1. Linear Motion. Table of Contents. 1.1 Linear Motion: Velocity Time Graphs (Multi Stage) 1.2 Linear Motion: Velocity Time Graphs (Up and Down)

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1 . LINEAR MOTION

2 Linear Motion Table of Contents. Linear Motion: Velocity Time Graphs (Multi Stage). Linear Motion: Velocity Time Graphs (Up and Down).3 Linear Motion: Common Initial Velocity.4 Linear Motion: Using.5 Linear Motion: Bodies Moing in Same Direction.6 Linear Motion: Bodies Moing Towards Each Other.7 Linear Motion: Vertical Motion Particle.8 Linear Motion: Vertical Motion Particles

graph into triangles and rectangles and sum the areas of each to find the total distance traelled.

3 elocity (m s ) If a graph is drawn of elocity(y-axis) against time (x-axis) then: Distance Traelled = Area Under Graph It s often quicker to use this than to use the equation s = ut + at Break the area under the graph into triangles and rectangles and sum the areas of each to find the total distance traelled. The slope of the line for a gien stage corresponds to the acceleration of the object during that stage. a = tan = This can also be quicker than using formula. t When the particle is accelerating the graph is diagonal upwards. When the particle is at constant speed the graph goes straight across. When the particle is decelerating the particle is diagonal downwards. If asked for the aerage speed of the journey use: Speed = Distance Time sum of areas under graph Speed = t + t + t Note that the area between the graph and the time axis represents the displacement s of the body. u Don t assume that we always start from rest. Acceleration is the slope of the line. a = tan = rise run a a t Deceleration β t t 3 a = tan β = rise run To sole complex problems we need to be comfortable expressing some ariables in terms of the others so that we can reduce the amount of ariables we hae to work with. (see 03 (b)). Show that = f(t t ) How would you sole the problem if the letters were numbers and proceed in the same way. time (s) If f is the acceleration then t t t t t Reision Checklist 06 (a) 03 (b) 0 (b) 007 (b) 999 (b) 998 (a) 997 (a) 996 (b) 99 (a) 990 (b) 986 (a) 979 (a) f = t t = f t t

www.mathspoints.ie A car has an initial speed of u m s. It moes in a straight line with constant acceleration f for 4 seconds. It traels 40 m while accelerating.

(ii) Find the alue of u. (iii) Find the total distance traelled. (i) (iii) Speed Speed-Time Graph Calculate f using information during the acceleration section.

Area of Rectangle + Area of Triangle = 40 4u + 4 u = 40 4u + 5 u = 40 4u + 30 u = 40 u = 0 u = 5 m/s Area of Rectangle 3 = 45 = 5 m/s (ii) Can be quicker to find

4 A car has an initial speed of u m s. It moes in a straight line with constant acceleration f for 4 seconds. It traels 40 m while accelerating. The car then moes with uniform speed and traels 45 m in 3 seconds. It is then brought to rest by a constant retardation f. (i) Draw a speed-time graph for the motion. (ii) Find the alue of u. (iii) Find the total distance traelled. (i) (iii) Speed Speed-Time Graph Calculate f using information during the acceleration section. Time u = 5 = 5 a = f t = 4 = u + at 5 = 5 + f 4 0 = 4f f =.5 m s Speed = Distance Time = 45 3 = 5 m/s or Traels 40 while accelerating. Area of Rectangle + Area of Triangle = 40 4u + 4 u = 40 4u + 5 u = 40 4u + 30 u = 40 u = 0 u = 5 m/s Area of Rectangle 3 = 45 = 5 m/s (ii) Can be quicker to find accelerations using: f = tan = 0 4 f =.5 m s Calculate t using information during the deceleration section. u = 5 = 0 a = f t = t = u + at 0 = 5 + f t 0 = t 0 = 5 5t t = 3 seconds Distance = Area Under Cure = = = 07.5 m

www.mathspoints.ie Some elocity time graphs hae no stage with constant elocity. The object accelerates and then immediately decelerates.

t : t = d: a The maximum acceleration of a body is 4 m/s and its maximum retardation is 8 m/s. What is the shortest time in which the body can trael a distance of 00 m from rest to rest?

5 Some elocity time graphs hae no stage with constant elocity. The object accelerates and then immediately decelerates. If the acceleration is equal to the deceleration then the time taken will be the same. If they are not equal then the ratio of acceleration to deceleration is opposite to the ratio of the times taken. t : t = d: a The maximum acceleration of a body is 4 m/s and its maximum retardation is 8 m/s. What is the shortest time in which the body can trael a distance of 00 m from rest to rest? Velocity Velocity Time Graph Alternate method to find ratio of times. tan = t 4 = t = 4t t = t tan β = t 8 = t = 8t Find expressions for in terms of T. Reision Checklist 009 (b) 006 (a) 00 (a) 994 (a) 987 (a) (b) To be sure of full marks generate this formula by getting the acceleration and deceleration in terms of and letting the s equal. See top right! tan = t = 4 = 4t = 4 3 T We can use this information to express each of the times t and t as a fraction of the total time T. β t t Time = 8 3 T We also try and express the elocity, in terms of the T. T Distance = Area Under Graph As with all topics in Applied Maths we need to be comfortable working with letters instead of numbers. The theory is the same but the finish can be trickier to spot (see 006 Q (a)). When labelling your graph include elocity, times for acceleration and deceleration t and t and a total time, T. Find expressions for t and t in terms of T. t : t = d: a t : t = 8: 4 = 3 : 3 t : t = d: a t = 3 T t = 3 T T = 00 T 8 T = T = 00 T = 900 T = 30 s Also include the angles and β.

6 A body starts from rest at p, traels in a straight line and then comes to rest at q which is km from p. The time taken is 66 seconds. For the first 0 seconds if has uniform acceleration a. It then traels at constant speed and is finally brought to rest by a uniform deceleration a acting for 6 seconds. (i) Calculate a and a. (ii) If the journey from rest at p to rest at q had been traelled with no interal of constant speed, but subject to a for time t followed by a for time t, show that time for the journey is 8 9 seconds. (i) elocity (ii) Find expressions for t and t in terms of T. t : t = d: a t : t = :. = 5: 3 t = 5 8 T Just in case generate these ratios. elocity First calculate. β time t = 3 8 T Find expressions for in terms of T. tan = =. t =.t =. 5 8 T t T t β time Distance = Area Under Graph = = 696 = 6 = 696 Calculate the accelerations using tan or formulae. Note: tan method quicker. = 3 4 T Distance = Area Under Graph 696 = T 696 = T 3 4 T tan = t =. =.t tan β = t = a = tan = 0 a =. m/s d = tan β = 6 a = m/s 696 = 3 8 T 856 = T T = 8 9 seconds = t.t = t t = t.

7 In this type of question the acceleration remains constant for the entire journey. We will be gien information about arious stages of a journey and need to sole for unknown ariables, usually a and u. The key is that the ariables for both stages must represent the same number. For example if we use u as the initial elocity from a to b it cannot also be used from b to c. The elocity has changed. To oercome this we use u as the initial elocity and find distance equations for the sections a to b and then a to c. Keep measuring eerything from a. A particle starts from rest and moes in a straight line with uniform acceleration. It passes three points a, b and c where ab = 05 m and bc = 63 m. If it takes 6 seconds to trael from a to b and seconds to trael from b to c find its acceleration. Use distance formula with A to B s = ut + at 05 = u 6 + a 6 05 = 6u + 8a 6u + 8a = 05 A 6 s B s C 05 m 63 m Use distance formula with A to C s = ut + at 68 = u 8 + a 8 68 = 8u + 3a 8u + 3a = 68 Then sole the simultaneous equation for u and a. Reision Checklist 05 (a) 00 (b) 003 (a) 00 (b) 996 (a) 995 (a) 993 (a) 988 (a) 986 (b) Another common question is when you are told that a particle traels a certain distance in a certain second. A particle moing in a straight line with uniform acceleration describes 3 m in the fifth second of its motion and 3 m in the seenth second. Calculate its initial elocity. S 5 S 4 = 3 S 7 S 6 = 3 The distance a particle traels in the 5 th second is equal to the distance that the particle traels in 5 seconds minus the distance the particle traelled in 4 seconds. u 5 + a 5 u 4 + a 4 = 3 5u +.5a 4u 8a = 3 u + 4.5a = 3 u 7 + a 7 u 6 + a 6 = 3 7u + 4.5a 6u 8a = 3 u + 6.5a = 3 Distance Traelled in 5 th Second = s 5 s 4. We use this to set up simultaneous equations so that we can sole for u and a. Then sole the simultaneous equation for u and a. a = 4 m/s u = 5 m/s

www.mathspoints.ie The points p, q and r all lie in a straight line. A train passes point p with speed u m/s.

The train takes 0 seconds to trael from p to q and 5 seconds to trael from q to r, where pq = qr = 5 metres.

(i) p 0 s q 5 s r The distance from p to q is 5 m. u = u a = f s = 5 t = 0 s = ut + at 5 = u 0 + 5 = 0u 50f f 0 The distance from p to r is 50 m.

5f Note: f is the retardation so f is the acceleration. Sole the simultaneous equation 5u 3.5f = 50 0u 50f = 5 5u 3.5f = 50 5u + 5f = 3.5 87.5 = 75.

8 The points p, q and r all lie in a straight line. A train passes point p with speed u m/s. The train is traelling with uniform retardation f m/s. The train takes 0 seconds to trael from p to q and 5 seconds to trael from q to r, where pq = qr = 5 metres. (i) Show that f = 3 (ii) The train comes to rest s metres after passing r. Find s, giing your answer correct to the nearest metre. (i) p 0 s q 5 s r The distance from p to q is 5 m. u = u a = f s = 5 t = 0 s = ut + at 5 = u = 0u 50f f 0 The distance from p to r is 50 m. Note we find pr rather than qr so we can use same initial elocity u. u = u a = f s = t = m s = ut + at 5 m 50 = u 5 + f 5 50 = 5u 3.5f Note: f is the retardation so f is the acceleration. Sole the simultaneous equation 5u 3.5f = 50 0u 50f = 5 5u 3.5f = 50 5u + 5f = = 75.5 f = 3 m/s (ii) Sole for u 0u 50f = 5 0u 50 3 = 5 0u 50 3 = 5 0u = 45 3 u = 85 6 m/s.5 The train will hae stopped when the final elocity is 0. u = 85 6 = 0 a = 3 s = s = u + as 0 = = s 75 s = 3 36 s = 75 4 = 30 4 m This is the distance traelled before the train had stopped. The question asks for how many metres after? Subtract 30 from the distance traelled 4 pr, 50m = = 5 4 m s

Pulling Force W = Weight R = Resistance Another useful formula to remember is Power P = T Power is measured in W so if gien kw make sure to conert before using the formula.

9 In this type of question we use the formula: F = Net Force m = mass of the object a = acceleration of the object F = T R (Object being pulled) or W R (Falling due to graity) where T = Tractie Pulling Force W = Weight R = Resistance Another useful formula to remember is Power P = T Power is measured in W so if gien kw make sure to conert before using the formula. Questions that hae been asked: Car Towing Trailer Up Hill 04, 004, 999 Mass Penetrating Soft Material 005, 98 Bullet Penetrating Block of Wood 970 For objects traelling up a hill we must resole the Forces Parallel and Perpendicular to the incline. A car of mass 500 kg moes up a hill. The force of the engine is 3000 N. There is air resistance of 00 N. Calculate the acceleration. Resole the downward force into components parallel and perpendicular to the incline g sin = 500a g 3 = 500a a = 38 5 m/s 500g = sin 3 A particle of mass 3 grammes falls from rest from a height of 0.4 m on to a soft material into which it sinks m. Neglecting air resistance, calculate the constant resistance of the material. Calculate the speed of the object at the moment in hits the soft material. = u + as = = 7.84 =.8 m/s Calculate the acceleration of the object in the soft material gien that it will stop when = 0. = u + as 0 =.8 + a = a a = 60 m/s Calculate the resistance of the material. W R = (9.8) R = R = 0.48 R = N Reision Checklist 04 (b) 005 (b) 004 (b) 999 (a) 994 (b) 98 (b) 970

A car of mass 00 kg tows a caraan of mass 900 kg first along a horizontal road with acceleration f and then up an incline to the horizontal road at uniform speed.

(ii) Calculate the alue of, to the nearest degree. T (i) First calculate the acceleration along horizontal road. 700 50 40 = 00a a =. m s Note: Uniform Speed means no acceleration!

10 A car of mass 00 kg tows a caraan of mass 900 kg first along a horizontal road with acceleration f and then up an incline to the horizontal road at uniform speed. The force exerted by the engine is 700 N. Friction and air resistance amount to 50 N on the car and 40 N on the caraan. (i) Calculate the acceleration, f, of the car along the horizontal road. (ii) Calculate the alue of, to the nearest degree. T (i) First calculate the acceleration along horizontal road = 00a a =. m s Note: Uniform Speed means no acceleration! OR ( taking the Car and Caraan separately) Up the incline there is uniform speed, therefore, acceleration is 0. (ii) Car T Caraan T 40 Total T + T 40 = 00a a =. m s Same Result. Car T 00g sin = 00 0 T 00g sin = 550 Caraan T 900g sin 40 = T 900g sin = 40 Sole the simultaneous equation. 900g 00g T 00g sin = 550 T 900g sin = 40 30g sin = 550 sin = g =

UNDERSTAND MOTION IN ONE AND TWO DIMENSIONS

SUBAREA I. COMPETENCY 1.0 UNDERSTAND MOTION IN ONE AND TWO DIMENSIONS MECHANICS Skill 1.1 Calculating displacement, aerage elocity, instantaneous elocity, and acceleration in a gien frame of reference