Kinematics of a particle moving in a straight line

Kinematics of a particle moving in a straight line

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

TeaChings for Exercise 2A

Kinematics of a Particle moving in a Straight Line
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 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 𝑎= 𝑣−𝑢 𝑡 Multiply by t 𝑎𝑡=𝑣−𝑢 Add u 𝑎𝑡+𝑢=𝑣 This is the usual form! 𝑣=𝑢+𝑎𝑡 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑=𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 ×𝑡𝑖𝑚𝑒 Replace with the appropriate letters 𝑠= 𝑢+𝑣 2 𝑡 2A

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 𝑣=𝑢+𝑎𝑡 If we are measuring displacements from O, and left to right is the positive direction… 𝑠= 𝑢+𝑣 2 𝑡 For particle P: For particle Q: The particle is to the right of the point O, which is the positive direction 𝑠=−4𝑚 𝑠=3𝑚 The particle is to the left of the point O, which is the negative direction 𝑣=2.5𝑚 𝑠 −1 𝑣=−6𝑚 𝑠 −1 The particle is moving at 2.5ms-1 in the positive direction The particle is moving at 6ms-1 in the negative direction 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: 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 𝑠=? 𝑢=2 𝑣=? 𝑎=3 𝑡=8 For part a) we need to calculate v, and we know u, a and t… 𝑣=𝑢+𝑎𝑡 Fill in the values you know 𝑣=𝑢+𝑎𝑡 𝑣=2+(3×8) Remember to include units! 𝑠= 𝑢+𝑣 2 𝑡 𝑣=26𝑚 𝑠 −1 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

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 𝑠=? 𝑢=2 𝑣=? 𝑣=26 𝑎=3 𝑡=8 For part b) we need to calculate s, and we know u, v and t… 𝑠= 𝑢+𝑣 2 𝑡 Fill in the values you know 𝑣=𝑢+𝑎𝑡 𝑠= ×8 𝑠= 𝑢+𝑣 2 𝑡 Show calculations 𝑠= 14 ×8 Remember the units! 𝑠=112𝑚 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: 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 𝑠=? 𝑢=4 𝑣=7.5 𝑎=? 𝑡=40 𝑠= 𝑢+𝑣 2 𝑡 We are calculating s, and we already know u, v and t… Sub in the values you know 𝑣=𝑢+𝑎𝑡 𝑠= ×40 𝑠= 𝑢+𝑣 2 𝑡 Remember units! 𝑠=230𝑚 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: 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 𝑠=230 𝑠=? 𝑢=4 𝑣=7.5 𝑎=? 𝑡=40 For part b, we are calculating a, and we already know u, v and t… 𝑣=𝑢+𝑎𝑡 Sub in the values you know 𝑣=𝑢+𝑎𝑡 7.5=4+40𝑎 𝑠= 𝑢+𝑣 2 𝑡 Subtract 4 7.5=40𝑎 Divide by 40 𝑎=0.0875𝑚 𝑠 −2 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: 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 𝑠=? 𝑢=8 𝑣=2 𝑎=−1.5 𝑡=? As the particle is decelerating, ‘a’ is negative 𝑣=𝑢+𝑎𝑡 Sub in the values you know 𝑣=𝑢+𝑎𝑡 2=8−1.5𝑡 Subtract 8 −6=−1.5𝑡 𝑠= 𝑢+𝑣 2 𝑡 Divide by -1.5 4=𝑡 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: 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 𝑠=? 𝑢=8 𝑣=2 𝑎=−1.5 𝑡=4 𝑡=? As the particle is decelerating, ‘a’ is negative 𝑠= 𝑢+𝑣 2 𝑡 Sub in the values you know 𝑣=𝑢+𝑎𝑡 𝑠= ×4 𝑠= 𝑢+𝑣 2 𝑡 Calculate the answer! 𝑠=20𝑚 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: 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 𝑠=? 𝑢=8 𝑣=? 𝑎=−1.5 𝑡=6 Write out ‘suvat’ using points A and C 𝑣=𝑢+𝑎𝑡 Sub in the values 𝑣=𝑢+𝑎𝑡 𝑣=8−(1.5×6) Work it out! 𝑠= 𝑢+𝑣 2 𝑡 𝑣=−1𝑚 𝑠 −1 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

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 𝑠=? 𝑢=8 𝑣=−1 𝑣=? 𝑎=−1.5 𝑡=6 Write out ‘suvat’ using points A and C 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 Sub in the values 𝑠= 8−1 2 ×6 𝑠= 𝑢+𝑣 2 𝑡 Work it out! 𝑠=21𝑚 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

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! 𝑣=𝑢+𝑎𝑡 45𝑘𝑚 ℎ −1 𝑠= 𝑢+𝑣 2 𝑡 Multiply by 1000 (km to m) 45,000𝑚 ℎ −1 Divide by 3600 (hours to seconds) 12.5𝑚 𝑠 −1 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: 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 𝑠=? 𝑢=0 𝑣=12.5 𝑎=? 𝑡=30 𝑣=𝑢+𝑎𝑡 Sub in the values 𝑣=𝑢+𝑎𝑡 12.5=0+30𝑎 Divide by 30 𝑠= 𝑢+𝑣 2 𝑡 5 12 𝑚 𝑠 −2 =𝑎 You can use exact answers! 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: 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 𝑎= 5 12 Write out ‘suvat’ and fill in what you know 𝑠=? 𝑢=0 𝑣=12.5 𝑎=? 𝑡=30 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 Sub in values 𝑠= ×30 𝑠= 𝑢+𝑣 2 𝑡 Work it out! 𝑠=187.5𝑚 2A

TeaChings for Exercise 2B

You can also use 3 more formulae linking different combination of ‘SUVAT’, for a particle moving in a straight line with constant acceleration 𝑣=𝑢+𝑎𝑡 Subtract u 𝑣−𝑢=𝑎𝑡 Divide by a 𝑣−𝑢 𝑎 =𝑡 𝑣=𝑢+𝑎𝑡 𝑠= 𝑢+𝑣 2 𝑡 𝑠= 𝑢+𝑣 2 𝑡 Replace t with the expression above 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠= 𝑢+𝑣 2 𝑣−𝑢 𝑎 Multiply numerators and denominators 𝑠= 𝑣 2 − 𝑢 2 2𝑎 Multiply by 2a 2𝑎𝑠= 𝑣 2 − 𝑢 2 Add u2 𝑢 2 +2𝑎𝑠= 𝑣 2 This is the way it is usually written! 𝑣 2 = 𝑢 2 +2𝑎𝑠 2B

𝑠= 𝑢+𝑣 2 𝑡 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’ 𝑠= 𝑢+𝑢+𝑎𝑡 2 𝑡 Group terms on the numerator 𝑣=𝑢+𝑎𝑡 𝑠= 2𝑢+𝑎𝑡 2 𝑡 Divide the numerator by 2 𝑠= 𝑢+𝑣 2 𝑡 𝑠= 𝑢+ 1 2 𝑎𝑡 𝑡 𝑣 2 = 𝑢 2 +2𝑎𝑠 Multiply out the bracket 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 2B

𝑣=𝑢+𝑎𝑡 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’ 𝑣−𝑎𝑡=𝑢 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣=𝑢+𝑎𝑡 Replace ‘u’ with ‘v - at’ from above’ 𝑠= 𝑢+𝑣 2 𝑡 𝑠=(𝑣−𝑎𝑡)𝑡+ 1 2 𝑎 𝑡 2 Multiply out the bracket 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠=𝑣𝑡−𝑎 𝑡 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Group up the at2 terms 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 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. 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 𝑠= 𝑢+𝑣 2 𝑡 Draw a diagram 𝑣=𝑢+𝑎𝑡 A B Write out ‘suvat’ with the information given 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=? 𝑢=3 𝑣=18 𝑎=5 𝑡=? 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 We are calculating s, using v, u and a 𝑣 2 = 𝑢 2 +2𝑎𝑠 Replace v, u and a 18 2 = (5)𝑠 Work out terms 324=9+10𝑠 Subtract 9 315=10𝑠 Divide by 10 31.5𝑚=𝑠 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: 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 𝑠= 𝑢+𝑣 2 𝑡 Draw a diagram 𝑣=𝑢+𝑎𝑡 Write out ‘suvat’ with the information given Pillar Box Lamp Post 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠=? 𝑢=8 𝑣=? 𝑎=0.75 𝑡=12 We are calculating s, using u, a and t 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Replace u, a and t 𝑠=(8×12)+ 1 2 (0.75×1 2 2 ) Calculate 𝑠=150𝑚 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: 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 𝑠= 𝑢+𝑣 2 𝑡 Draw a diagram 𝑣=𝑢+𝑎𝑡 Write out ‘suvat’ with the information given Pillar Box Lamp Post 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠=? 𝑢=8 𝑣=? 𝑎=0.75 𝑡=12 We are calculating v, using u, a and t 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑣=𝑢+𝑎𝑡 Replace u, a and t 𝑣=8+(0.75×12) Calculate 𝑣=17𝑚 𝑠 −1 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

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 𝑠= 𝑢+𝑣 2 𝑡 Draw a diagram 𝑣=𝑢+𝑎𝑡 13ms-1 Write out ‘suvat’ with the information given 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 O A 𝑠=20 𝑢=13 𝑣=? 𝑎=−4 𝑡=? We are calculating t, using s, u and a 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Replace s, u and a 20=(13)𝑡+ 1 2 (−4) 𝑡 2 Simplify terms 20=13𝑡−2 𝑡 2 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 2 𝑡 2 −13𝑡+20=0 Factorise (or use the quadratic formula…) (2𝑡−5)(𝑡−4)=0 𝑡=2.5 𝑜𝑟 4 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: 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 𝑠= 𝑢+𝑣 2 𝑡 Draw a diagram 𝑣=𝑢+𝑎𝑡 13ms-1 Write out ‘suvat’ with the information given 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 O A 𝑠=20 𝑢=13 𝑣=? 𝑎=−4 𝑡=? We are calculating t, using s, u and a 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 The particle passes through A at 2.5 seconds and 4 seconds, so it was beyond A for 1.5 seconds… 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: 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 𝑠= 𝑢+𝑣 2 𝑡 Draw a diagram 𝑣=𝑢+𝑎𝑡 13ms-1 Write out ‘suvat’ with the information given 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 O A 𝑠=20 𝑠=0 𝑢=13 𝑣=? 𝑎=−4 𝑡=? The particle returns to O when s = 0 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Replace s, u and a 0=(13)𝑡+(−2) 𝑡 2 Simplify 0=13𝑡−2 𝑡 2 Rearrange The particle is at O when t = 0 seconds (to begin with) and is at O again when t = 6.5 seconds 2 𝑡 2 −13𝑡=0 Factorise 𝑡(2𝑡−13)=0 𝑡=0 𝑜𝑟 6.5 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. 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 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 The total distance travelled will be double the distance the particle reaches from O (point X)  At X, the velocity is 0 O X 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠=? 𝑢=15 𝑣=0 𝑎=−2.5 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 Replace v, u and a We are calculating s, using u, v and a 0 2 = (−2.5)𝑠 Simplify 0=225−5𝑠 Add 5s 5𝑠=225 Divide by 5 𝑠=45𝑚 45m is the distance from O to X. Double it for the total distance travelled 𝑇𝑜𝑡𝑎𝑙 𝑠=90𝑚 2B

TeaChings for Exercise 2C

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

A ball is projected vertically upwards from a point O with a speed of 12ms-1. Find: The greatest height reached by the ball The total time the ball is in the air You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 𝑠=? Draw a diagram 0ms-1 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 𝑢=12 At its highest point, the velocity of the ball is 0ms-1 𝑣=0 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 As the ball has been projected upwards, gravity is acting in the opposite direction and hence the acceleration is negative 𝑎=−9.8 12ms-1 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 We are calculating s, using u, v and a 𝑣 2 = 𝑢 2 +2𝑎𝑠 Replace v, u and a 0 2 = (−9.8)𝑠 Simplify 0=144−19.6𝑠 Add 19.6s 19.6𝑠=144 Divide and round to 2sf (since gravity has been given to 2sf) 𝑠=7.4𝑚 (2𝑠𝑓) 2C

A ball is projected vertically upwards from a point O with a speed of 12ms-1. Find: The greatest height reached by the ball – 7.4m The total time the ball is in the air You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 𝑠=0 Draw a diagram 0ms-1 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 𝑢=12 For the total time the ball is in the air, the displacement (s) will be 0 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑎=−9.8 Also, we will not know v (yet!) when the ball strikes the ground 12ms-1 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 We are calculating t, using s, u and a 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Replace s, u and a 0=12𝑡−4.9 𝑡 2 Factorise 0=𝑡(12−4.9𝑡) Choose the appropriate answer! 𝑡=0 𝑜𝑟 2.4 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 (2𝑠𝑓) So the ball will be in the air for 2.4 seconds 2C

A book falls off the top shelf of a bookcase. The shelf is 1.4m above the ground. Find: The time it takes the book to reach the floor The speed with which the book strikes the floor You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 0ms-1 𝑠=1.4 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 The book’s initial speed will be 0 as it has not been projected to begin with 𝑢=0 1.4m 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 As the book’s initial movement is downwards, we take the acceleration due to gravity as positive 𝑎=9.8 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 We are calculating t, using s, u and a… 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Replace s, u and a 1.4=(0)𝑡+ 1 2 (9.8) 𝑡 2 Simplify 1.4=4.9 𝑡 2 Divide by 4.9 = 𝑡 2 Find the positive square root 𝑡=0.53𝑠 (2𝑠𝑓) 2C

A book falls off the top shelf of a bookcase. The shelf is 1.4m above the ground. Find: The time it takes the book to reach the floor – 0.53 seconds The speed with which the book strikes the floor You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 0ms-1 𝑠=1.4 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 The book’s initial speed will be 0 as it has not been projected to begin with 𝑢=0 1.4m 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 As the book’s initial movement is downwards, we take the acceleration due to gravity as positive 𝑎=9.8 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 We are calculating v, using s, u and a… 𝑣 2 = 𝑢 2 +2𝑎𝑠 Replace s, u and a 𝑣 2 = (9.8×1.4) Calculate 𝑣 2 =27.44 Find the positive square root 𝑣=5.2𝑚 𝑠 −1 (2𝑠𝑓) 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. You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity Draw a diagram 𝑠=−7 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 𝑢=21 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 𝑣=? 𝑎=−9.8 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 21ms-1 7m The ball is projected upwards, so the acceleration due to gravity is negative 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 We are calculating t, using s, u and a Replace s, u and a −7=(21)𝑡+ 1 2 (−9.8) 𝑡 2 Simplify −7=21𝑡−4.9 𝑡 2 Rearrange and set equal to 0 4.9 𝑡 2 −21𝑡−7=0 We will need the quadratic formula here, so write down a, b and c… 𝑎=4.9 𝑏=−21 𝑐=−7 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. You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity Draw a diagram 𝑠=−7 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 𝑢=21 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 𝑣=? 𝑎=−9.8 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 21ms-1 7m The ball is projected upwards, so the acceleration due to gravity is negative 𝑡=? 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝑎=4.9 𝑏=−21 𝑐=−7 𝑡= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Replace a, b and c (using brackets!) 𝑡= −(−21)± (−21) 2 −(4×4.9×−7) (2×4.9) Calculate and be careful with any negatives in the previous step!) 𝑡=4.6 𝑜𝑟−0.3 𝑡=4.6 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 (2𝑠𝑓) 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: The speed of projection The total time for which the ball is 50m or more above the ground You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 62.5m 𝑠=62.5 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 𝑢=? The maximum height is 62.5m  At this point the ball’s velocity is 0ms-1 𝑣=0 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 The ball is projected upwards, so the acceleration due to gravity is negative 𝑎=−9.8 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 u ms-1 𝑡=? We are calculating u, using s, v and a 𝑣 2 = 𝑢 2 +2𝑎𝑠 Replace v, a and s 0 2 = 𝑢 2 +2(−9.8×62.5) Simplify 0= 𝑢 2 −1225 Rewrite 𝑢 2 =1225 Find the positive square root 𝑢=35𝑚 𝑠 −1 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: The speed of projection – 35ms-1 The total time for which the ball is 50m or more above the ground You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 62.5m 𝑠=50 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 50m 𝑢=35 The ball will pass the 50m mark twice – we need to find these two times! 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑎=−9.8 We are calculating t, using s, u and a 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 u ms-1 𝑡=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Replace s, u and a 50=(35)𝑡+ 1 2 (−9.8) 𝑡 2 Simplify 50=35𝑡−4.9 𝑡 2 Rearrange, and set equal to 0 4.9 𝑡 2 −35𝑡+50=0 We will need the quadratic formula, and hence a, b and c 𝑎=4.9 𝑏=−35 𝑐=50 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: The speed of projection – 35ms-1 The total time for which the ball is 50m or more above the ground You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 62.5m 𝑠=50 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑣=𝑢+𝑎𝑡 50m 𝑢=35 The ball will pass the 50m mark twice – we need to find these two times! 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑎=−9.8 We are calculating t, using s, u and a 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 u ms-1 𝑡=? 𝑎=4.9 𝑏=−35 𝑐=50 𝑡= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Sub these into the Quadratic formula 𝑡= −(−35)± (−35) 2 −(4×4.9×50) 4.9×2 We get the two times the ball passes the 50m mark 𝑡=5.1686… 𝑜𝑟 𝑡=1.9742… Calculate the difference between these times! 𝑡=3.2𝑠 (2𝑠𝑓) 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. You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 𝐵𝑎𝑙𝑙 𝐴 𝐵𝑎𝑙𝑙 𝐵 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑠= 𝑠 1 𝑠= 𝑠 2 𝑣=𝑢+𝑎𝑡 In this case we need to consider each ball separately. We can call the two distances s1 and s2 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 𝑢=0 𝑢=21 s1 𝑣=? 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 63m 𝑎=9.8 𝑎=−9.8 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑡=𝑡 𝑡=𝑡 s2 21ms-1 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 𝐵𝑎𝑙𝑙 𝐴 𝐵𝑎𝑙𝑙 𝐵 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Sub in s, u, a and t for Ball A Sub in s, u, a and t for Ball B 𝑠 1 =(0)𝑡+ 1 2 (9.8) 𝑡 2 𝑠 2 =(21)𝑡+ 1 2 (−9.8) 𝑡 2 Simplify Simplify 𝑠 1 =4.9 𝑡 2 𝑠 2 =21𝑡−4.9 𝑡 2 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. You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 𝐵𝑎𝑙𝑙 𝐴 𝐵𝑎𝑙𝑙 𝐵 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑠= 𝑠 1 𝑠= 𝑠 2 𝑣=𝑢+𝑎𝑡 In this case we need to consider each ball separately. We can call the two distances s1 and s2 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 𝑢=0 𝑢=21 s1 𝑣=? 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 63m 𝑎=9.8 𝑎=−9.8 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑡=𝑡 𝑡=𝑡 s2 21ms-1 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 1) 𝑠 1 =4.9 𝑡 2 2) 𝑠 2 =21𝑡−4.9 𝑡 2 Add the two equations together (this cancels the 4.9t2 terms) 𝑠 1 +𝑠 2 =21𝑡 s1 + s2 must be the height of the tower (63m) 63=21𝑡 Divide by 21 3=𝑡 So the balls collide after 3 seconds… 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. You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence of gravity 𝐵𝑎𝑙𝑙 𝐴 𝐵𝑎𝑙𝑙 𝐵 Draw a diagram 𝑠= 𝑢+𝑣 2 𝑡 𝑠= 𝑠 1 𝑠= 𝑠 2 𝑣=𝑢+𝑎𝑡 In this case we need to consider each ball separately. We can call the two distances s1 and s2 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 𝑢=0 𝑢=21 s1 𝑣=? 𝑣=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 63m 𝑎=9.8 𝑎=−9.8 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑡=𝑡 𝑡=𝑡 s2 21ms-1 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 2) 𝑠 2 =21𝑡−4.9 𝑡 2 Sub in t = 3 (we use this equation since s2 is the height above the ground) 𝑠 2 =21(3)−4.9 (3) 2 𝑠 2 =18.9𝑚 (19𝑚 𝑡𝑜 2𝑠𝑓) 2C

TeaChings for Exercise 2D

You can represent the motion of an object on a speed-time graph, distance-time graph or an acceleration-time graph 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚= 𝑎+𝑏 2 ℎ 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒= 𝑢+𝑣 2 𝑡 Final velocity v 𝑠= 𝑢+𝑣 2 𝑡 v - u 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒=𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 Initial velocity u v On a speed-time graph, the Area beneath it is the distance covered! t u 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 O t t 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑣−𝑢 𝑡 Time taken 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡=𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 On a speed-time graph, the gradient of a section is its acceleration! 2D

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 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. Draw an acceleration-time graph for this information Draw a distance-time graph for this information Acceleration (ms-2) 5 For now, we assume the rate of acceleration jumps between different rates… 20 40 60 80 Time (s) -5 Distance (m) 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 20 40 60 80 Time (s) 2D

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 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: The distance travelled by the cyclist The rate of deceleration of the cyclist v(ms-1) 8 6 6 8 12 t(s) 12 𝐴𝑟𝑒𝑎= 𝑎+𝑏 2 ℎ Sub in the appropriate values for the trapezium above 𝐴𝑟𝑒𝑎= ×6 Calculate 𝐴𝑟𝑒𝑎=60 𝑇ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑖𝑠 60𝑚 2D

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 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: The distance travelled by the cyclist – 60m The rate of deceleration of the cyclist v(ms-1) 6 -6 8 12 t(s) 4 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 Sub in the appropriate values for the trapezium above 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡= −6 4 Calculate 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡=−1.5 𝑆𝑜 𝑡ℎ𝑒 𝑑𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 1.5𝑚 𝑠 −2 2D

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 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. Sketch a speed-time graph for this motion Given that the particle travels 600m, find the value of T Sketch an acceleration-time graph for this motion v(ms-1) 5T 8 𝐴𝑟𝑒𝑎= 𝑎+𝑏 2 ℎ Sub in values 8 600= 5𝑇+6𝑇 ×8 Simplify fraction 600= 5.5𝑇+20 ×8 T 5T 40 t(s) Divide by 8 6T + 40 75=5.5𝑇+20 Subtract 20 55=5.5𝑇 Divide by 5.5 10=𝑇 2D

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 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. Sketch a speed-time graph for this motion Given that the particle travels 600m, find the value of T – 10 seconds Sketch an acceleration-time graph for this motion v(ms-1) a(ms-2) 8 0.8 20 40 60 80 100 10 T 50 5T 40 t(s) t(s) -0.2 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 First section Last section 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= 8 10 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= −8 40 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛=0.8𝑚 𝑠 −2 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛=−0.2𝑚 𝑠 −2 2D

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 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. Sketch a speed-time graph to show the motion of both cars Calculate the distance between the lay-by and the road sign v(ms-1) 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 - 15 20 D 17.5 C 20 17.5 15 T T t(s) 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 𝐶 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 𝐷 𝑠= 𝑎+𝑏 2 ℎ 𝑠=𝑏×ℎ Sub in values Sub in values 𝑠=17.5×𝑇 𝑠= 𝑇+𝑇−15 2 ×20 Simplify fraction 𝑠=17.5𝑇 𝑠= 𝑇−7.5 ×20 Multiply bracket 𝑠=20𝑇−150 2D

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 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. Sketch a speed-time graph to show the motion of both cars Calculate the distance between the lay-by and the road sign v(ms-1) 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’ 20 D 17.5 C 15 T t(s) 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 𝐶 17.5𝑇=20𝑇−150 𝑠=17.5𝑇 Subtract 17.5T Sub in T 𝑠=17.5𝑇 0=2.5𝑇−150 𝑠=17.5(60) Add 150 𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 𝐷 Calculate! 𝑠=20𝑇−150 150=2.5𝑇 𝑠=1050𝑚 Divide by 2.5 Set these equations equal to each other! 60=𝑇 2D

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

Kinematics of a particle moving in a straight line

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Kinematics of a particle moving in a straight line

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