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**AS-Level Maths: Mechanics 2 for Edexcel**

M2.2 Kinematics 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. 1 of 26 © Boardworks Ltd 2006

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**Variable acceleration**

Motion in two or three dimensions Examination-style questions Contents 2 of 26 © Boardworks Ltd 2006

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**Variable acceleration**

Kinematics is the study of how objects move rather than of the forces that cause motion. 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.

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**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.

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**Variable acceleration**

The difference between constant and variable acceleration may not be discernible to the eye. Which ball is travelling with variable acceleration?

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**Variable acceleration question 1**

Question 1: The displacement of a particle moving in a straight line is given by x = t3 – 3t2 + 4t. Find a) the velocity of the particle when t = 4. b) the acceleration of the particle when t = 2. a) To find the velocity we must differentiate x: v = v = 3t2 – 6t + 4 When t = 4: v = 3 × 42 – 6 × = 28 Therefore the velocity of the particle when t = 4 is 28 ms-1.

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**Variable acceleration question 1**

b) To find the acceleration we must first differentiate v: v = 3t2 – 6t + 4 a = a = 6t – 6 When t = 2: a = 6 × 2 – 6 = 6 Therefore the acceleration of the particle when t = 2 is 6 ms-2.

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**Variable acceleration question 2**

Question 2: The velocity of a particle is given by v = t2 – 5t + 6. Initially its displacement from O is 2 m. Find a) the acceleration of the particle after 4 seconds. b) the displacement of the particle from O when t = 6. Differentiate v: a = a = 2t – 5 When the question says initially, it means when t = 0. When t = 4: a = 2 × 4 – 5 a = 3 Therefore the acceleration of the particle when t = 4 is 3 ms-2.

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**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: 2 = 0 – c c = 2 When t = 6: x = 72 – = 20 Therefore when t = 6 the displacement of the particle from O is 20 m.

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**Variable acceleration question 3**

Question 3: The velocity of a particle travelling in a straight line is given by v = –3t2 + 12t + 4. Find: the time when the particle is travelling at its maximum velocity. the value of this maximum velocity. Here v is a quadratic where the coefficient for t2 is negative. Therefore the maximum point on the velocity curve occurs where the gradient is 0. Differentiate to find the gradient function: In part a) it should be noted that the particle is travelling at its maximum speed when the acceleration is 0. Also mention that a positive quadratic has a minimum point when the gradient is 0. = –6t + 12 –6t + 12 = 0 t = 2

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**Variable acceleration question 3**

Question 3: The velocity of a particle travelling in a straight line is given by v = –3t2 + 12t + 4. Find: the time when the particle is travelling at its maximum velocity. the value of this maximum velocity. b) When t = 2: v = –3 × × 2 + 4 v = 16 Therefore the maximum velocity of the particle occurs when t = 2 seconds and the maximum velocity is 16 ms-1.

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**Motion in two or three dimensions**

Variable acceleration Motion in two or three dimensions Examination-style questions Contents 12 of 26 © Boardworks Ltd 2006

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**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 = xi + yj + zk 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. ci + dj + ek.

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3-D motion

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Vector question 1 Question 1: The position vector of a particle at time t is given by r = (2t – 4)i + (t2 – 3t + 1)j Find the velocity and acceleration of the particle at time t. Find the velocity and acceleration of the particle when t = 4. v = a) Differentiate to find v: v = 2i + (2t – 3)j Note that the acceleration is independent of time and is therefore constant. a = Differentiate to find a: a = 2j b) When t = 4: v = 2i + 5j ms-1 a = 2j ms-2

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Vector question 2 Question 2: The velocity of a particle at time t is given by v = 3t2i + (2t – 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 = When integrating a vector it is necessary to add a vector constant of integration. In this case this is represented by ci + dj a = 6ti + 2j

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Vector question 2 Question 2: The velocity of a particle at time t is given by v = 3t2i + (2t – 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. Integrate to find the position vector: r = (t3 + c)i + (t2 – 7t + d)j When t = 0: r = 3i – 2j ci + dj = 3i – 2j c = 3 and d = –2 r = (t3 + 3)i + (t2 – 7t – 2)j

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Vector question 3 Question 3: The position vector of a particle is given by r = 2t2i + 4tj + 3k Find the speed of the particle when t = 3. Differentiate to find the velocity: v = 4 12 v v = 4ti + 4j When t = 3: v = 12i + 4j The speed of the particle is given by the magnitude of the velocity. Speed = ( ) = 160 = 12.6 (3 s.f.) Therefore the speed of the particle when t = 3 is 12.6 ms-1.

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**Examination-style questions**

Variable acceleration Motion in two or three dimensions Examination-style questions Contents 19 of 26 © Boardworks Ltd 2006

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Exam question 1 Question 1: A particle has position vector at time t seconds given by r = (2t2 – 3t)i + (t3 – t2 + 4)j Find the velocity of the particle at time t. Find the acceleration of the particle at time t = 3. a) Differentiate to find v: v = v = (4t – 3)i + (3t2 – 2t)j

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Exam question 1 Find the velocity of the particle at time t. Find the acceleration of the particle at time t = 3. Question 1: A particle has position vector at time t seconds given by r = (2t2 – 3t)i + (t3 – t2 + 4)j b) Differentiate to find a: a = a = 4i + (6t – 2)j When t = 3: a = 4i + 16j

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Exam question 2 Question 2: A particle is moving with acceleration a = 3ti – 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:

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**Exam question 2 When t = 0: v = i – 4j 0 + 0 + ci + dj = i – 4j**

c = 1 and d = –4 b) Integrate to find r: When t = 0: r = 4i + 2j 4i + 2j = ci + dj c = 4 and d = 2

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Exam question 2 Question 2: A particle is moving with acceleration a = 3ti – 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. When t = 3: r = = 20.5i – 19j The distance from the origin = ( ) = 28.0 m (3 s.f.)

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Exam question 3 Question 3: The position vector of a particle at time t is given by r = (sin t)i + (cos t)j + k 0 ≤ t ≤ Find the speed of the particle when t = . Find the time when the velocity is parallel to j. v = a) Differentiate to find v: v = (cos t)i – (sin t)j When t =

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**Exam question 3 Speed =v= v = (½ + ½) = 1**

Therefore the speed when t = is 1 ms-1. b) v is parallel to j when the i component of v is 0. v = (cos t)i – (sin t)j cos t = 0 t = v is parallel to j when t = seconds.

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