Kinematics of a particle moving in a straight line or plane

Kinematics of a particle moving in a straight line or plane

Introduction In this chapter you will extend the knowledge of kinematics from M1 You will use the SUVAT equations to solve problems in 2D where a particle has vertical and horizontal motion (commonly referred to as projectiles) You will also see how to solve problems involving variable velocities and accelerations, and how calculus from Core Maths fits into Mechanics!

Teachings for Exercise 1A

u ms-1 You can use the constant acceleration formulae for a projectile moving in a vertical plane When a particle is projected with initial speed u, at an angle θ to the horizontal, it moves along a symmetrical curve The speed u is known as the speed of projection  The angle θ is known as the angle of projection (or angle of elevation) of the particle The initial projection speed can be resolved into two components as shown to the right θ u ms-1 uSinθ ms-1 θ uCosθ ms-1 Any particle projected at an angle will have an initial horizontal and vertical speed 1A

You can use the constant acceleration formulae for a projectile moving in a vertical plane When a particle is projected with initial speed u, at an angle θ to the horizontal, it moves along a symmetrical curve The speed u is known as the speed of projection  The angle θ is known as the angle of projection (or angle of elevation) of the particle The initial projection speed can be resolved into two components as shown to the right u ms-1 uSinθ ms-1 θ uCosθ ms-1 The horizontal speed will be constant as there is no acceleration in this direction 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒 The vertical speed will change over time because of the acceleration due to gravity (a = 9.8ms-2) Greatest height Range 1A

Start with a diagram! 28 You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30°. After projection, the particle moves freely under gravity until it strikes the plane at a point A. Find: The greatest height above the plane reached by P The time of flight of P The distance OA 28sin30 30° O A 28cos30 The greatest height will be reached when the vertical velocity is 0 (as the particle stops moving up and starts moving down)  Resolve vertically and use SUVAT 𝑠=? 𝑢=28𝑠𝑖𝑛30 𝑣=0 𝑎=−9.8 𝑡=? 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values 10𝑚 (0) 2 = (28𝑠𝑖𝑛30) 2 +2(−9.8)(𝑠) Calculate terms 0=196−19.6𝑠 Rearrange 19.6𝑠=196 Calculate 𝑠=10𝑚 1A

Start with a diagram! 28 You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30°. After projection, the particle moves freely under gravity until it strikes the plane at a point A. Find: The greatest height above the plane reached by P The time of flight of P The distance OA 28sin30 30° O A 28cos30 The time of flight of the particle will be when its vertical displacement is 0 (ie – it is at the same height it started at)  Resolve vertically and use SUVAT (again) 𝑠=0 𝑢=28𝑠𝑖𝑛30 𝑣=? 𝑎=−9.8 𝑡=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Sub in values 0=(28𝑠𝑖𝑛30)(𝑡)+ 1 2 (−9.8)( 𝑡 2 ) 10𝑚 Work out terms 0=28𝑡𝑠𝑖𝑛30−4.9 𝑡 2 Factorise 2.9𝑠 0=𝑡(28𝑠𝑖𝑛30−4.9𝑡) Either part could be 0 The answer of 0 corresponds to the particle’s starting position. The value 2.9 is the time it takes to hit the ground again 𝑡=0 𝑜𝑟 28𝑠𝑖𝑛30−4.9𝑡=0 Work out the second possibility 𝑡=2.9 1A

Start with a diagram! 28 You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30°. After projection, the particle moves freely under gravity until it strikes the plane at a point A. Find: The greatest height above the plane reached by P The time of flight of P The distance OA 28sin30 30° O A 28cos30 The distance OA will be based on the horizontal projection speed (which is constant)  We can use the time of flight calculated in part b) 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=28𝑐𝑜𝑠30×2.9 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=69𝑚 (2𝑠𝑓) 10𝑚 2.9𝑠 69𝑚 1A

Start with a diagram! 20 You can use the constant acceleration formulae for a projectile moving in a vertical plane A ball is thrown horizontally, with speed 20ms-1, from the top of a building of height 30m. Find: The time the ball takes to reach the ground The distance between the bottom of the building and the point where the ball strikes the ground 30m As the ball is projected horizontally, there is no initial velocity vertically Resolving vertically (with downwards as the positive direction) The ball must travel 30m to hit the ground 𝑡=2.5 𝑠=30 𝑢=0 𝑣=? 𝑎=9.8 𝑡=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Sub in values 30=(0)(𝑡)+ 1 2 𝑎(9.8) 𝑡 2 Work out terms 30=4.9 𝑡 2 Calculate the positive value 2.5=𝑡 (2sf) 1A

Start with a diagram! 20 You can use the constant acceleration formulae for a projectile moving in a vertical plane A ball is thrown horizontally, with speed 20ms-1, from the top of a building of height 30m. Find: The time the ball takes to reach the ground The distance between the bottom of the building and the point where the ball strikes the ground 30m 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=20×2.5 𝑡=2.5 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=49𝑚 (2sf) Make sure you use the exact value for t rather than the rounded on from part a)! 𝑠=49𝑚 1A

Start with a diagram! V You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point O with speed Vms-1 at an angle of elevation θ, where tanθ = 4/3. The point O is 42.5m above the horizontal plane. The particle strikes the plane 5 seconds after it is projected. Show that V = 20ms-1 Find the distance between O and A Find the values of Sinθ and Cosθ first, by labelling a triangle where Tanθ = 4/3 (That is, opposite = 4 and adjacent = 3) 𝑇𝑎𝑛𝜃= 4 3 Vsinθ θ O 𝑆𝑖𝑛𝜃= 4 5 Vcosθ 42.5m 𝐶𝑜𝑠𝜃= 3 5 A Resolving vertically using the information given (taking upwards as positive)  The ball will be 42.5m lower after 5 seconds 𝑠=−42.5 𝑢=𝑉𝑠𝑖𝑛θ 𝑣=? 𝑎=−9.8 𝑡=5 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Sub in values −42.5=(𝑉𝑠𝑖𝑛θ)(5)+ 1 2 (−9.8)( 5 2 ) Calculate terms (use sinθ = 4/5) Hyp −42.5=4𝑉−122.5 5 Use Pythagoras’ Theorem to find the missing side  Remember Sinθ = opp/hyp and Cosθ = adj/hyp Add 122.5 4 Opp 80=4𝑉 Divide by 4 3 20=𝑉 Adj 1A

Start with a diagram! V You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point O with speed Vms-1 at an angle of elevation θ, where tanθ = 4/3. The point O is 42.5m above the horizontal plane. The particle strikes the plane 5 seconds after it is projected. Show that V = 20ms-1 Find the distance between O and A 𝑇𝑎𝑛𝜃= 4 3 Vsinθ θ O 𝑆𝑖𝑛𝜃= 4 5 Vcosθ 42.5m 𝐶𝑜𝑠𝜃= 3 5 A 60m 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒 V = 20 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑉𝑐𝑜𝑠θ×5 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒= ×5 Make sure you read the question – we want the distance OA, not just the horizontal distance travelled! 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=60𝑚 1A

Start with a diagram! V You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point O with speed Vms-1 at an angle of elevation θ, where tanθ = 4/3. The point O is 42.5m above the horizontal plane. The particle strikes the plane 5 seconds after it is projected. Show that V = 20ms-1 Find the distance between O and A 𝑇𝑎𝑛𝜃= 4 3 θ O 𝑆𝑖𝑛𝜃= 4 5 42.5m 𝐶𝑜𝑠𝜃= 3 5 A 60m Use Pythagoras’ Theorem to calculate the distance V = 20 Distance = 74m =74𝑚 (2sf) 1A

Start with a diagram! You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point O with speed 35ms-1 at an angle of elevation of 30°. The particle moves freely under gravity. Find the length of time for which the particle is 15m or more above O  We want to find the times that the particle is exactly at 15m. There will be 2 of these, once as the particle is travelling up, and once as it is travelling down. 15m 35 35Sin30 30° 35Cos30 Resolving vertically, taking upwards as positive 𝑠=15 𝑢=35𝑆𝑖𝑛30 𝑣=? 𝑎=−9.8 𝑡=? 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 Sub in values 15=(35𝑠𝑖𝑛30)(𝑡)+ 1 2 (−9.8)( 𝑡 2 ) Work out each term 15=35𝑡𝑠𝑖𝑛30−4.9 𝑡 2 𝑎=4.9 Rearrange to a quadratic form 𝑏=−35𝑠𝑖𝑛30 4.9 𝑡 2 −35𝑡𝑠𝑖𝑛30+15=0 𝑐=15 This will be difficult to factorise so we should use the quadratic formula – hence we need a, b and c 1A

Start with a diagram! You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point O with speed 35ms-1 at an angle of elevation of 30°. The particle moves freely under gravity. Find the length of time for which the particle is 15m or more above O  We want to find the times that the particle is exactly at 15m. There will be 2 of these, once as the particle is travelling up, and once as it is travelling down. 15m 35 35Sin30 30° 35Cos30 𝑡= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑡= 35𝑠𝑖𝑛30± (−35𝑠𝑖𝑛30) 2 −(4×4.9×15) 2(4.9) 𝑡=1.43 𝑜𝑟 2.14 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 (2𝑑𝑝) The difference between these times will be the time spent above 15m  Subtract the smallest from the biggest 𝑎=4.9 𝑏=−35𝑠𝑖𝑛30 Time above 15m = 0.71 seconds (remember to use exact answers) 𝑐=15 1A

u You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point with speed u and an angle of elevation θ, and moves freely under gravity. When the particle has moved a horizontal distance x, its height above the point of projection is y. Show that: uSinθ θ uCosθ Finding the height in terms of t Resolving vertically (upwards as positive) Use g to represent acceleration and y to represent the height above the point of projection 𝑠=𝑦 𝑢=𝑢𝑠𝑖𝑛θ 𝑣=? 𝑎=−𝑔 𝑡=𝑡 𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑢 2 (1+𝑡𝑎 𝑛 2 𝜃) 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 In this type of question you need to form equations for the height, y, in terms of the time, t, as well as the horizontal distance, x, in terms of the time, t.  The equations can then be combined to eliminate t Sub in values from SUVAT 𝑦=(𝑢𝑠𝑖𝑛𝜃)(𝑡)+ 1 2 (−𝑔)( 𝑡 2 ) ‘Tidy up’ 𝑦=𝑢𝑡𝑠𝑖𝑛𝜃− 1 2 𝑔 𝑡 2 𝑦=𝑢𝑡𝑠𝑖𝑛𝜃− 1 2 𝑔 𝑡 2 We now have an expression for the height in terms of u, t and θ 1A

u You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point with speed u and an angle of elevation θ, and moves freely under gravity. When the particle has moved a horizontal distance x, its height above the point of projection is y. Show that: uSinθ θ uCosθ Finding the horizontal distance in terms of t 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒 Sub in values (remember x is to be used to represent the horizontal distance) 𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑢 2 (1+𝑡𝑎 𝑛 2 𝜃) 𝑥=𝑢𝑐𝑜𝑠𝜃×𝑡 𝑥=𝑢𝑡𝑐𝑜𝑠𝜃 In this type of question you need to form equations for the height, y, in terms of the time, t, as well as the horizontal distance, x, in terms of the time, t.  The equations can then be combined to eliminate t 𝑦=𝑢𝑡𝑠𝑖𝑛𝜃− 1 2 𝑔 𝑡 2 𝑥=𝑢𝑡𝑐𝑜𝑠𝜃 1A

Divide by cos2θ 𝑠𝑖 𝑛 2 𝜃+𝑐𝑜 𝑠 2 𝜃≡1 𝑡𝑎 𝑛 2 𝜃+1≡𝑠𝑒 𝑐 2 𝜃 You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point with speed u and an angle of elevation θ, and moves freely under gravity. When the particle has moved a horizontal distance x, its height above the point of projection is y. Show that: 𝑦=𝑢𝑡𝑠𝑖𝑛𝜃− 1 2 𝑔 𝑡 2 𝑥=𝑢𝑡𝑐𝑜𝑠𝜃 Divide by ucosθ 𝑥 𝑢𝑐𝑜𝑠𝜃 =𝑡 𝑦=𝑢𝑡𝑠𝑖𝑛𝜃− 1 2 𝑔 𝑡 2 Replace t with the expression above 𝑦=𝑢 𝑥 𝑢𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 − 1 2 𝑔 𝑥 𝑢𝑐𝑜𝑠𝜃 2 Group the first set of terms, square the bracket 𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑢 2 (1+𝑡𝑎 𝑛 2 𝜃) 𝑦= 𝑢𝑥𝑠𝑖𝑛𝜃 𝑢𝑐𝑜𝑠𝜃 − 𝑔 𝑥 2 𝑢 2 𝑐𝑜 𝑠 2 𝜃 Cancel u’s on the first term. Sin/Cos = Tan Now we have equations for y and x. The final answer we want has no ‘t’ terms, so this is an indication that you have to eliminate t from the equations The final answer is written as ‘y =‘, so it looks like we should rearrange the ‘x’ equation and substitute it into the ‘y’ equation Group terms except for Cos on the second − 𝑔 𝑥 2 2 𝑢 𝑐𝑜 𝑠 2 𝜃 𝑦=𝑥𝑡𝑎𝑛𝜃 1/cos = sec − 𝑔 𝑥 2 2 𝑢 2 𝑠𝑒 𝑐 2 𝜃 𝑦=𝑥𝑡𝑎𝑛𝜃 Use the trig identity above to replace sec − 𝑔 𝑥 2 2 𝑢 2 (𝑡𝑎 𝑛 2 𝜃+1) 𝑦=𝑥𝑡𝑎𝑛𝜃 1A

𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑢 2 (1+𝑡𝑎 𝑛 2 𝜃) You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point A on a horizontal plane, with initial speed 28ms-1 and an angle of elevation θ. The particle passes through a point B, which is 8m above the plane and a horizontal distance of 32m from A Find the two possible values of θ, giving your answers to the nearest degree. (Use the formula we have just calculated) Sub in the values we know 8=32𝑡𝑎𝑛𝜃− 9.8 (32) 2 2 (28) 2 (1+𝑡𝑎 𝑛 2 𝜃) Calculate the large fraction 8=32𝑡𝑎𝑛𝜃−6.4(1+𝑡𝑎 𝑛 2 𝜃) Multiply out the bracket 8=32𝑡𝑎𝑛𝜃− 𝑡𝑎 𝑛 2 𝜃 Rearrange to a quadratic form 6.4𝑡𝑎 𝑛 2 𝜃+32𝑡𝑎𝑛𝜃−14.4=0 You can solve this like a quadratic by using the quadratic formula 𝑎=6.4 𝑏=32 𝑐=−14.4 𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑢 2 (1+𝑡𝑎 𝑛 2 𝜃) 1A

6.4𝑡𝑎 𝑛 2 𝜃+32𝑡𝑎𝑛𝜃−14.4=0 You can use the constant acceleration formulae for a projectile moving in a vertical plane A particle is projected from a point A on a horizontal plane, with initial speed 28ms-1 and an angle of elevation θ. The particle passes through a point B, which is 8m above the plane and a horizontal distance of 32m from A Find the two possible values of θ, giving your answers to the nearest degree. (Use the formula we have just calculated) 𝑎=6.4 𝑏=32 𝑐=−14.4 𝑡𝑎𝑛𝜃= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Sub in values 𝑡𝑎𝑛𝜃= −32± −(4×6.4×−14.4) 2(6.4) Calculate each possibility 𝑡𝑎𝑛𝜃=0.5 𝑜𝑟 𝑡𝑎𝑛𝜃=4.5 Use inverse Tan (and round answers) 𝜃=27° 𝑜𝑟 𝜃=77° There will be two possibilities here: 𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑢 2 (1+𝑡𝑎 𝑛 2 𝜃) B If a projectile passes through point B, there are two possible angles it could have been launched through A 1A

(5i + 8j) You can use the constant acceleration formulae for a projectile moving in a vertical plane A ball is struck by a racket at a point A which is 2m above horizontal ground. Immediately after being struck, the ball has velocity (5i + 8j) ms-1, where i and j are unit vectors horizontally and vertically respectively. After being struck, the ball travels freely under gravity until is strikes the ground at a point B, as shown. Find: The greatest height above ground reached by the ball The speed of the ball as it reaches B The angle the velocity of the ball makes with the ground as the ball reaches B 8j A 5i 2m B You can use the vectors in each direction as the initial velocities  Resolve vertically to find the greatest height 𝑠=? 𝑢=8 𝑣=0 𝑎=−9.8 𝑡=? 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values (0) 2 = (8) 2 +2(−9.8)𝑠 𝑠=5.3𝑚 Work out terms 0=64−19.6𝑠 Calculate s 𝑠=3.3𝑚 Careful – the ball starts at a height of 2m, so this must be added on! 𝑠=5.3𝑚 1A

(5i + 8j) You can use the constant acceleration formulae for a projectile moving in a vertical plane A ball is struck by a racket at a point A which is 2m above horizontal ground. Immediately after being struck, the ball has velocity (5i + 8j) ms-1, where i and j are unit vectors horizontally and vertically respectively. After being struck, the ball travels freely under gravity until is strikes the ground at a point B, as shown. Find: The greatest height above ground reached by the ball The speed of the ball as it reaches B The angle the velocity of the ball makes with the ground as the ball reaches B 8j A 5i 2m B As the ball strikes B, its velocity will have both a horizontal and vertical component The horizontal speed is constant (5) so we do not need to calculate this The vertical speed however will vary as the ball travels so we need to work this out… At B the ball has travelled 2m down – resolve vertically again, taking downwards as the positive direction 𝑠=2 𝑢=−8 𝑣=? 𝑎=9.8 𝑡=? 𝑠=5.3𝑚 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values 𝑣 2 = (−8) 2 +2(9.8)(2) Work out right side 𝑣 2 =103.2 Square root 𝑣=10.2𝑚 𝑠 −1 So as it strikes B, the ball has a velocity of 10.2ms-1 downwards

(5i + 8j) You can use the constant acceleration formulae for a projectile moving in a vertical plane A ball is struck by a racket at a point A which is 2m above horizontal ground. Immediately after being struck, the ball has velocity (5i + 8j) ms-1, where i and j are unit vectors horizontally and vertically respectively. After being struck, the ball travels freely under gravity until is strikes the ground at a point B, as shown. Find: The greatest height above ground reached by the ball The speed of the ball as it reaches B The angle the velocity of the ball makes with the ground as the ball reaches B 8j A 5i 2m B As the ball strikes B, its velocity will have both a horizontal and vertical component The horizontal speed is constant (5) so we do not need to calculate this The vertical speed however will vary as the ball travels so we need to work this out… At B the ball has travelled 2m down – resolve vertically again, taking downwards as the positive direction 5 You can just use Pythagoras to work out the overall speed! 𝑠=5.3𝑚 10.2 = =11𝑚 𝑠 −1 =11𝑚 𝑠 −1 (2𝑠𝑓) B

(5i + 8j) You can use the constant acceleration formulae for a projectile moving in a vertical plane A ball is struck by a racket at a point A which is 2m above horizontal ground. Immediately after being struck, the ball has velocity (5i + 8j) ms-1, where i and j are unit vectors horizontally and vertically respectively. After being struck, the ball travels freely under gravity until is strikes the ground at a point B, as shown. Find: The greatest height above ground reached by the ball The speed of the ball as it reaches B The angle the velocity of the ball makes with the ground as the ball reaches B 8j A 5i 2m B We can use the same diagram to calculate the angle between the velocity and the ground… 𝑇𝑎𝑛𝜃= 𝑂𝑝𝑝 𝐴𝑑𝑗 5 Sub in values 𝑇𝑎𝑛𝜃= 26.2° θ 10.2 63.8° 𝑠=5.3𝑚 Calculate 𝑇𝑎𝑛𝜃=0.49 B =11𝑚 𝑠 −1 Inverse Tan 𝜃=26.2° Remember to work out the actual angle between the velocity and θ  (Subtract θ from 90°) 𝜃=63.8° =63.8°

Teachings for Exercise 1B

You can use calculus for a particle moving in a straight line, with acceleration that varies with time The SUVAT equations can be used when acceleration is constant and a particle is moving in a straight line. If the acceleration of a particle varies, you need to use Calculus Displacement (x) Differentiate Integrate Velocity (v) Differentiate Integrate Acceleration (a) Velocity is the rate of change of displacement with respect to time  Therefore, differentiating displacement gives velocity Acceleration is the rate of change of velocity with respect to time  Therefore, differentiating velocity gives acceleration In reverse, use integration! 1B

You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along the x-axis. At time t seconds, the displacement x metres from O is given by: Find: The speed of P when t = 3 The value of t for which P is instantaneously at rest The magnitude of acceleration when t = 1.5 𝑥= 𝑡 4 −32𝑡+12 Differentiate x with respect to t 𝑑𝑥 𝑑𝑡 = 4𝑡 3 −32 Displacement (x) dx/dt is the velocity Differentiate Integrate Velocity (v) 𝑣= 4𝑡 3 −32 Differentiate Integrate Acceleration (a) 𝑣= 4𝑡 3 −32 Sub in t = 3 𝑣= 4(3) 3 −32 𝑥= 𝑡 4 −32𝑡+12 Calculate 𝑣=76𝑚 𝑠 −1 1B

𝑥= 𝑡 4 −32𝑡+12 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along the x-axis. At time t seconds, the displacement x metres from O is given by: Find: The speed of P when t = 3 – 76ms-1 The value of t for which P is instantaneously at rest The magnitude of acceleration when t = 1.5 𝑣= 4𝑡 3 −32 Displacement (x) If P is instantaneously at rest, it is the time when the velocity is 0 Differentiate Integrate Velocity (v) Differentiate Integrate 𝑣= 4𝑡 3 −32 Acceleration (a) v = 0 0= 4𝑡 3 −32 Add 32 32= 4𝑡 3 Divide by 4 8= 𝑡 3 𝑥= 𝑡 4 −32𝑡+12 Cube root 2=𝑡 1B

𝑥= 𝑡 4 −32𝑡+12 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along the x-axis. At time t seconds, the displacement x metres from O is given by: Find: The speed of P when t = 3 – 76ms-1 The value of t for which P is instantaneously at rest – 2 seconds The magnitude of acceleration when t = 1.5 𝑣= 4𝑡 3 −32 Differentiate the velocity again to get the acceleration, in terms of t Displacement (x) Differentiate Integrate Velocity (v) Differentiate 𝑣= 4𝑡 3 −32 Integrate Differentiate v with respect to t Acceleration (a) 𝑑𝑣 𝑑𝑡 = 12𝑡 2 dv/dt is just acceleration 𝑎= 12𝑡 2 𝑥= 𝑡 4 −32𝑡+12 𝑎= 12𝑡 2 Sub in t = 1.5 𝑎= 12(1.5) 2 Calculate 𝑎=27𝑚 𝑠 −2 1B

Displacement (x) Velocity (v) Acceleration (a) You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle is moving along the x-axis. At time t = 0, the particle is at the point where x = 5. The velocity of the particle at time t seconds (where t ≥ 0) is (6t – t2)ms-1. Find: The acceleration of the particle when t = 2 and t = 4 An expression for the displacement of the particle from O at time t seconds The distance of the particle from its starting point when t = 6 𝑣=6𝑡− 𝑡 2 Differentiate the velocity to find acceleration 𝑑𝑣 𝑑𝑡 =6−2𝑡 dv/dt = a 𝑎=6−2𝑡 𝑎=6−2𝑡 𝑎=6−2𝑡 Sub in t = 2 Sub in t = 4 𝑎=6−2(2) 𝑎=6−2(4) Calculate Calculate 𝑎=2𝑚 𝑠 −2 𝑎=−2𝑚 𝑠 −2 𝑎=2𝑚 𝑠 −2 𝑎=−2𝑚 𝑠 −2 At t = 2, the acceleration is 2ms-2 in the direction of x-increasing At t = 4, the acceleration is 2ms-2 in the direction of x-decreasing Remember to consider the direction of the acceleration! 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑣=6𝑡− 𝑡 2 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle is moving along the x-axis. At time t = 0, the particle is at the point where x = 5. The velocity of the particle at time t seconds (where t ≥ 0) is (6t – t2)ms-1. Find: The acceleration of the particle when t = 2 and t = 4 An expression for the displacement of the particle from O at time t seconds The distance of the particle from its starting point when t = 6 Integrate v (remember to include C!) 𝑣 = 6 𝑡 2 2 − 𝑡 𝐶 Simplify terms if possible 𝑣 =3 𝑡 2 − 𝑡 𝐶 The integral of v is x (displacement) 𝑥=3 𝑡 2 − 𝑡 𝐶 We need to find C In the question, we were told at time t = 0, the displacement was 5 𝑎=2𝑚 𝑠 −2 𝑎=−2𝑚 𝑠 −2 𝑥=3 𝑡 2 − 𝑡 𝐶 Sub in values 5=3 (0) 2 − (0) 𝐶 𝑥=3 𝑡 2 − 𝑡 Calculate C 5=𝐶 Now we can write the formula for displacement with C = 5 in it 𝑥=3 𝑡 2 − 𝑡 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑥=3 𝑡 2 − 𝑡 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle is moving along the x-axis. At time t = 0, the particle is at the point where x = 5. The velocity of the particle at time t seconds (where t ≥ 0) is (6t – t2)ms-1. Find: The acceleration of the particle when t = 2 and t = 4 An expression for the displacement of the particle from O at time t seconds The distance of the particle from its starting point when t = 6 Sub in t = 6 𝑥=3 (6) 2 − (6) Calculate x 𝑥=41𝑚 Remember to read the question!  We need the particle’s displacement from its starting point. As its starting point was 5m from O, it has moved 36m (to get to 41m) 𝑎=2𝑚 𝑠 −2 𝑎=−2𝑚 𝑠 −2 𝑥=3 𝑡 2 − 𝑡 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑎=2−2𝑡 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along a straight line. At time t = 0, the particle is at a point A and is moving with velocity 8ms-1 towards a point B on the line, where AB = 30m. At time t seconds, where t ≥ 0, the acceleration of P is (2 – 2t)ms-2 in the direction AB. Find an expression, in terms of t, for the displacement of P from A at time t seconds Show that P does not reach B Find the value of t when P returns to A Find the total distance travelled by P in the interval between the two instances when it passes through A 𝑎 =2𝑡− 2 𝑡 𝐶 Integrate The integral of a is v, simplify the fraction 𝑣=2𝑡− 𝑡 2 +𝐶 We know at time t = 0, the velocity is 8  Use these to find the value of C 𝑣=2𝑡− 𝑡 2 +𝐶 Sub in values 8=2(0)− (0) 2 +𝐶 Calculate C 𝐶=8 We can now put the value C = 8 into our equation for velocity 𝑣=2𝑡− 𝑡 2 +8 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑎=2−2𝑡 𝑣=2𝑡− 𝑡 2 +8 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along a straight line. At time t = 0, the particle is at a point A and is moving with velocity 8ms-1 towards a point B on the line, where AB = 30m. At time t seconds, where t ≥ 0, the acceleration of P is (2 – 2t)ms-2 in the direction AB. Find an expression, in terms of t, for the displacement of P from A at time t seconds Show that P does not reach B Find the value of t when P returns to A Find the total distance travelled by P in the interval between the two instances when it passes through A 𝑣=2𝑡− 𝑡 2 +8 Integrate v now to get a formula for the displacement x 𝑣 = 2 𝑡 2 2 − 𝑡 𝑡+𝐷 The integral of v is x, simplify the fraction 𝑥= 𝑡 2 − 𝑡 𝑡+𝐷 At time t = 0, the particle is a point A  We can take A as the starting displacement – ie) 0 𝑥= 𝑡 2 − 𝑡 𝑡+𝐷 Sub in t = 0, x = 0 0= (0) 2 − (0)+𝐷 Calculate D 𝐷=0 So we do not need to include a constant in our formula for x 𝑥= 𝑡 2 − 𝑡 𝑡 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑥= 𝑡 2 − 𝑡 𝑡 𝑎=2−2𝑡 𝑣=2𝑡− 𝑡 2 +8 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along a straight line. At time t = 0, the particle is at a point A and is moving with velocity 8ms-1 towards a point B on the line, where AB = 30m. At time t seconds, where t ≥ 0, the acceleration of P is (2 – 2t)ms-2 in the direction AB. Find an expression, in terms of t, for the displacement of P from A at time t seconds Show that P does not reach B Find the value of t when P returns to A Find the total distance travelled by P in the interval between the two instances when it passes through A Show that P does not reach B Calculate the furthest distance P gets from A This will occur at the time where the velocity is instantaneously 0 𝑣=2𝑡− 𝑡 2 +8 Set v = 0 and start solving for t 0=2𝑡− 𝑡 2 +8 Rearrange to an easier quadratic form 𝑡 2 −2𝑡−8=0 Factorise 𝑡−4 𝑡+2 =0 2 values for t 𝑥= 𝑡 2 − 𝑡 𝑡 𝑡=4 𝑜𝑟 −2 Only the positive value is possible 𝑡=4 Now calculate the displacement at time t = 2 𝑥= 𝑡 2 − 𝑡 𝑡 As the maximum distance is 262/3m, and AB is 30m, P will never reach B Sub in t = 2 𝑥= (2) 2 − (2) Work out the displacement 𝑥= 𝑚 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑥= 𝑡 2 − 𝑡 𝑡 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along a straight line. At time t = 0, the particle is at a point A and is moving with velocity 8ms-1 towards a point B on the line, where AB = 30m. At time t seconds, where t ≥ 0, the acceleration of P is (2 – 2t)ms-2 in the direction AB. Find an expression, in terms of t, for the displacement of P from A at time t seconds Show that P does not reach B Find the value of t when P returns to A Find the total distance travelled by P in the interval between the two instances when it passes through A 𝑎=2−2𝑡 𝑣=2𝑡− 𝑡 2 +8 Find the value of t when P returns to A  This means the displacement from A will be 0 𝑥= 𝑡 2 − 𝑡 𝑡 Sub in x = 0 Multiply by 3 to remove the fraction 0= 3𝑡 2 − 𝑡 3 +24𝑡 Rearrange to have t3 as positive 0= 𝑡 3 −3 𝑡 2 −24𝑡 Factorise  Either inside the bracket or outside it must be 0 0=𝑡( 𝑡 2 −3𝑡−24) 𝑡=0 𝑡 2 −3𝑡−24=0 𝑎=1 𝑏=−3 𝑐=−24 𝑥= 𝑡 2 − 𝑡 𝑡 This is because the particle starts at A Use the Quadratic formula to solve this part 𝑡= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Max displacement = 262/3 Sub in values 𝑡= 3± (−3) 2 −(4×1×−24) 2(1) Calculate 𝑡=6.62 𝑜𝑟 𝑡=−3.62 So the particle returns to A 6.62 seconds after leaving it! 1B

Displacement (x) Velocity (v) Acceleration (a) 𝑥= 𝑡 2 − 𝑡 𝑡 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P is moving along a straight line. At time t = 0, the particle is at a point A and is moving with velocity 8ms-1 towards a point B on the line, where AB = 30m. At time t seconds, where t ≥ 0, the acceleration of P is (2 – 2t)ms-2 in the direction AB. Find an expression, in terms of t, for the displacement of P from A at time t seconds Show that P does not reach B Find the value of t when P returns to A Find the total distance travelled by P in the interval between the two instances when it passes through A 𝑎=2−2𝑡 𝑣=2𝑡− 𝑡 2 +8 The particle starts at A, moves 262/3m away and then comes back The total distance travelled in this time is therefore just given by: 2 x 262/3 = 531/3m 𝑥= 𝑡 2 − 𝑡 𝑡 Max displacement = 262/3 6.62 seconds 1B

Displacement (x) Velocity (v) Acceleration (a) You can use calculus for a particle moving in a straight line, with acceleration that varies with time A small metal ball, moving in a magnetic field, is modelled as a particle P of mass 0.2kg, moving in a straight line under the action of a single variable force F Newtons. At time t seconds, the displacement, x metres, of the ball (B) from A is given by: Find the magnitude of F when t = π/6  You know that F = ma, so we need to find the acceleration of the Ball, in terms of t 𝑥=3𝑠𝑖𝑛2𝑡  Differentiating Sin gives Cos  Remember to multiply by 2 (The differential of 2t) 𝑣=6𝑐𝑜𝑠2𝑡 Differentiating Cos gives –Sin Remember to multiply by 2 (The differential of 2t) 𝑎=−12𝑠𝑖𝑛2𝑡 𝑎=−12𝑠𝑖𝑛2𝑡 Sub in t = π/6 𝑎=−12𝑠𝑖𝑛2 π 6 Calculate a 𝑎=−6 3 𝑚 𝑠 −2 𝑥=3𝑠𝑖𝑛2𝑡 𝐹=𝑚𝑎 Sub in m and a 𝐹=(0.2)(−6 3 ) Calculate in surd form 𝐹= 𝑁 Write as a decimal if you need to! 𝐹=2.08𝑁 1B

Displacement (x) Velocity (v) Acceleration (a) You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 This function shows that the relationships between x, v and a vary depending on time  You need to consider each time interval carefully 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 1B

Displacement (x) Velocity (v) Acceleration (a) The velocity will be at a minimum or maximum when the acceleration is 0. Meaning the particle has either been accelerating and the acceleration is about to act in the opposite direction (maximum) Or, the particle has been decelerating, and the acceleration is increasing, about to cause the particle’s velocity to increase again (minimum) Either way, the acceleration at these is instantaneously at 0 Differentiate the velocity in the required time interval, and set it equal to 0 You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 Write in a differentiatable form 𝑣=𝑡+ 4 𝑡 2 𝑣=𝑡+4 𝑡 −2 Differentiate 𝑎=1+ −2 4 𝑡 −3 Rewrite 𝑎=1+ −8 𝑡 3 Set a = 0 0=1+ −8 𝑡 3 Rearrange 𝑡 3 =8 Calculate t 𝑡=2 So the velocity will be at a maximum at t = 2 (sometimes you will need to check this!) 1B

Displacement (x) Velocity (v) Acceleration (a) Sub t = 2 into the relevant equation for velocity You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 𝑣=𝑡+ 4 𝑡 2 Sub in t = 2 𝑣=2+ 4 (2) 2 Calculate 𝑣=3𝑚 𝑠 −1 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 As this was the only value it must be the answer with more than 1 answer you could sub both in and check which gives the biggest/smallest velocity 𝑡=2 𝑣=3 𝑚𝑠 −1 1B

Displacement (x) Velocity (v) Acceleration (a) Sketching a velocity-time graph  Consider each ‘section’ separately You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 v (ms-1) 5 4 3 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 2 1 1 2 3 4 5 6 t (seconds)  For the first second, the velocity increases with time in a linear way, reaching 5ms-1 after 1 second 𝑡=2 𝑣=3 𝑚𝑠 −1  After t = 3, the velocity is constant at 34/9ms-1 In the middle section, the velocity varies more, but we know from section a) that the velocity is at a minimum of 3ms-1 at t = 2 We can therefore draw a minimum through this point 1B

Displacement (x) Velocity (v) Acceleration (a) Calculating the distance at time t = 6  You will have to work out where the particle is after each section in order to find the overall distance You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 𝑣=5𝑡 Integrate the expression to get the displacement, remember to include C 𝑥= 5𝑡 𝐶 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 We are told at t = 0, P is at the origin (ie, x = 0) (0)= 5(0) 𝐶 Calculate C 𝐶=0 𝑥= 5𝑡 2 2 Using our formula for displacement, sub in t = 1 to find where the particle ends this section 𝑡=2 𝑣=3 𝑚𝑠 −1 𝑥= 5(1) 2 2 Calculate 𝑥=2.5𝑚 So after 1 second, the particle is 2.5m away from O 1B

Displacement (x) Velocity (v) Acceleration (a) So after 1 second, the particle is 2.5m from O  Now calculate the distance travelled in the next section You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 𝑣=𝑡+ 4 𝑡 2 Rewrite in an integratable form 𝑣=𝑡+4 𝑡 −2 Integrate, including a constant 𝑥= 𝑡 𝑡 −1 −1 +𝐷 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 Rewrite for substitution 𝑥= 𝑡 2 2 − 4 𝑡 +𝐷 We know that after 1 second, the displacement is 2.5m  Sub these in to find D 2.5= (1) 2 2 − 4 (1) +𝐷 Calculate D 𝐷=6 𝑡=2 𝑣=3 𝑚𝑠 −1 𝑥= 𝑡 2 2 − 4 𝑡 +6 Now we know D, sub in t = 3 to find the displacement of P at the end of this section 𝑥= (3) 2 2 − 4 (3) +6 Calculate 𝑥=9 1 6 𝑚 1B

Displacement (x) Velocity (v) Acceleration (a) So after 3 seconds, the particle is 91/6m from O  Now calculate the distance travelled in the last section You can use calculus for a particle moving in a straight line, with acceleration that varies with time A particle P moves on the x-axis. At time t seconds, the velocity of P is vms-1 in the direction of x-increasing, where v is given by. When t = 0, P is at the origin O. Find the lowest speed of P in the interval 1 ≤ t ≤ 3 b) Sketch a velocity-time graph to illustrate the motion of P in the interval 0 ≤ t ≤ 6 c) Find the distance of P from O when t = 6 This is easier!  As the velocity is constant, we can just use Distance = Speed x Time to find how far the particle travels from 3 to 6 seconds 𝑣= 5𝑡, ≤𝑡<1 𝑡+ 4 𝑡 2 , 1≤𝑡≤ , 𝑡>3 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑆𝑝𝑒𝑒𝑑×𝑇𝑖𝑚𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=3 4 9 ×3 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒= 𝑚 𝑡=2 𝑣=3 𝑚𝑠 −1 So add this to the current displacement to get the final displacement… = 𝑚 1B

Teachings for Exercise 1C

You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) When a particle is moving in a plane you can describe its position (r), its velocity (v) and its acceleration (a) using vectors. The relationships between these in two dimensions are the same as they are in one dimension Sometimes dot notation is used. A single dot means differentiate once and two dots means differentiate twice. When you integrate using vectors, the constant of integration (C) will also be a vector Position Vector (r) Differentiate Integrate Velocity Vector (v) Differentiate Integrate Acceleration Vector (a) 𝒓=𝑥𝒊+𝑦𝒋 Dot notation used for a single differentiation 𝒗= 𝒓 = 𝑥 𝒊+ 𝑦 𝒋 Dot notation used for a double differentiation 𝒂= 𝒓 = 𝑥 𝒊+ 𝑦 𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane 𝒗=3𝑡𝒊 𝑡 2 𝒋 You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane. At time t seconds, its velocity, vms-1, is given by: When t = 0, the position vector of P with respect to a fixed origin O is (2i – 3j)m. Find: The position vector of P at time t seconds The acceleration of P when t = 3 Integrate each term, remember to include C 𝒓= 3 𝑡 2 2 𝒊 𝑡 3 𝒋+𝐶 At t = 0, the position vector r = (2i – 3j) (2𝒊−3𝒋)= 3 (0) 2 2 𝒊 (0) 3 𝒋+𝐶 Calculate C (you might need to group i and j terms and rearrange!) 𝑣=3𝑡𝒊 𝑡 2 𝒋 (2𝒊−3𝒋)=𝐶 𝒓= 3 𝑡 2 2 𝒊 𝑡 3 𝒋+2𝒊−3𝒋 Putting the i and j terms together 𝒓= 3 𝑡 2 2 𝒊+2𝒊 𝑡 3 𝒋−3𝒋 Factorising the i and j terms separately 𝒓= 3 𝑡 𝒊 𝑡 3 −3 𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane 𝒓= 3 𝑡 𝒊 𝑡 3 −3 𝒋 𝒗=3𝑡𝒊 𝑡 2 𝒋 You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane. At time t seconds, its velocity, vms-1, is given by: When t = 0, the position vector of P with respect to a fixed origin O is (2i – 3j)m. Find: The position vector of P at time t seconds The acceleration of P when t = 3 𝒗=3𝑡𝒊 𝑡 2 𝒋 Differentiate to find a 𝒂=3𝒊+𝑡𝒋 Sub in t = 3 𝑣=3𝑡𝒊 𝑡 2 𝒋 𝒂=3𝒊+3𝒋 At time t = 3, the acceleration of the particle is (3i + 3j)ms-2 𝒓= 3 𝑡 𝒊 𝑡 3 −3 𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane so that, at time t seconds, its acceleration is: At t = 3, the velocity of P is 6ims-1 and the position vector of P is (20i + 3j)m with respect to a fixed origin O. Find: The angle between the direction of motion of P, and i, when t = 2 The distance of P from O when t = 0 𝒂=4𝒊−2𝑡𝒋 Integrate to get v, include C 𝒗=4𝑡𝒊− 𝑡 2 𝒋+𝐶 At t = 3, we know the velocity is 6i 6𝒊=4(3)𝒊− (3) 2 𝒋+𝐶 Calculate terms 𝒂= 4𝒊−2𝑡𝒋 𝑚 𝑠 −2 6𝒊=12𝒊−9𝒋+𝐶 Rearrange to get the missing vector C −6𝒊+9𝒋=𝐶 𝒗=4𝑡𝒊− 𝑡 2 𝒋−6𝒊+9𝒋 Group i and j terms and factorise 𝒗= 4𝑡−6 𝒊+(− 𝑡 2 +9)𝒋 Sub in t = 2 to find the velocity vector 𝒗= 4(2)−6 𝒊+(− (2) 2 +9)𝒋 Calculate terms 𝒗=2𝒊+5𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane 𝒗=2𝒊+5𝒋 You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane so that, at time t seconds, its acceleration is: At t = 3, the velocity of P is 6ims-1 and the position vector of P is (20i + 3j)m with respect to a fixed origin O. Find: The angle between the direction of motion of P, and i, when t = 2 The distance of P from O when t = 0 2𝒊+5𝒋 5𝒋 θ 𝒂= 4𝒊−2𝑡𝒋 𝑚 𝑠 −2 2𝒊 𝑇𝑎𝑛𝜃= 𝑂𝑝𝑝 𝐴𝑑𝑗 Sub in opp and adj 𝑇𝑎𝑛𝜃= 5 2 Inverse Tan 𝜃=𝑇𝑎 𝑛 − This angle is the one we want  It is between the direction of motion and the direction I (horizontal) 𝜃=68.2° 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane so that, at time t seconds, its acceleration is: At t = 3, the velocity of P is 6ims-1 and the position vector of P is (20i + 3j)m with respect to a fixed origin O. Find: The angle between the direction of motion of P, and i, when t = 2 The distance of P from O when t = 0 𝒂=4𝒊−2𝑡𝒋 𝒗= 4𝑡−6 𝒊+(− 𝑡 2 +9)𝒋 𝒗= 4𝑡−6 𝒊+(− 𝑡 2 +9)𝒋 Integrate to get r, include D 𝒓= 2 𝑡 2 −6𝑡 𝒊+ − 𝑡 𝑡 𝒋+𝐷 Sub in values we are told 𝒂= 4𝒊−2𝑡𝒋 𝑚 𝑠 −2 (20𝒊+3𝒋)= 2 (3) 2 −6(3) 𝒊+ − (3) 𝒋+𝐷 Calculate terms (20𝒊+3𝒋)=18𝒋+𝐷 Rearrange to find D (20𝒊−15𝒋)=𝐷 θ = 68.2° 𝒓= 2 𝑡 2 −6𝑡 𝒊+ − 𝑡 𝑡 𝒋+20𝒊−15𝒋 Group i and j terms and factorise 𝒓= 2 𝑡 2 −6𝑡+20 𝒊+ − 𝑡 𝑡−15 𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane so that, at time t seconds, its acceleration is: At t = 3, the velocity of P is 6ims-1 and the position vector of P is (20i + 3j)m with respect to a fixed origin O. Find: The angle between the direction of motion of P, and i, when t = 2 The distance of P from O when t = 0 𝒂=4𝒊−2𝑡𝒋 𝒗= 4𝑡−6 𝒊+(− 𝑡 2 +9)𝒋 𝒓= 2 𝑡 2 −6𝑡+20 𝒊+ − 𝑡 𝑡−15 𝒋 𝒓= 2 𝑡 2 −6𝑡+20 𝒊+ − 𝑡 𝑡−15 𝒋 𝒂= 4𝒊−2𝑡𝒋 𝑚 𝑠 −2 Sub in t = 0 𝒓= 2 (0) 2 −6(0)+20 𝒊+ − (0)−15 𝒋 Calculate r 𝒓=20𝒊−15𝒋 θ = 68.2° The initial position vector r, is just the value of the constant of integration! 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P is moving in a plane so that, at time t seconds, its acceleration is: At t = 3, the velocity of P is 6ims-1 and the position vector of P is (20i + 3j)m with respect to a fixed origin O. Find: The angle between the direction of motion of P, and i, when t = 2 The distance of P from O when t = 0 𝒂=4𝒊−2𝑡𝒋 𝒗= 4𝑡−6 𝒊+(− 𝑡 2 +9)𝒋 𝒓= 2 𝑡 2 −6𝑡+20 𝒊+ − 𝑡 𝑡−15 𝒋 𝒓=20𝒊−15𝒋 𝒂= 4𝒊−2𝑡𝒋 𝑚 𝑠 −2 20𝒊 −15𝒋 20𝒊−15𝒋 θ = 68.2° Use Pythagoras’ Theorem to calculate the actual distance – the vector just represents the displacement (20) 2 + (−15) 2 25𝑚 =25𝑚 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane y You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P of mass 0.5kg is moving under the action of a single force F Newtons. At time t seconds, the position vector of P, r metres, is given by: Find: The value of t when P is moving parallel to the vector i The magnitude of F when t = 3.5 j x O i P i 𝒓= 3 𝑡 2 2 − 𝑡 𝒊+ 2 𝑡 2 −8𝑡 𝒋 When the particle is travelling parallel to the vector i, its velocity has no j component  Find the velocity vector and set the j component to 0 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P of mass 0.5kg is moving under the action of a single force F Newtons. At time t seconds, the position vector of P, r metres, is given by: Find: The value of t when P is moving parallel to the vector i The magnitude of F when t = 3.5 𝒓= 3 𝑡 2 2 − 𝑡 𝒊+ 2 𝑡 2 −8𝑡 𝒋 Differentiate to get the velocity vector 𝒗= 3𝑡− 𝑡 2 𝒊+ 4𝑡−8 𝒋 We want the j term of the velocity to be equal to 0 4𝑡−8=0 Solve for t 𝒓= 3 𝑡 2 2 − 𝑡 𝒊+ 2 𝑡 2 −8𝑡 𝒋 𝑡=2 𝑡=2 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane 𝒓= 3 𝑡 2 2 − 𝑡 𝒊+ 2 𝑡 2 −8𝑡 𝒋 𝒗= 3𝑡− 𝑡 2 𝒊+ 4𝑡−8 𝒋 You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) A particle P of mass 0.5kg is moving under the action of a single force F Newtons. At time t seconds, the position vector of P, r metres, is given by: Find: The value of t when P is moving parallel to the vector i The magnitude of F when t = 3.5 We can use the formula F = ma to find the force, but first we will need to find the acceleration at the appropriate time! 𝒗= 3𝑡− 𝑡 2 𝒊+ 4𝑡−8 𝒋 Differentiate to get the acceleration vector 𝒂= 3−2𝑡 𝒊+4𝒋 Sub in t = 3.5 𝒂= 3−2(3.5) 𝒊+4𝒋 𝒓= 3 𝑡 2 2 − 𝑡 𝒊+ 2 𝑡 2 −8𝑡 𝒋 Calculate a 𝒂=−4𝒊+4𝒋 𝑭=𝑚𝒂 Sub in m and a 𝑭=(0.5)(−4𝒊+4𝒋) Calculate F in terms of i and j 𝑭=−2𝒊+2𝒋 Use Pythagoras’ to find the magnitude 𝑡=2 |𝑭|= (−2) 2 + (2) 2 Calculate F |𝑭|=2 2 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) The velocity of a particle at time t seconds is given by: When t = 0, the position vector of P with respect to a fixed origin is (2i – 4j)m Find the position vector of P after t seconds 𝒗= 3 𝑡 2 −8 𝒊+5𝒋 Integrate to find r, include C 𝒓= 𝑡 3 −8𝑡 𝒊+5𝑡𝒋+𝐶 Sub in the values given 2𝒊−4𝒋= (0) 3 −8(0) 𝒊+5(0)𝒋+𝐶 Find C 𝒗= 3 𝑡 2 −8 𝒊+5𝒋 2𝒊−4𝒋=𝐶 𝒓= 𝑡 3 −8𝑡 𝒊+5𝑡𝒋+𝐶 Sub in the value of C we found 𝒓= 𝑡 3 −8𝑡+2 𝒊+(5𝑡−4)𝒋 𝒓= 𝑡 3 −8𝑡 𝒊+5𝑡𝒋+2𝒊−4𝒋 Group i and j terms and factorise 𝒓= 𝑡 3 −8𝑡+2 𝒊+(5𝑡−4)𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane If the particles collide, there must be a time for which their i and j components are both the same Find the time for which either the i’s or j’s are equal, and check if it works for the other component Start by finding an equation for the position vector of Q in terms of t You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) The velocity of a particle at time t seconds is given by: When t = 0, the position vector of P with respect to a fixed origin is (2i – 4j)m Find the position vector of P after t seconds A second particle Q moves with constant velocity (8i + 4j)ms-1. When t = 0, the position vector of Q with respect to the origin O is 2i m. b) Prove that P and Q collide 𝒗=8𝒊+4𝒋 𝒗= 3 𝑡 2 −8 𝒊+5𝒋 Integrate to find r, include D 𝒓=8𝑡𝒊+4𝑡𝒋+𝐷 Sub in the values given 2𝒊=8(0)𝒊+4(0)𝒋+𝐷 Find D 2𝒊=𝐷 𝒓= 𝑡 3 −8𝑡+2 𝒊+(5𝑡−4)𝒋 𝒓=8𝑡𝒊+4𝑡𝒋+𝐷 Sub in the value of D we have 𝒓=8𝑡𝒊+4𝑡𝒋+2𝒊 Group i and j terms and factorise 𝒓=(8𝑡+2)𝒊+4𝑡𝒋 𝒓=(8𝑡+2)𝒊+4𝑡𝒋 1C

Position Vector (r) Velocity Vector (v) Acceleration Vector (a) Kinematics of a particle moving in a straight line or plane For Q For P You can use calculus with vectors for a particle moving in a plane (up to now was just along a straight line!) The velocity of a particle at time t seconds is given by: When t = 0, the position vector of P with respect to a fixed origin is (2i – 4j)m Find the position vector of P after t seconds A second particle Q moves with constant velocity (8i + 4j)ms-1. When t = 0, the position vector of Q with respect to the origin O is 2i m. b) Prove that P and Q collide 𝒓=(8𝑡+2)𝒊+4𝑡𝒋 𝒓= 𝑡 3 −8𝑡+2 𝒊+(5𝑡−4)𝒋 Find the value of t for which the j terms are equal (this is easier than comparing the i terms!) 4𝑡=5𝑡−4 Solve for t 𝑡=4 𝒗= 3 𝑡 2 −8 𝒊+5𝒋 So when t = 4 seconds, the particles have the same j component (16) This does not mean they collide, just that they are in the same position vertically See if they are also in the same horizontal position by subbing t = 4 into the i components 𝒓= 𝑡 3 −8𝑡+2 𝒊+(5𝑡−4)𝒋 For Q For P 8𝑡+2 𝑡 3 −8𝑡+2 8(4)+2 (4) 3 − 8(4)+2 =34 =34 𝒓=(8𝑡+2)𝒊+4𝑡𝒋 So the particles collide after 4 seconds at position (34i + 16j) 1C

Summary We have learnt how to solve problems involving projectiles
We have used calculus to solve problems involving displacements, velocities and accelerations that vary with time We have also seen how vectors are used to model movement in a 2D plane!

Kinematics of a particle moving in a straight line or plane

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Kinematics of a particle moving in a straight line or plane

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