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Kinematics

Kinematics. Definitions. Displacement Distance moved in a particular direction Scalar Quantity that only has magnitude (size) eg speed, distance, temperature, pressure Vector Quantity that has magnitude (size) and direction eg velocity, displacement, acceleration, force. Definitions.

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Kinematics

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  1. Kinematics

  2. Definitions • Displacement • Distance moved in a particular direction • Scalar • Quantity that only has magnitude (size) • eg speed, distance, temperature, pressure • Vector • Quantity that has magnitude (size) and direction • eg velocity, displacement, acceleration, force

  3. Definitions Examples • The aircraft flew due north at 300 ms-1 - velocity • The aircraft flew 600km due north – displacement • The ship sailed SW for 200 miles - ? • I averaged 7mph during the marathon - ? • The snail crawled at 2 mms-1 along the bench - ? • The sales rep’s round trip was 200km - ? displacement speed velocity distance

  4. Symbols and Units These symbols may be different to what you have been used to at GCSE – beware!

  5. Vector Questions • A spider runs along two sides of a table. What is its final displacement? 1.2 m B A 0.8 m θ O

  6. Vector Questions • You walk 3km due north, then 4km due east. • What is the total distance you have travelled? • Make a scale drawing of your walk, and use it to find your final displacement. Remember to give both distance and direction. • Check your answer to (b) by calculating your displacement • A boat leaves harbour and travels due north for a distance of 3km and then due west for a distance of 8km. What is the final displacement of the boat with respect to the harbour? The boat then travels a further distance of 1km due south. What is the new displacement with respect to the harbour?

  7. Using the components of a vector A vector quantity is a quantity with both magnitude (size) and direction. Sometimes it can be helpful to find the component (or effect) of a vector in a particular direction. In this activity, you will gain further practice in calculating vector components, using trigonometry, and with scale drawing. Part 1: Working out the component of a vector in a particular direction You will look at two ways of doing this: by drawing and by calculation (using trigonometry).

  8. Using the components of a vector Finding components by drawing You are walking at 4.0 m s–1 in a direction 30° N of E. What is the component of your velocity in an easterly direction? 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 scale. You need to find the component of velocity in the easterly direction. From the end of the velocity arrow, draw a line straight line down to the x-axis, point A. Measure from the origin O to A. Convert this distance using the scale into ms–1: This method is limited by the precision with which you can draw and measure. How precisely can you measure the angle? How precisely can you draw the velocity arrow? How accurately can you measure the length OA? All of these factors affect the precision of your final answer

  9. Using the components of a vector Finding components by calculation 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 one of the other sides. They are related by the cosine of the angle AOB. Now: Cos = adjacent hypotenuse so you have: cos 30° = OA = OA OB 4.0 ms-1

  10. Using the components of a vector Re-arranging gives: OA = 4.0 m s–1 × cos 30° and calculation gives: OA = 3.46 m s–1 This is the same answer as you found by scale drawing, but without the uncertainties introduced by drawing. Take care not to be misled by the apparent accuracy of this answer. (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 two significant figures, i.e. 3.5 m s–1.

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

  12. Fh 15° Fv Fh = cos 15 x 1800 N = 1739 N Fv = cos 75 x 1800 N = 466 N A boat is pulled along a canal using a rope tied to its bow. The rope makes an angle of 15 degrees with the centre line of the canal, and the force applied to the rope is 1800N. Using maths or scale drawing, calculate the force pulling the boat along the canal, and the force pulling the boat to the side of the canal.

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

  14. Exam Questions

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

  16. Solving problems using components 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

  17. Flying in a side wind A bird flies at a steady speed of 3 ms–1 through the air. It is pointing in the direction due north. However, there is a wind blowing from west to east at a speed of 2 ms–1. • What is the velocity of the bird relative to the ground? • What is the displacement of the bird, relative to its starting point, after it has flown for 20 seconds? • In what direction should the bird point if it is to travel in a northerly direction?

  18. Exam Question 1 • A boat’s speed through still water is 2ms-1. It heads due east across a river. The river runs north to south and is 20m wide. If the river flows south at 1ms-1, how far downstream does the boat reach the other shore? • In which direction should the boat aim in order to get straight across in the shortest possible distance? 2 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 of 30° with the canal bank. What force must be provided (by the rudder and keel) to keep the boat travelling parallel to the bank?

  19. Exam Question 1 • A boat’s speed through still water is 2ms-1. It heads due east across a river. The river runs north to south and is 20m wide. If the river flows south at 1ms-1, how far downstream does the boat reach the other shore? 10m downstream (the boat travels 1m downstream for every 2m across • In which direction should the boat aim in order to get straight across in the shortest possible distance? sin θ = 1 ÷ 2 = 30° Heading = 60° east of north 2 ms-1 1 ms-1 θ

  20. Exam Question • 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 of 30° with the canal bank. What force must be provided (by the rudder and keel) to keep the boat travelling parallel to the bank? 30° 600 N 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

  21. 500N 2m 10m Φ Fh 5000kg Question 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 bank. As the barge starts to move, the tension in the rope is 500N. Calculate the barge’s initial acceleration parallel to the bank. Fh = F x Cos Φ Sin Φ = 2m / 10m Φ = 11.5° Fh = F x Cos Φ = 500N x Cos 11.5° = 489N F = ma a = F / m = 489N / 5000kg = 0.098 ms-2

  22. Homework Page 13 Questions 1 to 4

  23. Answers • 1270N [1] at angle 19.3° [1] to the horizontal [1] • 256 ms-1 [1] at 20.6° [1] • a) [1] b) 2.0 ms-1 [1] at 90° to the bank [1] c) 32.5 s [1] • 954 N [1]

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