MEI Mechanics 1 - Woodhouse College
MEI Mechanics 1MotionSection 1: The language of motionNotes and ExamplesThese notes contain subsections on Definitions Notes on the textbookDefinitionsDisplacement is the shortest route between two points. A distance anddirection are needed. It is a vector quantity.Position describes the location of something relative to a fixed point. Thisfixed point is usually referred to as the origin. Like displacement, a distanceand direction are needed. It is a vector quantity.The distance between two points involves no direction. It is just the physicaldistance between the chosen points. It is a scalar quantity.The distance travelled between two points does not have to be the same asthe distance between two points, described above. If you take a route which isnot direct, them the distance travelled will be greater than the direct distancebetween the two points. Distance travelled is a scalar quantity.To highlight the differences in the concepts of displacement, position, distanceand distance travelled, look at Example 1 below.Example 1O5mA5mBAn object starts from A and travels to the right to B, then back through A to O andthen back through A to stop at B.Write down(i)The final displacement of the object(ii)The final position of the object(iii) The distance between its starting and finishing points(iv)The total distance travelled by the objectSolution(i)The final displacement from its initial position at A is 5 m to the right.(ii)Its final position is 10 m right of O.(iii) The distance between its starting and finishing points (A and B) is 5 m. MEI, 01/06/091/2
MEI M1 Motion Section 1 Notes and Examples(iv)The total distance travelled is 25 m: 5 m from A to B, 10 m from B to O andanother 10 m from O to B.Notes on the textbookFigure 1.4, page 3It is a common mistake to confuse the position - time graph with the path ofthe marble through the air. It is not. The path of the marble is straight up anddown. The position - time graph shows how the marble’s position, relative tothe origin (your hand, in this case) changes with time.Question mark, page 3The negative position means that the particle is below the level of your hand.The level of your hand has been taken as the origin in this example. MEI, 01/06/092/2
MEI Mechanics 1MotionSection 2: Speed and velocityNotes and ExamplesThese notes contain subsections on Definitions Notes on the textbookDefinitionsSpeed is a scalar quantity; it just has a particular size, e.g. 5 ms-1. Nodirection is given or implied. Speeds are always positive.Velocity is a vector quantity. It must have a size and a direction, e.g. 6 ms-1upwards.Velocities can be either positive or negative.Average speed total distance travelledtotal time takenAverage velocity total displacementtotal time takenFrom these formulae it is possible to see why average velocity can benegative if the resultant displacement is negative, whereas the total distancetravelled will always be positive.You can also find velocities as gradients of distance- time graphs (see p8).Example 1Mark gets off a bus, walks to the shop to buy some milk, and then walks home. Theposition-time graph below shows Mark’s journey.position (m)400300200100123456 MEI, 01/06/09789time (mins)1/4
MEI M1 Motion Section 2 Notes and hat is Mark’s final displacement?What is the total distance Mark has walked?In which part of the journey was Mark walking fastest?How far is the bus stop from Mark’s home?How far is the shop from the bus stop?What is Mark’s average speed for the whole journey?What is Mark’s average velocity for the whole journey?What is Mark’s average speed when he is actually walking?Draw a speed-time graph for Mark’s journey.Draw a velocity-time graph for Mark’s journey.Solution(i)Mark’s final displacement is -250 m.The distance between hisoriginal position and finalposition. Note that it isnegative because thedirection of the shop has beenchosen to be positive.(ii)The total distance walked is 150 400 550 m.(iii)Mark was walking fastest on the way home from the shop.(iv)The bus stop is 250 m from Mark’s house.(v)The shop is 150 m from the bus stop.(vi)Total distance travelled 550 m.Total time taken 9.5 60 570 seconds.550Average speed 0.965 ms-1 (3 s.f.)570(vii)Total displacement -250Total time taken 9.5 60 570 seconds.250 0.439 ms-1 (3 s.f.)Average velocity 570The last part of thegraph is steeper thanthe first part.(viii) Total distance travelled 550 m.Total time taken when walking 7.5 60 450 seconds.550Average speed 1.222 ms-1 (3 s.f.)450 MEI, 01/06/092/4
MEI M1 Motion Section 2 Notes and Examples150 1 ms-1.2.5 60400Speed during journey from shop to home 1.33 ms-1 (3 s.f.)5 60Speed during journey to shop (ix)speed (ms-1)211(x)23456789time (mins)-1Velocity during journey to shop 1 ms .Velocity during journey from shop to home -1.33 ms-1 (3 s.f.)velocity (ms-1)210123456789time (mins)-1-2Notice in Example 1 that if you want to know how fast Mark walks, then theaverage speed and velocity calculations are not very helpful as they includethe time he spent at the shop. The calculation of average speed when he waswalking in part (vii) gives you the best idea of Mark’s walking speed.Notes on the textbookQuestion mark, page 7It is impossible to start and stop or jump to any new velocity instantaneously.More realistic graphs would have no sharp corners. However, Amy will beable to change velocity very quickly on her bicycle, so it would not make muchdifference to the answers if we tried to be more realistic. This is an exampleof mathematical modelling. It is easier to deal with the situationmathematically if we assume Amy can change velocity instantaneously and itwill make little difference to our answers, so it is a reasonable modellingassumption to make. MEI, 01/06/093/4
MEI M1 Motion Section 2 Notes and ExamplesQuestion mark, page 8Velocity at H 5 ms-1Velocity at A 0 ms-1Velocity at B -5 ms-1Velocity at C -6 ms-1After the marble reaches the top, its velocity is negative so as the magnitudeof the velocity (i.e. the speed) increases, the velocity decreases (remember 2 1 ). MEI, 01/06/094/4
MEI Mechanics 1MotionSection 3: AccelerationNotes and ExamplesThese notes contain subsections on Acceleration Notes on the textbookAccelerationAcceleration is a measure of how much velocity is changing. This means itcan affect both the speed and direction of motion. In this section you onlylook at motion along a straight line, so only two directions are possible, eitherforwards or backwards.An acceleration of 2 ms-2 means that the velocity of a particle increases by2 ms-1 every second (by 2 metres per second per second).For example, if a car has an initial velocity of 6 ms-1 and an acceleration of2 ms-2, then after 1 second its velocity will be 8 ms-1, after 2 seconds 10 ms-1and after 3 seconds 12 ms–1 etc.If a particle has a negative acceleration but a positive velocity, then it will slowdown to a stop and then move in the opposite direction, with its speed steadilyincreasing.Take care with the word deceleration. It is probably better not to use it. Usenegative accelerations instead!Take care that the units of acceleration are ms-2. This is usually read as‘metres per second squared’, or sometimes as ‘metres per second persecond’.Accelerations can be found using the gradients of velocity – time graphs.Example 1When a local train leaves a station, it takes accelerates at a uniform rate of 3 ms-2 toits maximum speed of 60 ms-1. It then maintains this speed for 2 minutes beforeslowing down uniformly to a halt at the next station. The whole journey takes 3minutes.(i)Sketch a distance-time graph for the journey.(ii)Find the time the train takes to reach its maximum speed.(iii) Draw the velocity-time graph for the journey.(iv)What is the acceleration of the train in the last part of the journey?(v)Draw the acceleration-time graph for the journey. MEI, 01/06/091/3
MEI M1 Motion Section 3 Notes and ExamplesSolution(i)distanceHere the train isslowing downHere the train istravelling atconstant speedHere the train isacceleratingtime(ii)Acceleration change in velocitytime60 0t60t 203The train takes 20 seconds to reach its maximum speed.3 (iii)velocity (ms-1)601(iv)23time (mins)The last part of the journey takes 40 seconds, and the velocity changes from60 ms-1 to 0.change in velocityAcceleration time0 60 40 1.5The acceleration is -1.5 ms-2. MEI, 01/06/092/3
MEI M1 Motion Section 3 Notes and Examples(v)Acceleration (ms-2)420123Time (mins)-2Notes on the textbookQuestion mark, page 11(i)Distance travelled can never decrease (although you can decreasedisplacement). Graph D is the only graph that never decreases. Sothe distance-time graph can only be D.(ii)The velocity increases from zero to a maximum as the caraccelerates away from the first set of traffic lights, then decreasesfrom this maximum to zero as the car slows down to a stop for thesecond set of traffic lights. The velocity-time graph could be B, C orE as all these graphs start at zero, reach a maximum and thendecrease to zero again.(iii)As the car pulls away from the first set of traffic lights, its velocity isincreasing so its acceleration must be positive. Once it starts toslow down for the second set of traffic lights its velocity isdecreasing so its acceleration must be negative. Graph A is theonly graph that moves from positive to negative in this way. So theacceleration-time graph must be A. MEI, 01/06/093/3
MEI Mechanics 1MotionSection 4: Using areas to find distances anddisplacementsNotes and ExamplesThese notes contain subsections on Distances and displacements from velocity-time graphs Notes on the textbookDistances and displacements from velocity-time graphsThe area under a velocity–time graph is usually equal to the distance aparticle travels in the given time period, as long as the line does not cross thetime axis (which is the same as saying the velocity is always positive).However, if the graph crosses the time axis (which is the same as saying thevelocity becomes negative) as in example 1.2 on page 14, the situationchanges so that the distance travelled is equal to the sum of the areasbetween the graph and the time axis, disregarding the negative sign, whereasthe displacement is equal to the sum of the areas, incorporating the negativesign.In example 1.2 on page 14, the distance travelled is 14 m, whereas thedisplacement is 10 m.If the velocity-time graph is made up of a series of curves, then the area canbe approximated by counting squares or by approximating the curve bystraight lines to produce trapezia (as for the trapezium rule in Core 2).Example 1The diagram shows the velocity-time graph for the journey of a particle moving in astraight line.velocity (ms-1)4202051015time (s)-2-4 MEI, 01/06/091/3
MEI M1 Motion Section 4 Notes and Examples(i)(ii)(iii)(iv)(v)(vi)(vii)What is the acceleration of the particle during the first part of the journey?How far does the particle travel in the first 12 seconds of its motion?Estimate the distance travelled in the final 5 seconds of the motion.What is the total distance travelled by the particle?What is the final displacement of the particle?Find the average speed of the particle.Find the average velocity of the particle.Solution(i)During the first part of the journey the velocity of the particle increases from 0to 4 ms-1 in 5 seconds.change in velocityAcceleration time4 0 5 0.8Acceleration 0.8 ms-2.(ii)Distance travelled area under graph between t 0 and t 12.Area under graph between t 0 and t 5 is 12 5 4 10Area under graph between t 5 and t 8 is 3 4 12Area under graph between t 8 and t 12 is 12 4 4 8Total area 10 12 8 30Distance travelled 30 m.(iii)Splitting up the shape as shown in the diagram:Area of trapezium A 12 3 2 3 7.5Area of triangle B 12 2 2 2Total area 9.5Distance travelled 9.5 m.2015BA(iv)Distance travelled between t 12 and t 15 is 12 3 3 4.5Total distance travelled 30 4.5 9.5 44 m.(v)Final displacement 30 – (4.5 9.5) 16 m.(vi)Average speed (vi)Average velocity total distancetime44 2.2 ms -120total displacementtime16 0.8 ms -120 MEI, 01/06/092/3
MEI M1 Motion Section 4 Notes and ExamplesNotes on the textbookQuestion mark (near the top of page 15)The distance travelled by the dog is given by the area under its speed-timegraph. This can be estimated by approximating the area under the graphusing trapezia.In example 1.3, note that: When acceleration is uniform (constant) and positive, the speed - timegraph for this section of the journey is a straight line with a positivegradient (equal to the acceleration). When the speed is constant (so that the acceleration is zero), thespeed – time graph for this section of the journey is a horizontalstraight line with gradient 0 (equal to the acceleration). When acceleration is uniform (constant) and negative, the speed – timegraph for this section of the journey is a straight line with a negativegradient (equal to the acceleration).Question mark (bottom of page 15)If the journey is to take 2 minutes to cover 800 m and the maximum speed ismaintained for 90 seconds, the trapezium ABCD must have area 800, thedistance AD must be 120 m and the distance BC must be 90 m. From theformula for the area of a trapezium, this means that the value of v mustalways be the same.The area of the trapezium is determined by AD, BC and v only. It is notaffected by the gradients of AB or CD, so it does not matter how long the traintakes to speed up or slow down. MEI, 01/06/093/3
MEI Mechanics 1Modelling using constant accelerationSection 1: The constant acceleration formulaeNotes and ExamplesThese notes contain subsections on The constant acceleration formulae Applying the constant acceleration formulaeThe constant acceleration formulaeAll the equations used in this section are derived from the two properties ofthe velocity - time graph that you have used in the previous chapter: The gradient of the graph gives you the acceleration,v ua twhich can be rearranged to give the more usual formv u at The area under the graph, gives you the displacements 12 u v tIn these equationsu initial velocityv final velocitya accelerations displacementt timeFrom the above two equations, three others can be derived, as seen onpages 23, 24.These equations of motion or suvat equations are:v u ats 12 u v tv u 2 2as2s ut 12 at 2s vt 12 at 2These equations are very important. You should memorise them. MEI, 01/06/091/3
MEI M1 Constant acc. Section 1 Notes and ExamplesEach equation includes 4 of the 5 variables, and each variable is missing fromonly one of the five equations.Using the constant acceleration formulaeTo do the questions it is usually best to start by writing out a list of thevariables and filling in the ones that you know. In simple problems you will begiven details of three of the variables have to find a fourth. The easiest way tosolve such problems is to choose the correct suvat equation. This will be theequation that involves all of the variables that you are told in the question,together with the variable that you wish to calculate.Example 1A particle is accelerating at a constant 6ms-2. After 8 seconds its displacement is 5m.(i)What is its velocity after 8 seconds?(ii)What was its initial velocity?(iii) Describe the motion of the particle over these 8 seconds.SolutionUsing the information in the question you can write:s 5mu ?v ?a 6 ms-2t 8s(i)You know s, a and t and wish to know v. The suvat equation involving s, a, tand v is: s vt 12 at 2 , so substituting in the appropriate values gives15 8v 6 82 5 8v 1922 197 8v197 v8 v 24.6ms-1 (3s.f.) Note: As a general rule, round youranswers to 3 significant figures andstate clearly that you have done so.Avoid rounding until your final answer,to avoid rounding errors. MEI, 28/09/092/3
MEI M1 Constant acc. Section 1 Notes and Examples(ii)You know s, a and t and wish to know u. The suvat equation involving s, a, tand u is: s ut 12 at 2 , so substituting in the appropriate values gives5 8u 12 6 82 5 8u 192 187 8u 187 u8 u 23.4 ms -1 (3s.f.)(iii)Initially the particle has a speed of 23.4 ms-1 (3s.f.) and is moving in thenegative direction. Acceleration is constant at 6 ms-1 . During the next 8seconds the particle slows down until it is momentarily stationary and thenspeeds up to reach a speed of 24.6 ms-1 in the positive direction. At this timeits displacement is 5 m in the positive direction from its starting point.Important Note: Remember that the suvat equations can only be applied insituations where the acceleration is constant.To practice using the constant acceleration formulae, you can use theinteractive questions Choosing suvat equations and Solving suvatequations. MEI, 28/09/093/3
MEI Mechanics 1Modelling using constant accelerationSection 2: Further examplesNotes and ExamplesThese notes contain subsections on Worked examplesWorked examplesQuestions in this section usually involve a journey of two or more stages.Write down the values of u, v, a, s and t for each stage.Remember that the final velocity of one stage is the initial velocity of the nextstage and that the accelerations across the two stages are often the same(though NOT ALWAYS)Example 1A cyclist reaches the top of a hill moving at 2 ms-1, and accelerates uniformly so thatduring the sixth second after reaching the top he travels 13 m. Find his speed at theend of the sixth second.SolutionThe cyclist’s displacement has increased by 13 m during the sixth second.In the first 5 secondsu 2 ms-1v ?a a ms-2s xmt 5sIn the first 6 secondsu 2 ms-1v ?a a ms-2s (x 13) mt 613 m more thanafter 5 secondsUsing the equation which relates u, a, s and t, s ut 12 at 2For t 5 :For t 6 :x 2 5 12.5ax 13 2 6 18aSolving these equations simultaneously gives a 2 and x 35.v u at 2 2 6 14The speed at the end of the sixth second is 14 ms-1.To find the speed after 6 seconds: MEI, 28/09/091/3
MEI M1 Constant acc. Section 2 Notes and ExamplesIn Example 2, two methods of solution are given. Make sure you understandboth.Example 2A train travels from rest at station A to stop at station B, a distance of 2100 m. For thefirst 20 seconds it accelerates steadily, reaching a speed of 25 ms-1. It maintains thisspeed until the brakes are applied and the train brought to rest with uniformretardation over the last 125 m.Find the retardation and the total time for the journey between the two stations.Solution 1 (Using equations only)Consider the three stages of the journey separately:Stage 1 (accelerating)s xu 0v 25a at 20Stage 2 (constant speed)s yu 25v 25a 0t t1Stage 3 (decelerating)s 125u 25v 0a bt t2Also, from the total distance travelled we know that x y 125 2100.Stage 1:To find a use v u at:25 0 20aa 1.25To find x use s 12 u v t , giving x 12 0 25 20 x 250Stage 2:250 y 125 2100 y 1725 mUsing s ut 12 at 2 1725 25t1 0 t1 1725 6925Stage 3: 625 2.5250250 10and to find t2 use s 12 u v t 125 12 25 0 t2 t2 25To find b, use v 2 u 2 2as 0 625 2 125b b (Use this equation rather than v u at as ituses all given information rather than thequantities you have calculated – this is saferas you may have made errors in yourcalculations whereas given information mustbe assumed to be correct – this is a goodgeneral rule to follow for such questions) MEI, 28/09/092/3
MEI M1 Constant acc. Section 2 Notes and ExamplesThe retardation is 2.5 ms-2.The total time for the journey is 20 69 10 99 secondsSolution 2 (Using a velocity time graph)Begin by sketching a graph and marking on it what you know:Velocity ( ms-1 )25ABC
Jan 06, 2009 · MEI M1 Motion Section 1 Notes and Examples MEI, 01/06/09 2/2 (iv) The total distance travelled is 25 m: 5 m fr
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