Kinematics WORKSHEETS. KEEP IT SIMPLE SCIENCE Physics Module 1. Motion Graphs. KISS Resources for NSW Syllabuses & Australian Curriculum

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1 Worksheet 1 Fill in the blank spaces. KEEP IT IMPLE CIENCE Physics Module 1 Kinematics WORKHEET peed & Velocity The average speed of a moving object is equal to the a)... travelled, divided by... taken. On a Distance-Time graph, the c)... of the graph is equal to speed. A horizontal graph means d) tudent Name... The vector equivalent of distance is called j)..., and refers to distance in a particular k)... For example, if displacement was being measured in the north direction, then a distance southward would be considered as l)... displacement. On a peed-time graph, constant speed is shown by e)... on the graph. This does NOT mean stopped, unless the graph section is lined up with f)... peed and distance are both g)... quantities, because the direction doesn t matter. Often in Physics we deal with h)... quantities, which have both i)... and... On a displacement-time graph, movement south would result in the graph sloping m)... to the right and having a negative n)... The vector equivalent of speed is o)... The average velocity is equal to p)... divided by q)... Instantaneous velocity refers to r)... Worksheet 2 Motion Graphs Practice Problems A car travelled 200 km north in 0 hours, then stopped for 0 hr, and finally travelled south 100 km in 0 hr. a) What was the total distance travelled? tudent Name... g) Use your graph to find: i) average velocity for the first 3 hours. What was the total displacement? ii) velocity during the 4th hour. c) What was the total time for the whole journey? iii) velocity during the last hour. d) Calculate the average speed for the whole journey. e) Calculate the average velocity for the whole journey. f) Construct a Displacement - Time Graph for this trip. Displacement TIME h) Construct a Velocity- Time Graph for this trip. Velocity (km/hr) outh North Time (hr) page 1

2 An aircraft took off from town P and flew due north to town Q where it stopped to re-fuel. It then flew due south to town R. The trip is summarised by the graph. a) How far is it from towns P to Q? How long did the flight P to Q take? c) Calculate the average velocity for the flight from P to Q (include direction) Worksheet 3 Practice Problems Motion Graphs & Calculations Displacement north (km) P Q Time (hr) R tudent Name... UNIT OF MEAUREMENT o far all examples have used the familiar km/hr for speed or velocity. The correct.i. units are metres per second (ms -1 ). You need to be able to work in both, and convert from one to the other... here s how: 1 km/hr = 1,000 metres/hr = 1,000m/(60x60) seconds = 1,000/3,600 m/s = 1/6 o, to convert km/hr ms -1 divide by 6 to convert ms -1 km/hr multiply by 6 A car is travelling at 100 km/hr. a) What is this in ms -1? The driver has a micro-sleep for 00 s. How far will the car travel in this time? d) What is the value of the gradient of the graph from t=3 hr, to t=6 hr.? c) At this velocity, how long does it take (in seconds) to travel 00km (1,000m)? e) What part of the journey does this represent? f) Where is town R located compared to town P? g) What was the aircraft s position and velocity (including direction) at t=5 hr? h) What was the: i) total distance ii) average speed For this question consider north as (+), south as ( - ). A truck is travelling at a velocity of ms -1 as it passes a car travelling at -25 ms -1. a) What are these velocities in km/hr? (including directions?) What is the displacement (in m) of each vehicle in 30.0 s? iii) total displacement iv) average velocity (for the entire 6 hr journey) c) How long would it take each vehicle to travel 100 m? i) Construct a Velocity- Time Graph for the flight. Velocity (km/hr) outh North Time (hr) Where does this aircraft end up in relation to its starting point? Flight details: First flew west for 50 hr at 460 km/hr. Next, flew east at 105 ms -1 for 50.0 minutes. Next, flew west for 25 hours at 325 km/hr. Finally flew east for 50 hours at velocity 125 ms -1. page 2

3 Worksheet 4 Acceleration tarting from rest (i.e. u=0) a car reaches 25 ms -1 in 8.20 s. What is the rate of acceleration? Practice Problems tudent Name... a) Find the rate of acceleration of the racer in the first 5 sec. A truck decelerated at -60 ms -2. It came to a stop (v = 0) in 80 s. How fast was it going when the brakes were applied? What was the maximum velocity it achieved in km/hr? c) What was its total displacement (in metres) for the entire motion? A car was travelling at 10ms -1. How long would it take for it to reach 25ms -1, if it accelerated at 75ms -2? d) When was the car stationary? e) Describe the car s motion after t=8.0s. A spacecraft was travelling in space at 850ms -1 when its retro-rockets began to fire, producing a constant deceleration of 50.0ms -2 (i.e. acceleration of -50.0ms -2 ) The engines fire for 20.0s. What is the spacecraft s final velocity at the end of this time? Interpret the meaning of the mathematical answer. f) Find the rate of acceleration at time t=10s. g) ketch a graph of displacement-time for this motion. Values on the graph axes are NOT required. The graph shows the motion of a drag race car. Velocity (ms -1 ) Time (s) page 3

4 Worksheet 5 Equations of Motion Practice Problems 1 tudent Name... Note: It it essential in Kinematics problems to be careful about the directions involved and use +ve & -ve signs appropriately. To ensure consistency, we recommend always using +ve for upwards motion, -ve for downwards. On level ground, use +ve for north, -ve for south. (Or +ve for east, -ve for west, as applicable.) Use Equation 1 a) A car travelling at an initial velocity of 15ms -1 accelerated at 0ms -2 for 7.0s. What was its final velocity? How long would it take to slow down to 40ms -1 from an initial velocity of 200ms -1, if accelerating at -5ms -2? What is the acceleration rate if a rocket sled reaches 500ms -1 in 8.50s, starting from rest? (u = 0) Use Equation 3 7. Entering the Earth s atmosphere with an initial velocity of 22,000ms -1, a meteor decelerated at a constant rate for 17.0s over a distance of 350km. (Convert to metres!) What was the rate of deceleration? 8. An object accelerated at a constant 75ms -2 for 9.00s. In this time the displacement was 325m. What was its initial velocity? A bullet striking a soft target decelerates at -15,000ms -2 and stops (v = 0) within 0.05s. What was its initial velocity? Use Equation 4 9. From an initial velocity of 10ms -1, a train on a straight track accelerated at 50ms -2 over a displacement of 500m. What velocity did it reach? Use Equation 3 What distance would be covered in 15s by an object accelerating at 35ms -2, if its intial velocity was 25ms -1? 10. A rocket sled on rails accelerated from rest (u=0) at 100ms -2 and reached a final velocity of 500ms -1. What displacement was covered during the acceleration? 6. Accelerating at 75ms -2 from rest, (u = 0) a ball rolling down a smooth slope covered 66.0m. What time did this take? 1 A jet landing on an aircraft-carrier touched down with a velocity (u) of 30ms -1. Caught by the arresting gear, the jet was brought to a halt (v=0) in a distance of 45m. Find the (de-) acceleration rate. page 4

5 Worksheet 6 Equations of Motion Practice Problems 2 tudent Name... Note: It it essential in Kinematics problems to be careful about the directions involved and use +ve & -ve signs appropriately. To ensure consistency, we recommend always using +ve for upwards motion, -ve for downwards. On level ground, use +ve for north, -ve for south. (Or +ve for east, -ve for west, as applicable.) (Don t do this at home, kids!) Fred dropped a rock from the top of a very high building & used a stopwatch to measure that it took 70s to hit the ground. He knows that the acceleration due to gravity, g = -9.81ms -2. a) Assuming that he held the rock still before letting it go, what was the rock s initial velocity? A car is moving at a constant velocity of 100kmhr -1 on a level road when the driver sees a small child chase a ball onto the road exactly 70m ahead. It takes the driver 20s to react, then she applies emergency braking & safely stops the car 20m from the child. a) What was the car s initial velocity in ms -1? At what velocity did it hit the ground? How far did the car travel during the driver s reaction time? c) How high is the building? c) What distance was covered during braking? d) From the same height, Fred then threw another rock vertically downwards with an initial velocity of -25ms -1. At what rate did it accelerate downwards? d) What was the acceleration rate during braking? e) At what velocity did it hit the ground? e) What time elapsed during braking? f) How long did it take to hit the ground? An aircraft taking off accelerated from rest (u=0) at 8.50ms -2 and became airborne after 15s. a) What length of runway was used during the takeoff? Dangerous Fred is back! Armed with a catapult, he fired a golf ball vertically upwards. It took 75s for the ball to reach the apex of its flight, at which point it stopped momentarily (v = 0) before falling again. Don t forget, g = -9.81ms -2. a) What was the launch velocity of the ball? (u =?) How high did it go? What was its take-off velocity? (ms -1 and kmhr -1 ) c) How long does it take the golf ball to fall vertically back down? (What do you notice about your answer?) page 5

6 Worksheet 7 Practice Exercises Vectors in 1-D tudent Name... Town X is 20km NE of town Y. Town Z is 70km W of Y. a) Represent this information by displacement vectors. Where is X in relation to Z? Billy has a job as a delivery boy. One day his deliveries result in consecutive movements as follows: 5km east, 2km east, 4km west, 5km east, 7km west, 9km west. a) What is his total distance moved? What is his final displacement from his starting point? A boat is moving upstream at a velocity through the water of 5ms -1. There is a downstream water current with a velocity of 2ms -1. a) What is the boat s velocity compared to the river bank? (ie what is its actual velocity upstream?) How far upstream will the boat travel if it maintains this motion for 1 hour? c) After an hour, the boat turns around and travels back to its starting point. The water current and the boat s velocity through the water are unchanged. What is its downstream velocity compared to the river bank? Planes P&Q are both flying east, one behind the other as shown. P Q d) How long will it take to return to its starting point? They cannot see each other, but P s radar has detected Q directly in front. Plane P is flying at a velocity of 600kmhr -1. Plane Q s velocity is 500kmhr -1. ( ground speeds ) a) What is the relative velocity of Q, as seen by P s radar? Two cars are approaching each other head-on on a long straight road. Car J s velocity is 60kmhr -1 east. Car M is travelling at 90kmhr -1 west. a) Represent this situation with a vector diagram. To avoid disaster, plane P climbs to a safe altitude, over-flies Q, then drops back down when safely past. Velocities have not changed. Q s forward-facing radar suddenly picks up P. What is is the relative velocity of P, as seen by Q s radar? What is the velocity of M, relative to J? c) What is the velocity of J, relative to M? d) When they first notice each other, the cars are 500m apart. What time elapses until they meet? (afely passing each other in the road lanes.) page 6

7 Worksheet 8 Vectors in 2-D For each problem, sketch a vector diagram & show working. On an orienteering bushwalk, Kali uses her compass & pedometer to walk 4km west, then 8km north. Where does she end up, compared to her starting point? (displacement and compass bearing) Practice Exercises tudent Name... Town B is 150km east, and 400km south, of town A, as shown in the diagram. What is the displacement of B from A, (including direction) as the crow flies? A 150km 400km B A plane is flying with an air-speed of 350kmhr -1 west. There is a cross-wind blowing at 45kmhr -1 south. Use a vector diagram & appropriate working to calculate the plane s ground-speed velocity. (Velocity relative to a fixed point on the ground, including a compass-bearing direction.) Two aircraft are approaching the X 250kmhr-1 same airport from different directions, as shown. Plane X is flying east at 250kmhr -1. Plane Z is flying north at 400kmhr -1. a) What is the relative velocity of Z, seen from X? Z 400kmhr -1 A swimmer sets out to cross a river. Using breaststroke, she swims due north at 0.75ms -1. The tide is flowing east at 2ms -1. What is her velocity (inc.direction) compared to her starting point? What is the relative velocity of plane X, as seen from Z? 6. Bob is building a garden trellis at the side of his house. He needs to cut a piece of timber so it will reach from a point on the ground 50m horizontally away from the wall, to touch the vertical wall 80m above the ground. a) ketch a diagram to represent this task as vector displacements. To what length should he cut a piece of timber? c) When it is in position, at what angle will the timber meet the wall? page 7

8 Worksheet 9 Practice Problems Resolving Vectors tudent Name... Each of the diagrams show a displacement or velocity vector on a map. Resolve each vector into perpendicular components and give a magnitude and compass direction for each component. a) vector A = 20km 20 o B = 225ms o W N E A racing yacht is trying to sail due south, but is forced by a cross-wind to sail on compass bearing 240 o with a velocity of 22kmhr -1. What is its southward velocity? (ie what is the southerly component of its velocity?) c) Billy is pushing a trolley up a 25 o inclined ramp. He has moved it 10m along the ramp. d) C = 60kmhr o D = 800m 70 o a) How far has it moved vertically upwards? 25 o e) 10 o E = 4,200km How far has it moved horizontally? A cannon fired a shell with an initial velocity of 950ms -1 at an angle of 60 o above the horizontal. Resolve this velocity vector into horizontal and vertical components. 6. enemy An army observer W E spots an enemy tank on compass Ob. bearing 300 o. Range-finding equipment measures that the enemy is 6,000m from the observation post. 6,000m N How far a) north Using accurate equipment, a surveyor measures that a marker post is 382m away on compass bearing 30 o from his position. a) How many metres north of his position is the marker? west of the observation post is the enemy tank? 7. An aircraft is flying NE (bearing 45 o ) at 750kmhr -1. Calculate & compare the northerly & easterly components of its velocity. How many metres east of his position is the marker? page 8

9 Worksheet 10 Graphical Method of Vector Analysis Tutorial Worksheet tudent Name... This method of vector analysis relies on accurate scale diagrams, best done on graph paper. A scale must be assigned to the grid. Angles should be measured by protractor. Note that this method lacks the numerical accuracy & precision of algebraic methods, but may be more convenient in some cases. The grids provided here are 1cm, but may have been changed by photocopying. Answers given are approximate. A ship sails due north for 60km, then turns east and sails 50km. Where is now, compared to its starting point? The graph shows the vector diagram you can use to solve this problem. Complete it by constructing the Resultant displacement vector. Measure its length & use the scale to find its size. Measure an angle (protractor) to specify a direction bearing. cale: 1cm = 10km W N E On the grid is a vector representing the initial velocity of an artillery shell fired upwards at an angle. Use the graph to find: a) the magnitude (size) of the initial velocity. (use ruler) the firing angle, from the horizontal. c) the vertical component of its initial velocity. d) the horizontal component of initial velocity. cale: 1cm = 100ms -1 An small aircraft flies due east with an airspeed of 120kmhr -1. However, there is a gale-force cross-wind blowing due south at 60kmhr -1. Find the plane s true groundspeed (including direction) by graphical means. cale: 1cm = 20 kmhr -1 W N E N ADDING MORE THAN 2 VECTOR. tart in the centre (dot) of the grid. Construct vectors in the order given, head to tail. What is the final displacement from start point? (inc.direction) a) walked 30m north. crawled 40m east c) hopped 50m south d) ran 80m west cale: 1cm = 10m W E page 9

10 Answer ection Worksheet 1 a) distance time c) gradient d) stationary, not moving e) horizontal line f) zero on the speed scale g) scalar h) vector i) magnitude & direction j) displacement k) direction l) negative m) down n) gradient o) velocity p) displacement q) time r) velocity at a particular instant of time Worksheet 2 a) = 300km (-100) = 100km north c) 5hr d) peed = dist/time = 300/5 = 60km/hr e) V = /t = 100/5 = 20km/hr f) graph g) from graph: i) gradient = 200/3 67 km/hr ii) gradient = zero iii) gradient = -100/1 = -100km/hr (i.e. 100km/hr south) h) graph Displacement North (km) Velocity (km/hr) outh North Worksheet 3 a) 600km 5hr c) V = /t = 600/5 = 400km/hr north d) gradient = -900/3 = -300 e) Flight from Q to R f) R is 300km south of P g) Position = over town P. Velocity = 300km/hr south h) i) distance = 1,500km ii) peed = 1,500/6 = 250km/hr iii) Final displacement = 300 km south iv) V = /t = 300/6 = 50km/hr south i) graph Time (hr) Time (hr) Worksheet 3 (cont.) (i) a) 100/6 27.8ms -1. V = /t, so = V.t = 27.8x00 139m. c) V = /t, so t = /V = 1,000/ s. a) 20.5ms -1 = 78km/hr (north) -25ms -1 = -88.2km/hr (south) = V.t = 20.5x30,0 = 615m north -25x30.0 = -735m ( 735m south) c) t = /V = 100/20.5 = 88s 100/25 = 08s 1st leg: = V.t = 460x50 = 1,150km west 2nd leg: = 105x(50x60) =315,000m =315km east 3rd leg: = 325x25 1,056km west 4th leg: Velocity (km/hr) outh North = 125x(50x60x60) = 2,475,000m = 2,475km east Let east be (+ve), west be ( -ve) Final displacement = -1, , ,475 = +575 km (east) of starting point. Worksheet 4 a = (v - u)/t = (25-0)/8.20 = 74ms -2 u = v - at = 0-( -60x80) = 15ms -1 a = (v - u)/t t = (v - u)/a = (25-10)/75 = 6.00s v = u + at = (-50.0)x20.0 = -150ms -1 Time (hr) The final negative velocity means it is moving backwards, compared to its original direction. page 10

11 Answer ection Worksheet 4 (cont.) a) in first 0 seconds, gradient = 70/0 = 14 acceleration = 14ms -2. reached 70ms -1 c) Displacement = area under graph (2 triangles + rectangle) Area = = 560m d) At t=0 and t=13s e) It was decelerating to a stop. f) Acceleration = gradient = -70/0 = -10ms -2. g) rough sketch Deceleration: curves down to horizontal Constant Velocity: straight line Acceleration: curves up from horizontal Note: Although slowing down, the vehicle continues to move away from the start, so the Displacement-Time graph never shows a negative gradient. Worksheet 5 v = u + at = x7 = 36ms -1 v=u+at, so t= (v-u)/a = (40-200) / -5 = 64s. a = (v-u) / t = (500-0)/8.50 = 58.8ms -2. v=u+at, so u = v - at = 0 - (-15,000)x0.05 = 750ms -1. =ut+0.5at 2 = 25x x35x15 2 = 889m 6. = ut + 0.5at = x 75 x t 2 t 2 = 66 / 75x0.5 t = 8.68s t Worksheet 5 (cont.) 7. =ut + 0.5at 2 so a = (s-ut) / 0.5t 2 = (350,000-(22,000x17.0)) / 0.5x17 2 = -24,000 / 145 a = -166ms =ut + 0.5at 2 so u = ( - 0.5at 2 ) / t = ( x75x9 2 ) / 9 = ( ) / 9 u = 27ms V 2 = U 2 + 2a = x5x500 V 2 = 1725 V = 45ms V 2 = U 2 + 2a 1 V 2 = U 2 + 2a Worksheet 6 a) u = 0 so = (V 2 - U 2 ) / 2a = ( ) / 2x100 = 250,000 / 200 = 1,250m. so a = (V 2 - U 2 ) / 2 = (0 -(-35 2 ) / 2x45 = / 85 a = -14ms -2. v = u + at = 0 + (-9.81)x70 = -59ms -1. (down) c) =ut+0.5at 2 = x(-9.81)x7 2 = -159m. (negative answer indicates a downward displacement. The appropriate answer is that the building is 159m high) d) ince it is accelerating due to gravity, a = g = -9.81ms -2. e) V 2 = U 2 + 2a = (-25 2 ) + 2x(-9.81)x(-159) = V 2 = V = - 64ms -1. (note: mathematically V is derived from a square root, so it can be + or - The appropriate interpretation is that it is negative because it is a downward vector.) f) v=u+at, so t= (v-u)/a = (-64 - (-25)) / = -39 / t = 66s. page 11

12 Answer ection Worksheet 6 (cont.) a) =ut+0.5at 2 = x15x6.5 2 = 264m v = u + at = x6.50 = 83ms -1 Conversion: 825 x 6 = 293kmhr -1. a) Conversion: 100 / 6 = 27.8ms -1. = ut = 27.8x2 = 33m. c) = 37.5m d) V 2 = U 2 + 2a e) v=u+at, so a = (V 2 - U 2 ) / 2 = (0 - (27.8) 2 ) / 2x37.5 = -778 / 75 a = -10.3ms -2. so t= (v-u)/a = (0-27.8) / = 70s. a) v = u + at so u = v - at = 0 - (-9.81)x75 = 56.4ms -1. =ut+0.5at 2 = 56.4x x(-9.81)x(75) 2 = = 161m c) (The downward movement begins with the ball stopped at the apex of its flight, so u=0. We know the height of fall is -161m (down). Acceleration is -9.81ms -2. We want to find time.) = ut + 0.5at 2 u=0, so = 0.5at 2 so t 2 = 2/a = 2x(-161) / = 305 t = 75s (Notice that the time to fall down under gravity is the same as the time to climb up against gravity. The ball will also achieve the same speed (downwards) as its launch speed upwards.) Worksheet 7 a) diag. X is 90km NE of Z. Z Y 70km 20km X Worksheet 7 (cont.) a) 12 km Total east = 6.2km. Total west = 9km Final displacement = 8km west. a) Rel.Vel = = -100 kmhr -1 (west) (Q appears to be moving west toward P) 100km -1 (east) a) 3ms -1 upstream W v = /t so = v.t = 3 x (60x60) = 4,600m c) velocity downstream, v = = 7ms -1. d) t = /v = 4,600 / 7 = 1,243s (about 21 minutes) a) N Rel. Vel of M = (-60) = -150kmhr -1. (J sees M moving west at 150kmhr -1 ) c) Rel.Vel of J = = 150kmhr -1. (east) d) approach velocity = 150kmhr -1 = 47ms -1. Time to cover 500m, t = /v = 500/47 = 12s Worksheet 8 E car J R 2 = = 9.0 R = 9 = 0 km 60 km/hr Tan = 8 / 4 = o N of W (this angle is 307 o clockwise from north, bearing = 307 o ) 90 km/hr 8km car M R 4km Kali is 0km from starting point, on bearing 307 o. page 12

13 Worksheet 8 (cont.) Ground speed is 353kmhr -1 on bearing 263 o. R 2 = = 124,525 R = 124,525 = 353 kmhr kmhr -1 Tan = 45 / 350 = o of W (this angle is 263 o clockwise from north, bearing = 263 o ) R 2 = = 0025 R = 0025 = 42 ms -1 Tan = 2 / 0.75 = 6 58 o E of N bearing = 58 o 0.75ms -1 Actual velocity is 42ms -1 on bearing 58 o. Answer ection 2ms kmhr -1 R R 6. a) diag. R 2 = = R = = 18 m He needs to cut the timber to be 18m long. c) Tan = 50 / 80 = o It will meet the wall at an angle of 28 o. Worksheet 9 a) A y = 20.sin20 = 6.8km north A x = 20.cos20 = 18.8km east B y = 22sin55 = 184ms -1 south 80m timber = R 50m vector A = 20km 20 o A x A y R 2 = = 182,500 A 150km B x = 22cos55 = 129ms -1 east c) Careful! Angle measured differently. R = 182,500 = 427 km Tan = 400 / 150 = o of E (this angle is 159 o clockwise from north, bearing = 159 o ) Town B is 427km from A, on bearing 159 o. R 2 = = 222,500 R = 222,500 = 472 kmhr -1 Tan = 250 / 400 = o W of N (this angle is 328 o clockwise from north, bearing = 328 o ) (-V x ) = 250kmhr -1 een from X, plane Z is approaching at 472kmhr -1 flying on bearing 328 o. (X sees Z approx E) R R 400km B V Z = 400kmhr -1 C y = 60.cos40 = 46kmhr -1 north C x = 60.sin40 = 39kmhr -1 west d) Check how angle is measured! D y = 800.cos70 = 273m south D x = 800.sin70 = 752m west e) E y = 4200.sin10 = 729km south E x = 4200.cos10 = 4,136km east V y = 950.sin60 = 823ms -1 V x = 950.cos60 = 475ms -1 D = 800m D x V = 950ms o V x 70 o D y V y page 13

14 Answer ection Worksheet 9 (cont.) a) P y = 38cos30 = 331m north P x = 38sin30 = 191m east Bearing 240 o is 60 o W of as shown. y = 2cos60 = 11 kmhr -1 south P y P x 30 o P = 382m N W E 60 o Worksheet 10 (cont.) cale: 1cm = 20 kmhr kmhr -1 Resultant 60 kmhr -1 a) T y = 10.sin25 = 2m up T x = 10.cos25 = 9.1m horizontally The resultant measures approx. 6.8cm. Applying the scale gives the plane s groundspeed as 136kmhr -1. Angle measures about 25 o. This gives a compass bearing of 115 o. 6. Bearing 300 o gives an angle as shown. a) E y = 6,000.sin30 = 3,000m north E x = 6,000.cos30 = 5,196m west 7. V y = 750.sin45 = 530ms -1 north E y E = 6,000m E x 30 o measured from graph: a) approx 9.3cm = 930ms o c) 600ms -1 d) 700ms -1 cale: 1cm = 10m V x = 750.cos45 = 530ms -1 east Comparison: they are equal. Learn this: whenever the angle is 45 o, the x & y components will be equal. a) c) Worksheet 10 Resultant d) Construct & measure this resultant vector. Measure the angle shown by protractor By measurement, the ship is approx. 78km from its starting point, on compass bearing 40 o. Resultant displacement vector = approx 5cm. Applying the scale = 45m from start point. Angle = approx 63 o. This gives a compass bearing of 243 o. page 14