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Nuffield Free-Standing Mathematics Activity Vectors © Nuffield Foundation 2011 Vectors • Is the water skier moving in the same direction as the rope? • What forces are acting on the water skier? • Which directions are the forces acting in? Vectors Scalar quantities have magnitude but no direction Examples mass distance speed temperature Vectors have magnitude and direction Examples displacement force momentum velocity acceleration Unit vectors Suppose the velocity of a yacht has an easterly component of 12 ms–1 and a northerly component of 5 ms–1 The velocity is v ms–1 where v = 12i + 5j i represents a unit vector to the east and j represents a unit vector to the north Column vector notation v= 12 5 Magnitude and direction of a vector v1 v= v2 v v2 v1 Magnitude Direction v= v12 v22 v2 tan = v1 –1 v2 = tan v1 Think about How can you use the triangle to find the magnitude and direction of v? N Example v= Speed Direction 12 5 bearing Think about v How can you find the 5 speed and direction of the yacht? 12 v = 122 52 = 13 tan = 5 = 0.416 12 = 22.6 The yacht is sailing at 13 ms–1 on bearing 067 (nearest ) To add or subtract vectors Add or subtract the components Example Forces acting on an object (in newtons) 7 5 - 4 3 where i is a horizontal unit vector to the right and j is a vertical unit vector upwards Think about how to find the total force Total force acting on the object 7 - 4 3 5 3 8 To multiply a vector by a scalar Multiply each component by the scalar Example 1 Displacement s = - 2 (in metres) s 3s Think about What do you get if you multiply both components of the vector by 3? 3s = 3 - 6 Multiplying by 3 gives a displacement 3 times as big in the same direction Constant acceleration equations Equation 1 v u at where u = initial velocity Equation 2 Equation 3 Momentum s ut 1at 2 2 s 1(u v) t 2 mv is a vector v = final velocity a = acceleration t = time taken s = displacement Forces and acceleration F1 Resultant force is the sum of the forces acting on a body, in this case F1 + F2 + F3 F3 Newton’s First Law A particle will remain at rest or continue to move uniformly in a straight line unless acted upon by a non-zero resultant force. Newton’s Second Law Resultant force causes acceleration F = ma Newton’s Third Law Action and reaction are equal and opposite. This means if a body A exerts a force on a body B, then B exerts an equal and opposite force on A. F2 Swimmer Find the magnitude and direction of the swimmer’s resultant velocity. v= S Resultant velocity Speed 2.4 0.5 2.9 1.5 - 2.1 - 0.6 vR = 2.92 0.62 = 2.96 Direction v= C i = unit vector to the east j = unit vector to the north ms–1 tan = 0.6 = 0.2068 … 2.9 = 11.7 2.4 1.5 0.5 - 2.1 (ms–1) N bearing 2.9 vR= vR The swimmer will travel at 2.96 ms–1 on bearing 102 (nearest ) 0.6 25 u= 16 Golf ball Find a the velocity at time t b the velocity when t = 2 c the ball’s displacement from O, when t = 2 a c 0 a= - 9.8 O i = horizontal unit vector j = vertical unit vector 25 25 0 v u at t 16 - 9.8t 16 - 9.8 b When t = 2 (ms–1) 25 25 –1 v (ms ) - 3.6 16 - 9.82 25 0 2 25t s ut 1at 2 t 1 t 2 2 16 2 - 9.8 16t - 4.9t When t = 2 252 50 s (m) 2 162 - 4.92 12.4 (m) - 8 u= 6 Skier a Find the skier’s acceleration. b Find the speed and direction of the skier 20 seconds later. 60 kg i = unit vector to the east j = unit vector to the north a) F = ma b) 24 = 60a -15 0.4 a= (ms–2) - 0.25 v u at - 8 0.4 v 20 6 - 0.25 0 v (ms–1) 1 The skier is travelling at 1 ms–1 to the north. 24 F= -15 Ship Ship travels at a constant velocity u ms–1 a What is the force, F, from the tug? b 300 Ship’s initial position vector r 500 - 5600 2.5 R= u = 1200 -1 i Find the position vector of the boat at time t. x ii The ship is aiming for a buoy which has position vector 100 Assuming the ship reaches the buoy, find x. Ship a Ship travels at a constant velocity u ms–1 This means there is no acceleration - 5600 2.5 R= 1200 u = -1 5600 F= -1200 1 2.5 t 2.5 0 2 1 b i Displacement s ut at 2 t t 2 2 1 0 -t 300 Ship’s initial position vector r 500 At time t, b ii 300 2.5 t 300 2.5t r 500 - t 500 - t When ship reaches 500 – t = 100 x 100 300 500 = 1300 2.5t - t r O t = 400 x = 300 + 2.5t = 300 + 2.5 400 x 100 Reflect on your work • How have you used the fact that i and j are perpendicular unit vectors? • Are there any similarities between the problems or the techniques you have used? • Can you think of other scenarios which could be tackled using vectors in component form?