2. Introduction This chapter you will learn the SUVAT equations These are the foundations of many of the Mechanics topics You will see how to use them to use many types of problem involving motion

4. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A Replace with the appropriate letters.  Change in velocity = final velocity – initial velocity Multiply by t Add u This is the usual form! Replace with the appropriate letters

5. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A You need to consider using negative numbers in some cases PQ Positive direction O 4m3m 2.5ms -1 6ms -1 If we are measuring displacements from O, and left to right is the positive direction… For particle P:For particle Q: The particle is to the left of the point O, which is the negative direction The particle is moving at 2.5ms -1 in the positive direction The particle is to the right of the point O, which is the positive direction The particle is moving at 6ms -1 in the negative direction

6. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A particle is moving in a straight line from A to B with constant acceleration 3ms -2. Its speed at A is 2ms -1 and it takes 8 seconds to move from A to B. Find: a)The speed of the particle at B b)The distance from A to B AB 2ms -1 Start with a diagram Write out ‘suvat’ and fill in what you know For part a) we need to calculate v, and we know u, a and t… Fill in the values you know Remember to include units! You always need to set up the question in this way. It makes it much easier to figure out what equation you need to use (there will be more to learn than just these two!)

7. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A particle is moving in a straight line from A to B with constant acceleration 3ms -2. Its speed at A is 2ms -1 and it takes 8 seconds to move from A to B. Find: a)The speed of the particle at B – 26ms -1 b)The distance from A to B AB 2ms -1 For part b) we need to calculate s, and we know u, v and t… Fill in the values you know Show calculations Remember the units!

8. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A cyclist is travelling along a straight road. She accelerates at a constant rate from a speed of 4ms -1 to a speed of 7.5ms -1 in 40 seconds. Find: a)The distance travelled over this 40 seconds b)The acceleration over the 40 seconds 4ms -1 7.5ms -1 Draw a diagram (model the cyclist as a particle) Write out ‘suvat’ and fill in what you know We are calculating s, and we already know u, v and t… Sub in the values you know Remember units!

9. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A cyclist is travelling along a straight road. She accelerates at a constant rate from a speed of 4ms -1 to a speed of 7.5ms -1 in 40 seconds. Find: a)The distance travelled over this 40 seconds – 230m b)The acceleration over the 40 seconds 4ms -1 7.5ms -1 Draw a diagram (model the cyclist as a particle) Write out ‘suvat’ and fill in what you know For part b, we are calculating a, and we already know u, v and t… Sub in the values you know Subtract 4 Divide by 40

10. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A particle moves in a straight line from a point A to B with constant deceleration of 1.5ms -2. The speed of the particle at A is 8ms -1 and the speed of the particle at B is 2ms -1. Find: a)The time taken for the particle to get from A to B b)The distance from A to B 8ms -1 2ms -1 Draw a diagram Write out ‘suvat’ and fill in what you know As the particle is decelerating, ‘a’ is negative Sub in the values you know Subtract 8 Divide by -1.5 AB

11. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A particle moves in a straight line from a point A to B with constant deceleration of 1.5ms -2. The speed of the particle at A is 8ms -1 and the speed of the particle at B is 2ms -1. Find: a)The time taken for the particle to get from A to B – 4 seconds b)The distance from A to B 8ms -1 2ms -1 Draw a diagram Write out ‘suvat’ and fill in what you know As the particle is decelerating, ‘a’ is negative Sub in the values you know Calculate the answer! AB

12. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A After reaching B the particle continues to move along the straight line with the same deceleration. The particle is at point C, 6 seconds after passing through A. Find: a)The velocity of the particle at C b)The distance from A to C 8ms -1 2ms -1 ABC ? Update the diagram Write out ‘suvat’ using points A and C Sub in the values Work it out! As the velocity is negative, this means the particle has now changed direction and is heading back towards A! (velocity has a direction as well as a magnitude!) The velocity is 1ms -1 in the direction C to A…

13. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A After reaching B the particle continues to move along the straight line with the same deceleration. The particle is at point C, 6 seconds after passing through A. Find: a)The velocity of the particle at C - -1ms -1 b)The distance from A to C 8ms -1 2ms -1 ABC ? Update the diagram Write out ‘suvat’ using points A and C Sub in the values Work it out! It is important to note that 21m is the distance from A to C only…  The particle was further away before it changed direction, and has in total travelled further than 21m…

14. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A car moves from traffic lights along a straight road with constant acceleration. The car starts from rest at the traffic lights and 30 seconds later passes a speed trap where it is travelling at 45 kmh -1. Find: a)The acceleration of the car b)The distance between the traffic lights and the speed-trap. 0ms -1 45kmh -1 LightsTrap Standard units to use are metres and seconds, or kilometres and hours  In this case, the time is in seconds and the speed is in kilometres per hour  We need to change the speed into metres per second first! Draw a diagram Multiply by 1000 (km to m) Divide by 3600 (hours to seconds)

15. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A car moves from traffic lights along a straight road with constant acceleration. The car starts from rest at the traffic lights and 30 seconds later passes a speed trap where it is travelling at 45 kmh -1. Find: a)The acceleration of the car b)The distance between the traffic lights and the speed-trap. 0ms -1 45kmh -1 LightsTrap Draw a diagram = 12.5ms -1 Write out ‘suvat’ and fill in what you know Sub in the values Divide by 30 You can use exact answers!

16. Kinematics of a Particle moving in a Straight Line You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2A A car moves from traffic lights along a straight road with constant acceleration. The car starts from rest at the traffic lights and 30 seconds later passes a speed trap where it is travelling at 45 kmh -1. Find: a)The acceleration of the car b)The distance between the traffic lights and the speed-trap. 0ms -1 45kmh -1 LightsTrap Draw a diagram = 12.5ms -1 Write out ‘suvat’ and fill in what you know Sub in values Work it out!

18. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B Subtract u Divide by a Replace t with the expression above Multiply numerators and denominators Multiply by 2a Add u 2 This is the way it is usually written!

19. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B Replace ‘v’ with ‘u + at’ Group terms on the numerator Divide the numerator by 2 Multiply out the bracket

20. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B Subtract ‘at’ Replace ‘u’ with ‘v - at’ from above’ Multiply out the bracket Group up the at 2 terms

21. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A particle is moving in a straight line from A to B with constant acceleration 5ms -2. The velocity of the particle at A is 3ms -1 in the direction AB. The velocity at B is 18ms -1 in the same direction. Find the distance from A to B. 3ms -1 18ms -1 AB Draw a diagram Write out ‘suvat’ with the information given Replace v, u and a Work out terms Subtract 9 Divide by 10 We are calculating s, using v, u and a

22. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A car is travelling along a straight horizontal road with a constant acceleration of 0.75ms -2. The car is travelling at 8ms -1 as it passes a pillar box. 12 seconds later the car passes a lamp post. Find: a)The distance between the pillar box and the lamp post b)The speed with which the car passes the lamp post 8ms -1 Pillar Box Lamp Post Draw a diagram Write out ‘suvat’ with the information given We are calculating s, using u, a and t Replace u, a and t Calculate

23. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A car is travelling along a straight horizontal road with a constant acceleration of 0.75ms -2. The car is travelling at 8ms -1 as it passes a pillar box. 12 seconds later the car passes a lamp post. Find: a)The distance between the pillar box and the lamp post – 150m b)The speed with which the car passes the lamp post 8ms -1 Pillar Box Lamp Post Draw a diagram Write out ‘suvat’ with the information given We are calculating v, using u, a and t Replace u, a and t Calculate Often you can use an answer you have calculated later on in the same question. However, you must take care to use exact values and not rounded answers!

24. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A particle is moving in a straight horizontal line with constant deceleration 4ms -2. At time t = 0 the particle passes through a point O with speed 13ms -1, travelling to a point A where OA = 20m. Find: a)The times when the particle passes through A b)The total time the particle is beyond A c)The time taken for the particle to return to O 13ms -1 OA Draw a diagram Write out ‘suvat’ with the information given We are calculating t, using s, u and a Replace s, u and a Simplify terms Rearrange and set equal to 0 Factorise (or use the quadratic formula…) We have 2 answers. As the acceleration is negative, the particle passes through A, then changes direction and passes through it again!

25. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A particle is moving in a straight horizontal line with constant deceleration 4ms -2. At time t = 0 the particle passes through a point O with 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 seconds b)The total time the particle is beyond A c)The time taken for the particle to return to O 13ms -1 OA Draw a diagram Write out ‘suvat’ with the information given We are calculating t, using s, u and a The particle passes through A at 2.5 seconds and 4 seconds, so it was beyond A for 1.5 seconds…

26. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A particle is moving in a straight horizontal line with constant deceleration 4ms -2. At time t = 0 the particle passes through a point O with 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 seconds b)The total time the particle is beyond A – 1.5 seconds c)The time taken for the particle to return to O 13ms -1 OA Draw a diagram Write out ‘suvat’ with the information given The particle returns to O when s = 0 Replace s, u and a Simplify Rearrange Factorise The particle is at O when t = 0 seconds (to begin with) and is at O again when t = 6.5 seconds

27. Kinematics of a Particle moving in a Straight Line You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 2B A particle is travelling along the x-axis with constant deceleration 2.5ms -2. At time t = O, the particle passes through the origin, moving in the positive direction with speed 15ms -1. Calculate the distance travelled by the particle by the time it returns to the origin. 15ms -1 OX Draw a diagram The total distance travelled will be double the distance the particle reaches from O (point X)  At X, the velocity is 0 Replace v, u and a Simplify Add 5s Divide by 5 45m is the distance from O to X. Double it for the total distance travelled We are calculating s, using u, v and a

29. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight 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. As gravity 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 is another name for the object’s initial speed (u) 2C

30. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A ball is projected vertically upwards from a point O with a speed of 12ms -1. Find: a)The greatest height reached by the ball b)The total time the ball is in the air 12ms -1 0ms -1 Draw a diagram At its highest point, the velocity of the ball is 0ms -1 As the ball has been projected upwards, gravity is acting in the opposite direction and hence the acceleration is negative Replace v, u and a Simplify Add 19.6s Divide and round to 2sf (since gravity has been given to 2sf) We are calculating s, using u, v and a

31. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A ball is projected vertically upwards from a point O with a speed of 12ms -1. Find: a)The greatest height reached by the ball – 7.4m b)The total time the ball is in the air 12ms -1 0ms -1 Draw a diagram For the total time the ball is in the air, the displacement (s) will be 0 Also, we will not know v (yet!) when the ball strikes the ground We are calculating t, using s, u and a Replace s, u and a Factorise Choose the appropriate answer! So the ball will be in the air for 2.4 seconds

32. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A book falls off the top shelf of a bookcase. The shelf is 1.4m above the ground. Find: a)The time it takes the book to reach the floor b)The speed with which the book strikes the floor 0ms -1 Draw a diagram 1.4m The book’s initial speed will be 0 as it has not been projected to begin with As the book’s initial movement is downwards, we take the acceleration due to gravity as positive We are calculating t, using s, u and a… Replace s, u and a Simplify Divide by 4.9 Find the positive square root

33. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A book falls off the top shelf of a bookcase. The shelf is 1.4m above the ground. Find: a)The time it takes the book to reach the floor – 0.53 seconds b)The speed with which the book strikes the floor 0ms -1 Draw a diagram 1.4m The book’s initial speed will be 0 as it has not been projected to begin with As the book’s initial movement is downwards, we take the acceleration due to gravity as positive We are calculating v, using s, u and a… Replace s, u and a Calculate Find the positive square root

34. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A ball is projected upwards from a point X which is 7m above the ground, with initial speed 21ms -1. Find the time of flight of the ball. 21ms -1 7m Draw a diagram The ball’s flight will last until it hits the ground  We want the ball to be 7m lower than it starts (in the negative direction)  Hence, s = -7 The ball is projected upwards, so the acceleration due to gravity is negative We are calculating t, using s, u and a Replace s, u and a Simplify Rearrange and set equal to 0 We will need the quadratic formula here, so write down a, b and c…

35. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A ball is projected upwards from a point X which is 7m above the ground, with initial speed 21ms -1. Find the time of flight of the ball. 21ms -1 7m Draw a diagram The ball’s flight will last until it hits the ground  We want the ball to be 7m lower than it starts (in the negative direction)  Hence, s = -7 The ball is projected upwards, so the acceleration due to gravity is negative Replace a, b and c (using brackets!) Calculate and be careful with any negatives in the previous step!)

36. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A particle is projected vertically upwards from a point O with initial speed u ms -1. The greatest height reached by the particle is 62.5m above the ground. Find: a)The speed of projection b)The total time for which the ball is 50m or more above the ground u ms -1 62.5m Draw a diagram The maximum height is 62.5m  At this point the ball’s velocity is 0ms -1 The ball is projected upwards, so the acceleration due to gravity is negative We are calculating u, using s, v and a Replace v, a and s Simplify Rewrite Find the positive square root

37. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A particle is projected vertically upwards from a point O with initial speed u ms -1. The greatest height reached by the particle is 62.5m above the ground. Find: a)The speed of projection – 35ms -1 b)The total time for which the ball is 50m or more above the ground u ms -1 62.5m Draw a diagram The ball will pass the 50m mark twice – we need to find these two times! 50m We are calculating t, using s, u and a Replace s, u and a Simplify Rearrange, and set equal to 0 We will need the quadratic formula, and hence a, b and c

38. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 2C A particle is projected vertically upwards from a point O with initial speed u ms -1. The greatest height reached by the particle is 62.5m above the ground. Find: a)The speed of projection – 35ms -1 b)The total time for which the ball is 50m or more above the ground u ms -1 62.5m Draw a diagram The ball will pass the 50m mark twice – we need to find these two times! 50m We are calculating t, using s, u and a Sub these into the Quadratic formula We get the two times the ball passes the 50m mark Calculate the difference between these times!

39. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 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 projected vertically upwards from the bottom of the tower with velocity 21ms -1. The balls collide. Find the height at which this happens. 63m s1s1 s2s2 21ms -1 Draw a diagram In this case we need to consider each ball separately.  We can call the two distances s 1 and s 2  The time will be the same for both when they collide, so we can just use t  Make sure that acceleration is positive for A as it is travelling downwards and negative for B as it is travelling upwards Sub in s, u, a and t for Ball B Simplify Sub in s, u, a and t for Ball A Simplify

40. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 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 projected vertically upwards from the bottom of the tower with velocity 21ms -1. The balls collide. Find the height at which this happens. 63m s1s1 s2s2 21ms -1 Draw a diagram In this case we need to consider each ball separately.  We can call the two distances s 1 and s 2  The time will be the same for both when they collide, so we can just use t  Make sure that acceleration is positive for A as it is travelling downwards and negative for B as it is travelling upwards 1) 2) Add the two equations together (this cancels the 4.9t 2 terms) s 1 + s 2 must be the height of the tower (63m) Divide by 21 So the balls collide after 3 seconds…

41. Kinematics of a Particle moving in a Straight Line You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 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 projected vertically upwards from the bottom of the tower with velocity 21ms -1. The balls collide. Find the height at which this happens. 63m s1s1 s2s2 21ms -1 Draw a diagram In this case we need to consider each ball separately.  We can call the two distances s 1 and s 2  The time will be the same for both when they collide, so we can just use t  Make sure that acceleration is positive for A as it is travelling downwards and negative for B as it is travelling upwards 2) Sub in t = 3 (we use this equation since s 2 is the height above the ground)

43. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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! v u t On a speed-time graph, the Area beneath it is the distance covered!

44. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled 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, then decelerates uniformly for the final 20 seconds until it is at rest again. a)Draw an acceleration-time graph for this information b)Draw a distance-time graph for this information 20406080 5 Acceleration (ms -2 ) 0 -5 For now, we assume the rate of acceleration jumps between different rates… Time (s) 20406080 Time (s) As the speed increases the curve gets steeper, but with a constant speed the curve is straight. Finally the curve gets less steep as deceleration takes place Distance (m)

45. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled during that period 2D The diagram below shows a speed-time graph for the motion of a cyclist moving along a straight road for 12 seconds. For the first 8 seconds, she moves at a constant speed of 6ms -1. She then decelerates at a constant rate, stopping after a further 4 seconds. Find: a)The distance travelled by the cyclist b)The rate of deceleration of the cyclist v(ms -1 ) t(s) 0 6 812 8 6 Sub in the appropriate values for the trapezium above Calculate

46. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled during that period 2D The diagram below shows a speed-time graph for the motion of a cyclist moving along a straight road for 12 seconds. For the first 8 seconds, she moves at a constant speed of 6ms -1. She then decelerates at a constant rate, stopping after a further 4 seconds. Find: a)The distance travelled by the cyclist – 60m b)The rate of deceleration of the cyclist v(ms -1 ) t(s) 0 6 812 4 -6 Sub in the appropriate values for the trapezium above Calculate

47. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled during that period 2D A particle moves along a straight line. It accelerates uniformly from rest to a speed of 8ms -1 in T seconds. The particle then travels at a constant speed for 5T seconds. It then decelerates to rest uniformly over the next 40 seconds. a)Sketch a speed-time graph for this motion b)Given that the particle travels 600m, find the value of T c)Sketch an acceleration-time graph for this motion v(ms -1 ) t(s) 0 8 T5T40 5T 8 6T + 40 Sub in values Simplify fraction Divide by 8 Subtract 20 Divide by 5.5

48. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled during that period 2D A particle moves along a straight line. It accelerates uniformly from rest to a speed of 8ms -1 in T seconds. The particle then travels at a constant speed for 5T seconds. It then decelerates to rest uniformly over the next 40 seconds. a)Sketch a speed-time graph for this motion b)Given that the particle travels 600m, find the value of T – 10 seconds c)Sketch an acceleration-time graph for this motion v(ms -1 ) t(s) 0 8 T5T405010 First sectionLast section t(s) a(ms -2 ) 20406080100 0.8 -0.2

49. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled during that period 2D A car C is moving along a straight road with constant speed 17.5ms -1. At time t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves the lay-by. Car D accelerates from rest to a speed of 20ms -1 in 15 seconds and then maintains this speed. Car D passes car C at a road sign. a)Sketch a speed-time graph to show the motion of both cars b)Calculate the distance between the lay-by and the road sign v(ms -1 ) t(s) 0 20 17.5 15 C D At the road sign, the cars have covered the same distance in the same time  We need to set up simultaneous equations using s and t…  Let us call the time when the areas are equal ‘T’ T 17.5 T T - 15 20 Sub in values Simplify fraction Multiply bracket

50. Kinematics of a Particle moving in a Straight Line You can represent the motion of an object 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 = distance travelled during that period 2D A car C is moving along a straight road with constant speed 17.5ms -1. At time t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves the lay-by. Car D accelerates from rest to a speed of 20ms -1 in 15 seconds and then maintains this speed. Car D passes car C at a road sign. a)Sketch a speed-time graph to show the motion of both cars b)Calculate the distance between the lay-by and the road sign v(ms -1 ) t(s) 0 20 17.5 15 C D At the road sign, the cars have covered the same distance in the same time  We need to set up simultaneous equations using s and t…  Let us call the time when the areas are equal ‘T’ T Subtract 17.5T Add 150 Divide by 2.5 Sub in T Calculate! Set these equations equal to each other!

51. 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 motion involving gravity We have also seen how to solve problems involving the motion of two particles We have also used graphs to solve some more complicated problems