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© Boardworks Ltd 2006 1 of 26 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the.

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Presentation on theme: "© Boardworks Ltd 2006 1 of 26 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the."— Presentation transcript:

1 © Boardworks Ltd of 26 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. © Boardworks Ltd of 26 AS-Level Maths: Mechanics 2 for Edexcel M2.2 Kinematics

2 © Boardworks Ltd of 26 Variable acceleration Motion in two or three dimensions Examination-style questions Contents © Boardworks Ltd of 26 Variable acceleration

3 © Boardworks Ltd of 26 Variable acceleration In M1 most situations involved constant acceleration. However, the acceleration of a particle is not always constant. In cases of variable acceleration the displacement, velocity and acceleration will be functions of time. Calculus must be used to change between displacement, velocity and acceleration. x = f ( t ) v = f ’( t ) a = f ’’( t ) where x is displacement, v is velocity and a is acceleration. Kinematics is the study of how objects move rather than of the forces that cause motion.

4 © Boardworks Ltd of 26 Variable acceleration To find a where acceleration is variable, differentiate v with respect to t. To find x, integrate v with respect to t. To find v, either differentiate x or integrate a with respect to t. When integrating, remember to include the constant, c. If we know the initial velocity or position of the particle we can use this information to find c.

5 © Boardworks Ltd of 26 Variable acceleration The difference between constant and variable acceleration may not be discernible to the eye. Which ball is travelling with variable acceleration?

6 © Boardworks Ltd of 26 Variable acceleration question 1 Question 1: The displacement of a particle moving in a straight line is given by x = t 3 – 3 t t. Find a) the velocity of the particle when t = 4. a) To find the velocity we must differentiate x : Therefore the velocity of the particle when t = 4 is 28 ms -1. When t = 4: v = 3 × 4 2 – 6 × = 28 v = 3 t 2 – 6 t + 4 v = b) the acceleration of the particle when t = 2.

7 © Boardworks Ltd of 26 Variable acceleration question 1 Therefore the acceleration of the particle when t = 2 is 6 ms -2. b) To find the acceleration we must first differentiate v : a = 6 t – 6 When t = 2: a = 6 × 2 – 6 = 6 a = v = 3 t 2 – 6 t + 4

8 © Boardworks Ltd of 26 Variable acceleration question 2 Question 2: The velocity of a particle is given by v = t 2 – 5 t + 6. Initially its displacement from O is 2 m. Find a) the acceleration of the particle after 4 seconds. a) Differentiate v : Therefore the acceleration of the particle when t = 4 is 3 ms -2. When t = 4: a = 2 × 4 – 5 a = 2 t – 5 a = a = 3 b) the displacement of the particle from O when t = 6.

9 © Boardworks Ltd of 26 Variable acceleration question 2 b) To find the displacement we need to integrate v : The initial displacement from O is 2 m, i.e. when t = 0, x = 2: When t = 6: x = 72 – Therefore when t = 6 the displacement of the particle from O is 20 m.  c = 2 = 20 2 = 0 – c

10 © Boardworks Ltd of 26 Variable acceleration question 3 Question 3: The velocity of a particle travelling in a straight line is given by v = –3 t t + 4. Find: a)Here v is a quadratic where the coefficient for t 2 is negative. Therefore the maximum point on the velocity curve occurs where the gradient is 0. a)the time when the particle is travelling at its maximum velocity. b)the value of this maximum velocity.  t = 2 Differentiate to find the gradient function: = –6 t + 12 –6 t + 12 = 0

11 © Boardworks Ltd of 26 Variable acceleration question 3 Therefore the maximum velocity of the particle occurs when t = 2 seconds and the maximum velocity is 16 ms -1. b) When t = 2: v = 16 Question 3: The velocity of a particle travelling in a straight line is given by v = –3 t t + 4. Find: a)the time when the particle is travelling at its maximum velocity. b)the value of this maximum velocity. v = –3 × × 2 + 4

12 © Boardworks Ltd of 26 Variable acceleration Motion in two or three dimensions Examination-style questions Contents © Boardworks Ltd of 26 Motion in two or three dimensions

13 © Boardworks Ltd of 26 Motion in two or three dimensions The motion discussed so far has taken place in one dimension. However, motion can also occur in two or three dimensions. In these cases displacement, velocity and acceleration are represented by the unit vectors i, j and k. r = x i + y j + z k v = x `i + y `j + z `k a = x ``i + y ``j + z ``k where x, y and z are all functions of time. When integrating v or a as unit vectors, we need to remember the constants c, d and e, one for each dimension. i.e. c i + d j + e k.

14 © Boardworks Ltd of 26 3-D motion

15 © Boardworks Ltd of 26 Question 1: The position vector of a particle at time t is given by Vector question 1 a) Differentiate to find v: b) When t = 4: r = (2 t – 4)i + ( t 2 – 3 t + 1)j Differentiate to find a: v = 2i + (2 t – 3)j a = 2j a = 2j ms -2 v = 2i + 5j ms -1 v =a = a)Find the velocity and acceleration of the particle at time t. b)Find the velocity and acceleration of the particle when t = 4.

16 © Boardworks Ltd of 26 Vector question 2 Question 2: The velocity of a particle at time t is given by v = 3 t 2 i + (2 t – 7)j Find the position vector and acceleration of the particle at time t if the position vector of the particle at time t = 0 is 3i – 2j. Differentiate to find the acceleration: a = 6 t i + 2j a =

17 © Boardworks Ltd of 26 Vector question 2 Integrate to find the position vector: r = ( t 3 + c )i + ( t 2 – 7 t + d )j When t = 0: r = 3i – 2j c i + d j = 3i – 2j  c = 3 and d = –2 r = ( t 3 + 3)i + ( t 2 – 7 t – 2)j Question 2: The velocity of a particle at time t is given by v = 3 t 2 i + (2 t – 7)j Find the position vector and acceleration of the particle at time t if the position vector of the particle at time t = 0 is 3i – 2j.

18 © Boardworks Ltd of 26 Vector question 3 Question 3: The position vector of a particle is given by r = 2 t 2 i + 4 t j + 3k Find the speed of the particle when t = 3. Differentiate to find the velocity: Therefore the speed of the particle when t = 3 is 12.6 ms -1. Speed =  ( ) When t = 3:v = 12i + 4j v = 4 t i + 4j =  160 = 12.6 (3 s.f.) v = 4 12 v

19 © Boardworks Ltd of 26 Variable acceleration Motion in two or three dimensions Examination-style questions Contents © Boardworks Ltd of 26 Examination-style questions

20 © Boardworks Ltd of 26 Exam question 1 a)Find the velocity of the particle at time t. b)Find the acceleration of the particle at time t = 3. Question 1: A particle has position vector at time t seconds given by r = (2 t 2 – 3 t )i + ( t 3 – t 2 + 4)j a) Differentiate to find v: v = (4 t – 3)i + (3 t 2 – 2 t )j v =

21 © Boardworks Ltd of 26 Exam question 1 b) Differentiate to find a: a = 4i + (6 t – 2)j When t = 3: a = 4i + 16j a = a)Find the velocity of the particle at time t. b)Find the acceleration of the particle at time t = 3. Question 1: A particle has position vector at time t seconds given by r = (2 t 2 – 3 t )i + ( t 3 – t 2 + 4)j

22 © Boardworks Ltd of 26 Exam question 2 Question 2: A particle is moving with acceleration a = 3 t i – 2j at time t seconds. Given that at time t = 0 the particle is at 4i + 2j and has velocity i – 4j, find: a) the velocity of the particle at time t. b) the position vector of the particle at time t. c) the distance of the particle from the origin when t = 3. a) Integrate to find v:

23 © Boardworks Ltd of 26 Exam question 2 b) Integrate to find r: When t = 0: r = 4i + 2j   c = 4 and d = 2  When t = 0: v = i – 4j  c = 1 and d = – c i + d j = i – 4j 4i + 2j = c i + d j

24 © Boardworks Ltd of 26 Exam question 2 c)When t = 3:  The distance from the origin =  ( ) = 28.0 m (3 s.f.) Question 2: A particle is moving with acceleration a = 3 t i – 2j at time t seconds. Given that at time t = 0 the particle is at 4i + 2j and has velocity i – 4j, find a) the velocity of the particle at time t. b) the position vector of the particle at time t. c) the distance of the particle from the origin when t = 3. = 20.5i – 19jr =

25 © Boardworks Ltd of 26 Exam question 3 Question 3: The position vector of a particle at time t is given by r = (sin t )i + (cos t )j + k0 ≤ t ≤  a) Differentiate to find v: v = (cos t )i – (sin t )j v = a)Find the speed of the particle when t =. b)Find the time when the velocity is parallel to j. When t =

26 © Boardworks Ltd of 26 Exam question 3 b) v is parallel to j when the i component of v is 0.  cos t = 0 Speed =  v  =  v  =  (½ + ½) = 1 Therefore the speed when t = is 1 ms -1. v = (cos t )i – (sin t )j t =  v is parallel to j when t = seconds.


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