.e. 3.5 m s–1. two significant figures, i.e. 3.5 m s–1.


You are running at a velocity of 8 msYou are running at a velocity of 8 ms–1–1 in a north-easterly in a north-easterly direction (i.e. at 45° to both N and E). Find the components of direction (i.e. at 45° to both N and E). Find the components of your velocity in an easterly direction, and in a northerly your velocity in an easterly direction, and in a northerly direction, firstly by drawing, and then by calculation using direction, firstly by drawing, and then by calculation using trigonometry.trigonometry.

Explain why these two components have the same magnitude.Explain why these two components have the same magnitude.

As the velocity vector makes an angle of 45° with north, the component north is

8 ms-1 x cos 45° = 5.7 ms-1

The component east can be calculated in the same way or by using the 45° angle between the vector and north to calculate the component east from the sine function:

8 ms-1 x sin 45° = 5.7 ms-1

A boat is pulled along a canal using a rope tied to A boat is pulled along a canal using a rope tied to its bow. The rope makes an angle of 15 degrees its bow. The rope makes an angle of 15 degrees with the centre line of the canal, and the force with the centre line of the canal, and the force applied to the rope is 1800N. Using maths or applied to the rope is 1800N. Using maths or scale drawing, calculate the force pulling the scale drawing, calculate the force pulling the boat along the canal, and the force pulling the boat along the canal, and the force pulling the boat to the side of the canal.boat to the side of the canal.

15°

Fh

Fv

Fh = cos 15 x 1800 N = 1739 N

Fv = cos 75 x 1800 N = 466 N

Solving problems using componentsSolving problems using components

You set off to run across an empty supermarket parking strip, You set off to run across an empty supermarket parking strip, 100 m wide. You set off at 55° to the verge, heading 100 m wide. You set off at 55° to the verge, heading towards the entrance to the supermarket. Your speed is 8 towards the entrance to the supermarket. Your speed is 8 m sm s–1–1. How long will it take you to reach the far entrance?. How long will it take you to reach the far entrance?

8 m s–1

55º

100 m

Exam QuestionsExam Questions


A train is gradually travelling up a long gradient. The speed of A train is gradually travelling up a long gradient. The speed of the train is 20 m sthe train is 20 m s–1–1 and the slope makes an angle of 2° and the slope makes an angle of 2° with the horizontal. The summit is 200 m above the starting with the horizontal. The summit is 200 m above the starting level. How long will it take to reach the summit?level. How long will it take to reach the summit?

200 m2º


200 m2º

To find the time taken to travel to the top of the hill you need first to find the distance along the slope. From the diagram in the question, it is clear that sin2° = 200 / slope, so that the length of the slope is:

slope = 200 / sin 2° = 5731m

Then the time taken is:

Time taken = distance / velocity = 5731m / 20 ms-1 = 4 mins 47 secs

Flying in a side windFlying in a side wind

A bird flies at a steady speed of 3 msA bird flies at a steady speed of 3 ms–1–1 through the air. It is through the air. It is pointing in the direction due north. However, there is a pointing in the direction due north. However, there is a wind blowing from west to east at a speed of 2 mswind blowing from west to east at a speed of 2 ms–1–1..

1.1. What is the velocity of the bird relative to the ground?What is the velocity of the bird relative to the ground?

2.2. What is the displacement of the bird, relative to its What is the displacement of the bird, relative to its starting point, after it has flown for 20 seconds?starting point, after it has flown for 20 seconds?

3.3. In what direction should the bird point if it is to travel in a In what direction should the bird point if it is to travel in a northerly direction?northerly direction?

Exam QuestionExam Question

11a)a) A boat’s speed through still water is 2msA boat’s speed through still water is 2ms-1-1. It heads due . It heads due

east across a river. The river runs north to south and is east across a river. The river runs north to south and is 20m wide. If the river flows south at 1ms20m wide. If the river flows south at 1ms-1-1, how far , how far downstream does the boat reach the other shore?downstream does the boat reach the other shore?

b)b) In which direction should the boat aim in order to get In which direction should the boat aim in order to get straight across in the shortest possible distance?straight across in the shortest possible distance?

22A rope is used to pull a narrow boat along a canal. The rope A rope is used to pull a narrow boat along a canal. The rope

is pulled by a horse. The tension in the rope is 600N is pulled by a horse. The tension in the rope is 600N and the rope makes an angle of 30and the rope makes an angle of 30° with the canal ° with the canal bank. What force must be provided (by the rudder and bank. What force must be provided (by the rudder and keel) to keep the boat travelling parallel to the bank?keel) to keep the boat travelling parallel to the bank?

Exam QuestionExam Question11

a)a) A boat’s speed through still water is 2msA boat’s speed through still water is 2ms-1-1. It heads due east across . It heads due east across a river. The river runs north to south and is 20m wide. If the river a river. The river runs north to south and is 20m wide. If the river flows south at 1msflows south at 1ms-1-1, how far downstream does the boat reach the , how far downstream does the boat reach the other shore?other shore?

10m downstream (the boat travels 1m downstream for 10m downstream (the boat travels 1m downstream for every 2m acrossevery 2m across

a)a) In which direction should the boat aim in order to get straight In which direction should the boat aim in order to get straight across in the shortest possible distance?across in the shortest possible distance?

1 ms1 ms-1-12 ms2 ms-1-1

θ

sin θ = 1 ÷ 2 = 30°Heading = 60° east of north

Exam QuestionExam Questionb)b) A rope is used to pull a narrow boat along a canal. The rope is pulled by A rope is used to pull a narrow boat along a canal. The rope is pulled by

a horse. The tension in the rope is 600N and the rope makes an angle a horse. The tension in the rope is 600N and the rope makes an angle of 30of 30° with the canal bank. What force must be provided (by the rudder ° with the canal bank. What force must be provided (by the rudder and keel) to keep the boat travelling parallel to the bank?and keel) to keep the boat travelling parallel to the bank?

600 N

30°

Force pulling the boat towards the bank is opposed by the rudder. So, resolve to find the vertical component= cos 60° x 600N= 300 N

QuestionQuestion

A horse pulls a barge of mass 5000 kg along a canal using a rope 10m A horse pulls a barge of mass 5000 kg along a canal using a rope 10m long. The rope is attached to a point on the barge 2m from the long. The rope is attached to a point on the barge 2m from the bank. As the barge starts to move, the tension in the rope is 500N. bank. As the barge starts to move, the tension in the rope is 500N. Calculate the barge’s initial acceleration parallel to the bank.Calculate the barge’s initial acceleration parallel to the bank.

Fh = F x Cos Φ

Sin Φ = 2m / 10mΦ = 11.5°

Fh = F x Cos Φ= 500N x Cos 11.5°= 489N

F = maa = F / m

= 489N / 5000kg= 0.098 ms-2

5000kg

500N

10m2m

Fh

Φ

HomeworkHomework

Page 13Page 13

Questions 1 to 4Questions 1 to 4

AnswersAnswers

1.1. 1270N [1] at angle 19.31270N [1] at angle 19.3° [1] to the horizontal [1]° [1] to the horizontal [1]

2.2. 256 ms256 ms-1-1 [1] at 20.6° [1] [1] at 20.6° [1]

3.3. a) [1]a) [1]

b) 2.0 msb) 2.0 ms-1-1 [1] at 90° to the bank [1] [1] at 90° to the bank [1]

c) 32.5 s [1]c) 32.5 s [1]

4.4. 954 N [1]954 N [1]

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Kinematics