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Chapter 3

Chapter 3. Acceleration and Newton’s Second Law of Motion. Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object.

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Chapter 3

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  1. Chapter 3 Acceleration and Newton’s Second Law of Motion

  2. Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object. Kinematics – Mechanics that describes how objects move, no reference to the force or mass. Dynamics– Mechanics that deals with force: why objects move.

  3. Describing Motion Involves: • Reference frames • Reference Direction • Displacement • Velocity • Acceleration

  4. Reference frames and coordinates When we say a car is moving at 45 km/hr, we usually mean “with respect to the earth” although is is not explicitly stated. An escalator is moving at 3 m/s relative to the ground. A lady walks on the escalator at 2 m/s relative to the escalator. The lady’s speed relative to the ground is 5 m/s.

  5. Reference Direction • Can use compass directions: North, East, South, West. • In physics, we use coordinate axes 1-D coordinate system: • Specify the origin. • Show where x axis is pointing: to the right is positive direction of x. • Label the axis with the relevant units.

  6. + y - + 0 x - Reference Direction:2-D coordinate system: • Specify the origin. • Show x and y axes. To the right is positive direction of x. Vertically upward is positive y. • Label the axis with the relevant units.

  7. Distance Just a length. A Scalar quantity. SI unit = meter (m).

  8. y 3 A x 3 -3 B -3 Position • A vector quantity describing where you are relative to an “origin”. • Point A is located at x =3, y =1 or (3,1). Point B is located at (-1,-2). 10

  9. Displacement (Dr) • A vector quantity. • Change in position relative to the starting point. • SI unit = meter (m). • Dr = rf – ri The displacement from A to B is x-direction: -1 – 3 = -4 y-direction: -2 – 1 = -3 • r = (-4, -3), |Dr| = (42 + 32) = 5 y 3 A x 3 -3 B -3 13

  10. Example • You drive 40 km north, then turn east and drive 30 km east. What is your net displacement? 30 km E 40 km N

  11. 2. A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of itsdisplacement after this movement? • 8 km • 5.7 km • 16 km • 332 km • 0 km

  12. Average Speed and Average Velocity Average Speed = (distance traveled)/time taken = d/t Speed: - specified only by magnitude. It is a scalar quantity. SI unit = m/s - always a positive number Average velocity = (displacement)/time taken = (x2 – x1)/(t2 – t1) = x/t Velocity: - specified by both magnitude and direction. - It is a vector quantity. - units – m/s - Positive/ negative sign used to indicate direction

  13. Magnitude of average speed and average velocity: not always equal. Charles walks 120 m due north in 40s. He then walks another 60 m still due north in another 60s. Average speed? Average velocity? Average speed = Total distance/time Average velocity = Displacement/time

  14. Charles walks 120 m due north. He then walks another 60 m due south. He took a total of 100 s for the journey. Average speed? Average velocity? Average speed = Total distance/time Average velocity = Displacement/time

  15. Charles walks 120 m due north. He then walks another 120 m due south. He took a total of 100 s for the journey. Average speed? Average velocity?

  16. Instantaneous Velocity x(t) Dx Dt t x(t) t 23 • The average velocity is displacement divided by the change in time. • Instantaneous velocity is limit of average velocity as Dt gets small. It is the slope of the x(t) versus t graph. • v = lim t0x/ t.

  17. Instantaneous Velocity • Magnitude of instantaneous speed and instantaneous velocity are equal. • If velocity is uniform (constant) in a motion, then magnitude of Instantaneous velocity = Average speed. uniform (constant) velocity velocity (m/s) velocity (m/s) t2 t1 time (s) time (s)

  18. Determine the velocity of the car at times A, B, C and D. A Positive Zero Negative B Positive Zero Negative C Positive Zero Negative D Positive Zero Negative x(t) A t B C D

  19. Uniform Velocity v(t) x(t) t t • An object moving with uniform velocity – means velocity stays uniform (constant) throughout the motion. • Its magnitude and direction stays the same. • Its displacement changes uniformly. • Moving along a straight line with constant speed. v = slope

  20. A car moves at a constant velocity of magnitude 20 m/s. At time t = 0, its position is 50 m from a reference point. What is its position at (i) t = 2s? (ii) t = 4s (iii) t = 10s?

  21. Non-uniform Velocity • An object moving with non-uniform velocity – means velocity does not stays uniform (constant) during the motion. • Either its magnitude or direction stays the same. • Moving along a straight line with varying speed, or moving in a curved path. x(t) v(t) t t

  22. Area Under velocity-time Graph Area under the velocity-time graph = magnitude of the displacement over the time interval. uniform (constant) velocity velocity (m/s) Area = v(t2-t1) = (x2 – x1) = x time (s) t2 t1 velocity (m/s) time (s)

  23. A stray dog ran 3 km due north and then 4 km due east. What is the magnitude of its average velocity if it took 45 min?

  24. v(t) D C Dv B E A Dt t Acceleration (a) • The average acceleration is the change in velocity divided by the change in time. • SI unit = m/s2 • Slope of velocity-time graph.

  25. v(t) t Acceleration (a) • Instantaneous acceleration is limit of average acceleration as Dt gets small. It is the slope of the v(t) versus t graph. • a = lim t0v/ t.

  26. Uniform Acceleration a(t) v(t) v t t t • An object moving with uniform acceleration – means acceleration stays uniform (constant) throughout the motion. • Its magnitude and direction stays the same. • Its velocity changes uniformly. • Moving along a straight line. a = slope

  27. A car moves at a constant acceleration of magnitude 5 m/s2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at (i) t = 2s? (ii) t = 4s? (iii) t = 10s?

  28. Acceleration (a) Average acceleration =v/ t T/F? If the acceleration of an object is zero, it must be at rest. T/F? If an object is at rest, its acceleration must be zero.

  29. Example Train A moves due east along a straight line with a velocity 8 m/s. Within 7 seconds, it’s velocity increases to 22 m/s. What is its average acceleration? Train B is moves due east along a straight line at a velocity of 18 m/s. Within 10 s, its velocity drops to 3 m/s. What is its average acceleration? When an object slows down, we say it is decelerating. In that case, the direction of the acceleration will be opposite to that of the velocity.

  30. Example Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 1 - Yes 2 - No “Yes, because an object can be going forward but at the same time slowing down which would give it a negative acceleration.” 37

  31. Graphical Representation Slope of the line = average velocity Position, x (m) Time, t (s) If the slope is zero, the object is at rest

  32. Graphical Representation Slope of the line = average acceleration velocity, v (m/s) O Time, t (s) If the slope is zero, the object is moving with zero acceleration (constant velocity)

  33. velocity, v (m/s) C D E B A O Time, t (s) Slope of the line = average acceleration = v/t • OA • AB • BC • CD • DE 1. When is a = 0? 2. When is a < 0? 3. When is a = maximum?

  34. 1. The area under a velocity-time graph represents • Average velocity • Average acceleration • Total displacement • Instantaneous velocity

  35. 2. A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of itsdisplacement after this movement? • 8 km • 5.7 km • 16 km • 332 km • 0 km

  36. 3. Which car has a southward acceleration? A car traveling • Southward at constant speed • Northward at constant speed • Southward and slowing down • Northward and speeding up • Northward and slowing down

  37. 4. A train moves due north along a straight path with a uniform acceleration of 0.18 m/s2. Ifits velocity is 2.4 m/s, what will its velocity be after 1 minute? • 13.2 m/s north • 2.58 m/s north • 10.8 m/s north • 144 m/s north • None of these

  38. 5. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was the average speed of the bird? • 5.00 m/s • 0.300 m/s • 300 m/s • 4.12 m/s • 247 m/s

  39. 6. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was magnitude of the average velocity of the bird? • 5.00 m/s • 0.300 m/s • 300 m/s • 4.12 m/s • 247 m/s

  40. 7. A train moving along a straight path due north has a velocity of 20 m/s. Within 5.0 seconds,its velocity became 5.0 m/s. What was the train's average acceleration? • 15 m/s2 north • 15 m/s2 south • 3.0 m/s2 north • 3.0 m/s2 south • 5.0 m/s2 north • 5.0 m/s2 south

  41. VECTORS tip Vector A tail +y A Equal Vectors: A = B only if they have equal magnitudes and same directions. B +x Displacing a vector parallel to itself does not change it

  42. The negative of a vector: Two vectors with equal magnitude but opposite directions are negatives of each other A -A Vectors can be multiplied by a scalar: A 2A ½ A

  43. Components of a vector A vector can be expressed as a sum of 2 vectors called “components” that are parallel to the x and y axes: +y A = Ax + Ay Ax = x- component of A. Ay = y-component of A. A Ay +x Ax

  44. h = hypotenus o = opposite  a = adjacent SOH CAH TOA sin  = opposite/hypotenuse = o/h cos  = adjacent/hypotenuse = a/h tan  = opposite/adjacent = o/a Pythagorean theorem: h2 = a2 + o2

  45. A Ay  Ax sin  = o/h = Ay/A or y-component:Ay = A . sin  cos  = a/h = Ax/A or x-component: Ax = A . cos  A2 = Ax2 + Ay2 or magnitude of vectorA = (Ax2 + Ay2 ) tan  = o/a = Ay/Ax or  = tan-1(Ay/Ax)

  46. Addition of vectors Three methods: Component method (analytical). Tip-to-tail (graphical). Parallelogram (graphical).

  47. Component method Add vectors V1 + V2 + V3: • Find x and y components of each vector. {V1=V1x + V1y}, {V2 = V2x + V2y}, {V3 = V3x + V3y 2. Add x-components and y-components separately: {Vx = V1x + V2x +V3x} and {Vy = V1y + V2y +V3y} 3. Find the magnitude of the resultant vector using pythagorean theorem: V = {V2x + V2y} 4. Find the angle of the resultant measured from the +x axis:  = tan-1(Vy/Vx)

  48. Example y A B C 30o x Three displacement vectors are shown in the figure below. Their magnitudes are A = 20 cm, B = 16 cm and C = 12 cm. Find the magnitude and angle of the resultant vector.

  49. Tip-to-tail Method Draw the vectors such that the “tail” of the second vector connects to the tip of the first vector. The resultant is from the tail of the first to the tip of the second. A B B C A -B A D C = A + B D = A - B

  50. Parallelogram Method Draw the vectors such that their “tails” are joined to a common origin. Construct a parallelogram with the two vectors as adjacent sides. The resultant vector is the diagonal line of the parallelogram drawn from the common origin. A B C = A + B C A B

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