1. Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 11) Core Mathematics for Mechanics M1 Mechanics M1

2. Copyright © 2015 Chris J Jewell 2 Mechanics M1 Introduction This section covers very specifically those areas of Core Mathematics which are most useful for the M1 course.. It is not intended to supersede or supplant any teaching of the Core syllabus. A basic competence in mathematics at GCSE level is assumed. The topics begin with short sections on Number, Algebra followed by Geometry and Trigonometry and finally with Calculus, including, Differentiation and Integration of Polynomials.

3. Copyright © 2015 Chris J Jewell 3 Mechanics M1 Number Surds occur throughout M1, in static and dynamic questions involving inclines and vector components; some of their basic properties are illustrated below : Multiplication and Division Addition and Subtraction Continued on next slide

4. Copyright © 2015 Chris J Jewell 4 Mechanics M1 Number Surds continued : Rationalising the denominator:

5. Copyright © 2015 Chris J Jewell 5 Mechanics M1 Algebra Solving Quadratic Equations (i) Factorising: Continued on next slide These are generally solved by one of three methods. Such equations occur commonly in kinematics and vectors in M1. (5 goes into 30 and 5) (+3t + -2t = +t, -2 x +3 = - 6 )

6. Copyright © 2015 Chris J Jewell 6 Mechanics M1 Algebra Solving Quadratic Equations (ii) Formula: Continued on next slide (Substitute the values) (Separate the t values)

7. Copyright © 2015 Chris J Jewell 7 Mechanics M1 Algebra Completing the square: Continued on next slide Solving Quadratic Equations (iii) (Halve the 6 in 6x, form (x - 3) and square, subtract the resulting 9)

8. Copyright © 2015 Chris J Jewell 8 Mechanics M1 Algebra The method can also be used to find minimum values, for example a minimum distance value in M1 vector collision questions. (Half the 6 in 6x, form (x - 3) and square, subtract the resulting 9) d Example with two boats

9. Copyright © 2015 Chris J Jewell 9 Mechanics M1 Algebra Simultaneous Equations: Simultaneous equations can be formed to solve for speed, acceleration or tension A B C 32 m s -1 16 m s -1 v c m s -1 16 m s -1 v c m s -1 32 m s -1 (Eliminate “a”)

10. Copyright © 2015 Chris J Jewell 10 Trigonometry Interlude 1 1 45º “CAST” ALL Sin Tan Cos Positive Angles Mechanics M1 2 2 1 1 30º 60º 30º

11. Copyright © 2015 Chris J Jewell 11 Mechanics M1 Geometry and Trigonometry Cosine and Sine Rule - Definitions Cosine Rule : Sine Rule: c a b C A B

12. Copyright © 2015 Chris J Jewell 12 Mechanics M1 Geometry and Trigonometry Cosine and Sine Rule - Applications Example: A boat moves 8 km from A to B on a bearing of 028° then 12 km from B to C on a bearing of 067°. Find the resultant R, the bearing of R and the bearing of A from C Cosine Rule: Sine Rule: E N A C B R = 12 km 8 km θ°θ° 028° 067° 113° 28° Angles given: 028°and 067° To find R use the cosine rule To find bearings use the sine rule for θ° first: Continued on next slide 18.9 km 15.5°

13. Copyright © 2015 Chris J Jewell 13 Mechanics M1 Geometry and Trigonometry Cosine and Sine Rule E N A C B 18.9 km 12 km 8 km 15.5° 028° 067° 141° 68° Required bearings are the angles shown by red curved lines: α and β α β

14. Copyright © 2015 Chris J Jewell 14 Mechanics M1 Calculus Calculus is used in the analysis of problems involving variable acceleration, but it is not included in all syllabuses, at least one only considers constant acceleration at AS level. The gradient of the hypotenuse in the triangle (Fig. 1) represents the gradient of the curve at point P v t 0 t1t1 v1v1 v = t 2 P Other triangles could be used to find acceleration at different times – however, calculus allows us to find the gradient analytically, using the formula: Differentiation (Fig. 1)

15. Copyright © 2015 Chris J Jewell 15 Mechanics M1 Calculus

16. Copyright © 2015 Chris J Jewell 16 Mechanics M1 Calculus The result gives us the gradient of the curve (rate of change) as a function of time. We simply substitute the time we are interested in to obtain the rate of change. Differentiation gives the gradient of the curve and can be used to find locations of “stationary points”: maxima, minima and inflection points (IP) – where gradient is exactly zero. See below: t s Max Min Max IP Example: If the displacement of a particle as a function of time is given by s = t 3, find the velocity at time t = 2 seconds and acceleration at time t = 1 second.

17. Copyright © 2015 Chris J Jewell 17 Mechanics M1 Calculus The type of stationary point (max, min or inflection point) can be determined using differentiation, for example: Stationary points – literally where the particle would be stationary, at least for a short time interval, exist wherever:

18. Copyright © 2015 Chris J Jewell 18 Mechanics M1 Calculus The stationary point at t = 1 changes from positive for t 1, so it is a maxima. The nature of the stationary points of each location can examined in turn. At t = 1: + - t = 1 t > 1 t < 1

19. Copyright © 2015 Chris J Jewell 19 Mechanics M1 Calculus At the point where t = 2: The stationary point at t = 2 changes from negative for t 1, so it is a minima + - t = 2 t > 2 t < 2

20. Copyright © 2015 Chris J Jewell 20 Mechanics M1 Calculus 2 nd Derivative – If the displacement function is differentiated twice the acceleration is obtained – this is the rate of change of the rate of change of displacement with time. The 2 nd derivatives have important diagnostic properties. If a stationary points exist on a curve then the 2 nd derivative will indicate one or all of the following: When the 2 nd derivative is 0 it is necessary to test values either side of the location to see if a maximum or minimum occurs

21. Copyright © 2015 Chris J Jewell 21 Mechanics M1 Calculus 2 nd derivative example: Continued on next slide (Divide by 3)1 st Derivative 2 nd Derivative Find the nature of the stationary points on the curve s = t 3 - 4.5t 2 + 6t at points t = 1 and t = 2 seconds.

22. Copyright © 2015 Chris J Jewell 22 M1 Mechanics Calculus 2 nd derivative example: 0 1 2 2 t (s) S (m) 4 S = t 3 – 4.5t 2 + 6t max min Maxima and minima determined by differentiation

23. Copyright © 2015 Chris J Jewell 23 Mechanics M1 Calculus The significance of the of the maxima, minima and inflection point in M1: (a) When the displacement curve is a maximum or minimum the velocity is zero S max S min or v = 0 + + - - (b) When the velocity curve is a maximum or minimum the acceleration is zero S max a = 0 or S min a = 0 + + - - (c) Inflection points on the displacement curve correspond to zero acceleration a = 0 + + - -

24. Copyright © 2015 Chris J Jewell 24 Mechanics M1 Calculus Integration – Indefinite Integral Integration is the opposite of differentiation, with one caveat; when a function is integrated it does not produce a unique result, and so has to be modified by control values. See below: Then you might expect that integration of 3t 2 would give t 3, and indeed it would, but this is not unique, for example, t 3 + 5 would also give 3t 2 when differentiated, since any constants would be zero. In fact, the same result would occur whatever the constant. When integration is used a constant is added and evaluated from control values, these are usually provided in the question: Continued on next slide

25. Copyright © 2015 Chris J Jewell 25 Mechanics M1 Calculus Integration – Indefinite Integral Continued. The general formula for integration is given by: Continued on next slide

26. Copyright © 2015 Chris J Jewell 26 Mechanics M1 Calculus Integration – Indefinite Integral Continued.

27. Copyright © 2015 Chris J Jewell 27 Mechanics M1 Calculus Integration – The Definite Integral - can be used to find distance travelled by a particle in a straight line. This technique is equivalent to finding the area under a velocity-time graph and is defined below: Note: that for positive and negative parts the total area is calculated from the individual parts which are then added together without regard for sign. t1t1 t 3 t (s) v (ms-1) v = f (t) t2t2

28. Copyright © 2015 Chris J Jewell 28 Mechanics M1 Calculus Application Example: The maximum velocity occurs when a = 0: A car starts from rest and moves for 5 seconds with an acceleration a ms -2 at time t given by the formula a = 12 – 3t. Find the maximum velocity. Velocity Formula Continued on next slide

29. Copyright © 2015 Chris J Jewell 29 Mechanics M1 Calculus Continued on next slide Example: A force of (24 – t 2 ) N acts on a particle of mass 3 kg. At time t = 0 the particle has a velocity of 2 m s -1 in the direction of the force. Find the velocity of the particle after 2 seconds and the distance travelled since t = 0. Substitute for t = 4 in the velocity formula gives Maximum velocity is 24 ms -1 Divide by 3

30. Copyright © 2015 Chris J Jewell 30 Mechanics M1 Calculus The velocity after 2 seconds is given by: Continued on next slide (Given)

31. Copyright © 2015 Chris J Jewell 31 Mechanics M1 Calculus Example: Find the velocity of the particle after 2 seconds and the distance travelled since t = 0. To find the distance travelled in 2 seconds the velocity is integrated to find the distance travelled using the Definite Integral The distance travelled is then the difference between the value at t = 2 and at t = 0