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Kinematics of a particle moving in a straight line

Mechanics 1 6677 (m1). Kinematics of a particle moving in a straight line. 21ms -1. Jan 27, 14. 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.

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Kinematics of a particle moving in a straight line

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  1. Mechanics 1 6677 (m1) Kinematics of a particle moving in a straight line 21ms-1 Jan 27, 14

  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

  3. TeaChings for Exercise 2A

  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 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 2A

  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 You need to consider using negative numbers in some cases Positive direction 2.5ms-1 6ms-1 P Q 4m O 3m 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 right of the point O, which is the positive direction 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 moving at 6ms-1 in the negative direction 2A

  6. Kinematics of a Particle moving in a Straight Line 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: The speed of the particle at B The distance from A to B You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2ms-1 Start with a diagram A B 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!) 2A

  7. Kinematics of a Particle moving in a Straight Line 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: The speed of the particle at B – 26ms-1 The distance from A to B You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 2ms-1 A B 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! 2A

  8. Kinematics of a Particle moving in a Straight Line 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: The distance travelled over this 40 seconds The acceleration over the 40 seconds You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 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! 2A

  9. Kinematics of a Particle moving in a Straight Line 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: The distance travelled over this 40 seconds – 230m The acceleration over the 40 seconds You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 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 2A

  10. Kinematics of a Particle moving in a Straight Line 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: The time taken for the particle to get from A to B The distance from A to B You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 8ms-1 2ms-1 Draw a diagram A B 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 2A

  11. Kinematics of a Particle moving in a Straight Line 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: The time taken for the particle to get from A to B – 4 seconds The distance from A to B You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 8ms-1 2ms-1 Draw a diagram A B 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! 2A

  12. Kinematics of a Particle moving in a Straight Line 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: The velocity of the particle at C The distance from A to C You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 8ms-1 2ms-1 ? Update the diagram A B C 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… 2A

  13. Kinematics of a Particle moving in a Straight Line 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: The velocity of the particle at C - -1ms-1 The distance from A to C You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 8ms-1 2ms-1 ? Update the diagram A B C 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… 2A

  14. Kinematics of a Particle moving in a Straight Line 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: The acceleration of the car The distance between the traffic lights and the speed-trap. You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 0ms-1 45kmh-1 Draw a diagram Lights Trap • 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! Multiply by 1000 (km to m) Divide by 3600 (hours to seconds) 2A

  15. Kinematics of a Particle moving in a Straight Line 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: The acceleration of the car The distance between the traffic lights and the speed-trap. You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 0ms-1 45kmh-1 = 12.5ms-1 Draw a diagram Lights Trap Write out ‘suvat’ and fill in what you know Sub in the values Divide by 30 You can use exact answers! 2A

  16. Kinematics of a Particle moving in a Straight Line 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: The acceleration of the car The distance between the traffic lights and the speed-trap. You will begin by learning two of the SUVAT equations s = Displacement (distance) u = Starting (initial) velocity v = Final velocity a = Acceleration t = Time 0ms-1 45kmh-1 = 12.5ms-1 Draw a diagram Lights Trap Write out ‘suvat’ and fill in what you know Sub in values Work it out! 2A

  17. TeaChings for Exercise 2B

  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 Subtract u Divide by a Replace t with the expression above Multiply numerators and denominators Multiply by 2a Add u2 This is the way it is usually written! 2B

  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 Replace ‘v’ with ‘u + at’ Group terms on the numerator Divide the numerator by 2 Multiply out the bracket 2B

  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 Subtract ‘at’ Replace ‘u’ with ‘v - at’ from above’ Multiply out the bracket Group up the at2 terms 2B

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

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

  23. Kinematics of a Particle moving in a Straight Line 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: The distance between the pillar box and the lamp post – 150m The speed with which the car passes the lamp post You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 8ms-1 Draw a diagram Write out ‘suvat’ with the information given Pillar Box Lamp Post 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! 2B

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

  25. Kinematics of a Particle moving in a Straight Line 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: The times when the particle passes through A – 2.5 and 4 seconds The total time the particle is beyond A The time taken for the particle to return to O You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration Draw a diagram 13ms-1 Write out ‘suvat’ with the information given O A 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… 2B

  26. Kinematics of a Particle moving in a Straight Line 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: The times when the particle passes through A – 2.5 and 4 seconds The total time the particle is beyond A – 1.5 seconds The time taken for the particle to return to O You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration Draw a diagram 13ms-1 Write out ‘suvat’ with the information given O A The particle returns to O when s = 0 Replace s, u and a Simplify Rearrange The particle is at O when t = 0 seconds (to begin with) and is at O again when t = 6.5 seconds Factorise 2B

  27. Kinematics of a Particle moving in a Straight Line 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. You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 15ms-1 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 O X Replace v, u and a We are calculating s, using u, v and a Simplify Add 5s Divide by 5 45m is the distance from O to X. Double it for the total distance travelled 2B

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