Chapter 10: Linear Kinematics Distance and Displacement Used to describe the extent of a body’s motion Distance – length of the path that a body follows.

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Presentation transcript:

Chapter 10: Linear Kinematics Distance and Displacement Used to describe the extent of a body’s motion Distance – length of the path that a body follows when moving from one location to the next Distance – length of the path that a body follows when moving from one location to the next Displacement – length of (imaginary) line joining starting position and finishing position and noting direction. (i.e. as the crow flies.) Displacement – length of (imaginary) line joining starting position and finishing position and noting direction. (i.e. as the crow flies.) Therefore, distance is a scalar quantity, and displacement is a vector quantity. Therefore, distance is a scalar quantity, and displacement is a vector quantity.

Example: Boston Marathon Hopkinton to Downtown Boston 42.2 km 38.6 km Hopkinton Downtown Boston Dist = 42.2 km (26 miles, 385 yds). Displacement = 38.6 km ENE Ironman triathlon:Dist = 3.87 km km+42.2 km = km Displ = nearly zero

Speed and Velocity The rate analogs of distance and displacement The rate analogs of distance and displacement Speed = distance /change in timeSpeed = distance /change in time Velocity = (displacement /change in time) & directionVelocity = (displacement /change in time) & direction Units are m/s, km/h, miles/hr Units are m/s, km/h, miles/hr Examples: 100m sprint in 10.32s: S = avg. speed = 100m/10.32s = 9.69 m/s S = avg. speed = 100m/10.32s = 9.69 m/s V = avg. velocity = 100m / 10.32s = 9.69 m/s in an easterly direction or 90 o from due north V = avg. velocity = 100m / 10.32s = 9.69 m/s in an easterly direction or 90 o from due north Note: Speed is simply the magnitude of velocity when motion is in a straight line Note: Speed is simply the magnitude of velocity when motion is in a straight line

Instantaneous speed: average speed over a very small distance or unit of time so that the speed does not have time to change

Av speed 100m/9.79s = m/s 100m/9.92s = m/s

ACCELERATION The rate of change of velocity ā= V Final – V Initial = ∆V ā = V Final – V Initial = ∆V ∆T ∆T ∆T ∆T  0 – 60 mph in 6.6 seconds 60 mph = 27 m/s  0 – 60 mph in 6.6 seconds 60 mph = 27 m/s ā = V F – V I 27 m/s – 0 m/s ā = V F – V I 27 m/s – 0 m/s ∆T 6.6 s ∆T 6.6 s = 4.1 m/s 2 or 4.1 ms -2 = 4.1 m/s 2 or 4.1 ms -2

Negative and Positive Acceleration

Velocity changes during the running stride m/s m/s m/s ∆t =0.06 s ā = (9.5 m/s – 9.7 m/s)/0.06 s = (- 0.2 m/s)/0.06 s = m/s 2 Decreasing speed in a positive direction: negative acceleration ā = (9.7 m/s – 9.5 m/s)/0.06 s = (0.2 m/s)/0.06 s = +3.3 m/s 2 Increasing speed in a positive direction: positive acceleration Weight-acceptancePush-off

- 9.7 m/s- 9.5 m/s- 9.7 m/s ∆t =0.06 s Push-offWeight-acceptance ā = (- 9.7 m/s – (-9.5 m/s))/0.06 s = (-0.2 m/s)/0.06 s = -3.3 m/s 2 Increasing speed in a negative direction: negative acceleration ā = (- 9.5 m/s – (-9.7 m/s))/0.06 s = (0.2 m/s)/0.06 s = +3.3 m/s 2 Decreasing speed in a negative direction: positive acceleration

Vertical acceleration in the vertical jump V = 0 +V peak V = 0 - V V = 0 1 to 2: ā = (Vf – Vi)/∆t = (V – 0)/∆t = + acceleration increasing speed in a positive direction 2 to 3: ā = (Vf – Vi)/∆t = (0 – V)/∆t = - acceleration decreasing speed in a positive direction 3 to 4: ā = (Vf – Vi)/∆t = (-V – 0)/∆t = - acceleration increasing speed in a negative direction 4 to 5: ā = (Vf – Vi)/∆t = (0 – (-V))/∆t = + acceleration decreasing speed in a negative direction

∆V ∆t m/s ā= ā= ∆V/∆t