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External (6 credits)

Mechanics 2.4. External (6 credits). Scalars and Vectors. Vector quantities have magnitude (size) and direction e.g . displacement, velocity. Scalar quantities only have magnitude, e.g . distance, speed. Displacement. Displacement.

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External (6 credits)

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  1. Mechanics 2.4 External (6 credits)

  2. Scalars and Vectors • Vector quantities have magnitude (size) and direction e.g. displacement, velocity. • Scalar quantities only have magnitude, e.g. distance, speed.

  3. Displacement

  4. Displacement • Displacement is a vector quantity. It is the distance and direction from a known origin (starting point). Displacement is measured in m. e.g. if you travel 4 km North, then 3 km West, your displacement (from your starting position) is 5 km at 37° West of North. 3 km 5 km 4 km 37°

  5. Scalars and Vectors This process of adding two (or more) vector quantities is called vector addition. The vectors must be of the same quantity e.g. both displacement vectors or both velocity vectors Vectors are added head-to-tail.

  6. Vector Addition watch how the ORDER of vector addition does not matter!

  7. HMWK pg 37-38

  8. Displacement-Time Graphs A displacement-time graph has displacement on the y-axis and time on the x-axis. Displacement (m) Time (s)

  9. Displacement-Time Graphs Here is a displacement-time graph of a person who walks forward 3 m in a time of 3 s, stands still for 3 s, then turns around and takes 6 s to walk the 3 m back to their starting point. Displacement (m) Time (s)

  10. Displacement-Time Graphs  The gradient of a displacement time graph gives the ________of the object. velocity Displacement (m) Time (s)

  11. Velocity

  12. Starter Apollo 11, the first spacecraft to land people on the Moon, took 102 hours to reach its destination. The average speed of the craft was 5500 km/h. What is the distance to the Moon? distance = speed x time = 5500 km x 102 hr hr = 561 000 km

  13. Velocity Velocity is a vector quantity. It is the speed and the direction from a known origin. Velocity is calculated by the change in displacement divided by the time taken. v = Dd Dt v = velocity, m.s-1 Dd = change in displacement, m Dt = change in time, s

  14. Motion of an object can be described using the vector quantities: • displacement • distance and direction from a known origin • velocity • the speed and direction in which an object moves • acceleration • the change in velocity per unit time a.= Δv Δt

  15. e.g. 8 km A car travels 6 km north in 4 minutes and then 8 km east in 5 minutes. Find their displacement, average velocity and their average speed. 6 km 10 km N 53° tanθ = Δd = √62 + 82 Displacement is 10 km at a bearing of 053° = √100 θ = tan-1 = 10 θ = 53.13° Average Velocity Average Speed vavg. = Δd Δt .= Δdistance Δtime .= 14 km 0.15 hr Δt = 9 min = 10/.15 = 0.15 hr = 93.3 km/hr = 66.7 km/hr

  16. Recall: Displacement-Time Graph  The gradient of a displacement time graph gives the ________of the object. velocity Displacement (m) Time (s)

  17. Velocity -Time Graphs  A velocity time graph has velocity on the y-axis and time on the x-axis velocity (m/s) Time (s)

  18. We can draw one using information from the other Displacement (m) Time (s) velocity (m/s) Time (s)

  19. Motion GraphsVelocity - Time  The gradient of a velocity - time graph gives the ___________ of the object. acceleration Velocity (m/s) Time (s)   Astraight linerepresents constant acceleration

  20. Motion GraphsVelocity - Time a. = Δv Δt   The acceleration is equal to the change in velocity divided by the change in time a. = 7− 2 10 7 Velocity (m/s) a. = 0.5 m/s2 2 Time (s) 10   Astraight linerepresents constant acceleration

  21. Motion GraphsVelocity - Time  The area under a velocity - time graph gives the ________________of the object. total displacement Velocity (m/s) Time (s)   Astraight linerepresents constant acceleration

  22. Motion GraphsVelocity - Time Distance travelled = (2 × 14) + ½ (14 × 5) = 63 m 7 Velocity (m/s) 2 Time (s) 14 initial velocity

  23. Motion GraphsVelocity - Time  The gradient of a velocity - time graph gives the ___________ of the object.  The area under a velocity - time graph gives the ______________ . acceleration distance travelled   Astraight linerepresents constant acceleration

  24. Acceleration Acceleration is a vector quantity. It is the change in velocity per unit time. It is measured in Acceleration is calculated by the change in velocity divided by the change in time. a = Dv Dt a = acceleration, m.s-2 Dv = change in velocity, m.s-1 Dt = change in time, s

  25. Acceleration Acceleration due to gravity near the surface of the earth is 9.81 Acceleration due to gravity near the surface of the moon is 1.63

  26. Starter A car starts moving from rest at a stop sign and accelerates at 1.5 m/s2. a.) Calculate the speed of the bus after 7 seconds. b.) After how many seconds is the bus travelling at a speed of 18 m/s? c.) If the car is traveling at 20m/s and comes to a stop in 5 s, what is the braking deceleration? a = Dv Dt v = a × t v = 1.5 m/s2× 7 s = 10.5 m/s t = v a t = 18 1.5 t = 12 s a = 0 - 20 5 a = - 4 m/s2

  27. Motion GraphsVelocity - Time   Draw a velocity time graph for a car initially moving at a constant velocity of 20 m/s for 3 seconds and then decelerates at a rate of 2.5 m/s2. At what time is the car at rest? Calculate the total displacement of the car from your graph. 3 x 20 + ½ (11-3) x 20 = 60 + 80 = 140 m Velocity (m/s) 30 20 10 Time (s) 3 11

  28. Kinematics The geometry of motion

  29. Kinematic Equations The kinematic equationsare a series of equations linking displacement, initial velocity, final velocity, time taken, and acceleration. These equations only hold in situations where the acceleration is constant e.g. constant acceleration due to gravity. The kinematic equations can be derived from the equations for velocity and acceleration.

  30. Kinematic Equations vf = vi + at d = ½ (vi + vf)t d = vit + ½ at2 vf2 = vi2+ 2ad d = displacement, m vi (or u) = initial velocity, m.s-1 vf (or v) = final velocity, m.s-1 a = acceleration, m.s-2 t = time, s.

  31. Example A motorcycle accelerates from rest to 25m/s in 22 seconds. a.) Calculate the acceleration. b.) Calculate the distance travelled. c.) The bike brakes to a stop in 5 seconds. Calculate is deceleration. d.) Calculate the distance travelled while braking. vf - vi = a t a = 1.14m/s2 vf = vi + at d = 275m d = ½ (vi + vf)t vf - vi = a t 0 - 25= a 5 vf = vi + at a = -5 m/s2 d = vit + ½ at2 d =25(5) + ½ (-5)(52) d = 62.5m

  32. Gravity g = 9.81 ms-2 = 9.81 Nkg-1 ‘ each kg feels a weight force of 9.81 N’

  33. Weight m = mass (kg) g = -9.81 ms-2 W = mg * we often assume the direction ‘up’ is positive the length of time it takes an object to fall is independent of it mass (assuming there is no air resistance)

  34. Example vi = 0 m/s This boy drops a ball off a 80 m cliff. How long will it take to reach the bottom? d = vit + ½ at2 -80 = 0 + ½ (-9.81)t2 -80x2 = t2 -9.81 What is the balls speed when it reaches the bottom? 16.31 = t2 t = 4.04 s vf = vi + at = 0– 9.81 x 4.04 vf = - 39.6 m/s

  35. Example vi = 0 m/s How far has the ball fallen after only 2.5 seconds? d = vit + ½ at2 = 0 + ½ (-9.81)(2.5)2 = - 30.7 m How long does it take to reach a speed of -25 m/s? vf = vi + at -25 = 0 – 9.81 t t = -25_ -9.81 t = 2.55 s

  36. Explain vi = 0 m/s Explain the forces that are acting on the ball as it falls towards the ground. Fair Air Resistance – acting upwards Force of gravity – acting downwards Fg W = mg

  37. pg 32-33 Starter A 1.5kg firecracker is fired vertically upwards with an initial speed of 40ms-1. Calculate the time to reach maximum height. vf = vi + at 0 = 40 – 9.81 t t = -40_ -9.81 t = 4.08 s What is its acceleration at the top of its flight? g = -9.81 m/s2 Calculate the weight force, W, on the firecracker just before it lands. W = -14.7 N W = mg W = 1.5 x -9.81

  38. Starter A parachutist free falls from a plane through a vertical distance of 530m. Her initial vertical velocity is zero and g = 9.81 ms-2 Calculate the time of her free fall, ignoring air resistance? d = vit + ½ at2 t2 = 108.053 s2 -530 = 0 + ½ (-9.81) t2 t = 10.4 s t2 = -530 × 2_ -9.81 What is the final velocity at the end of the free fall? vf2 = vi2 + 2ad a = -9.81 m/s2 vf2 = 0 x 2(-9.81m/s2)(-530 m) vf = 102 m/s vf2 = 10398.6 m2/s2 vf = -102 m/s (downwards)

  39. Starter A parachutist free falls from a plane through a vertical distance of 530m. Her initial vertical velocity is zero and g = 9.81 ms-2 At the end of the free fall the parachutist opens her parachute. This action slows her speed down to 10 m/s in 3.4s. Calculate her average deceleration during this action. vf = vi + at -10 m/s = -102 m/s + a (3.4 s) -10 + 102 = a (3.4 s) 92 m/s = a 3.4 s a = 27.1 m/s2

  40. Projectile Motion A projectile is an object that is travelling through the air under the influence of gravity only (i.e. we assume there is no friction/air resistance). We can consider the horizontal and vertical components of projectile motion separately.

  41. Projectile Motion Since there are no horizontal forces acting on the object, the horizontal component of the velocity remains constant. The horizontal displacement can be calculated using v = Dd Dt The vertical component of the velocity changes at the acceleration due to gravity, so the vertical displacement and velocity can be calculated using the kinematics equations.

  42. Projectile Motion Since there are no horizontal forces acting on the object, the horizontal component of the velocity remains constant. The horizontal displacement can be calculated using v = Dd Dt The vertical component of the velocity changes at the acceleration due to gravity, so the vertical displacement and velocity can be calculated using the kinematics equations. The resultant displacement and resultant velocity can be calculated by vector addition of the horizontal and vertical components.

  43. Example A skier jumping down a slope is an example of projectile motion. initial horizontal velocity Vh projectile path The horizontal component of the velocity remains constant The vertical component of the velocity increases by g each second

  44. Example Two seconds after his jump the skier has travelled a horizontal distance of 10 m. What is the magnitude of the initial horizontal velocity? 5 m/s Vh projectile path What is the magnitude of his vertical component of velocity at a.) the start of the jump? b.) 2 seconds after starting his jump? 0 m/s 19.6 m/s vf = vi + at vf = 0– 9.81 x 2

  45. Example A golf ball is hit with an initial speed of 50 m/s at an angle of 55° to the ground. Resolve into vector components 55° Vertical vvert = v x sinθ 50m/s the vertical component decreases by 9.81 m/s each second = 50 x sin55 55° = 41.0 m/s vhoriz = v x cosθ Horizontal the horizontal component stays the same throughout the entire flight = 50 x cos55 = 28.7 m/s

  46. Example A golf ball is hit with an initial speed of 50 m/s at an angle of 55° to the ground. At the top of the projectiles motion the vertical component is 0 55° Find the time to the maximum height vf = vi + at vf= 0 m/s t = Δvvert a = -41.0 -9.8 = 4.2 seconds Find the time of the entire flight 2 x 4.2 s = 8.4 s

  47. Example A golf ball is hit with an initial speed of 50 m/s at an angle of 55° to the ground. h What is the maximum height? 55° d = ½ (vi + vf)t vf= 0 m/s t = 4.2 s d = ½ (41 x 4.2) = 86.1 m What is the range (horizontal distance) travelled? r = Vhorizx time in flight = Vhorizx 8.4 s = 241 m

  48. Starter A baseball player hits a home run out of the park a horizontal distance of 120 m. The ball is in the air for 3.2 seconds. Calculate the horizontal component of the initial velocity? h v = Dd Dt 120/3.2 = 37.5 m/s How long does it take to reach its greatest height? 3.2/2 = 1.6 seconds What is the vertical component of the balls initial velocity? vf = vi + at Calculate the max height reached by the ball. 0 = Vi + (-9.81)(1.6) Vi = 15.7 m/s (up) d = vit + ½ at2 = 15.7(1.6) + ½ (-9.81)(1.6)2 = 12.6 m(3sf)

  49. Starter How long will it take the cannonball to reach the top of its flight? A cannonball is fired at 75° to the horizontal and takes 2.8 seconds to return to the ground h 2.8/2 = 1.4 seconds Calculate the vertical component of its initial velocity vf = vi + at 0 = vi+ (-9.81)(1.4) vi= 13.7 m/s Calculate the max height reached by the ball. d = vit + ½ at2 74 vi = ? = 13.7(1.4) + ½ (-9.81)(1.4)2 vvert = 13.7 m/s = 9.61 m(3sf) 75° Using trig, calculate the shell’s initial velocity from the gun. vi = 13.7 sin 75° sin 75° = 13.7 vi = 14.2 m/s

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