Chapter 10: Linear Chapter 10: Linear Kinematics Kinematics Distance and Displacement Distance and Displacement Used to describe the extent of a body’s motion Used to describe the extent of a body’s motion Distance Distance – length of the path that a body – length of the path that a body follows when moving from one location to the follows when moving from one location to the next next Displacement Displacement – length of (imaginary) line – length of (imaginary) line joining starting position and finishing joining starting position and finishing position and noting position and noting direction direction . (i.e. as the . (i.e. as the crow flies.) crow flies.) Therefore, distance is a Therefore, distance is a scalar scalar quantity, and quantity, and displacement is a displacement is a vector vector quantity. quantity.
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Chapter 10: Linear KinematicsChapter 10: Linear Kinematics
Distance and DisplacementDistance and Displacement Used to describe the extent of a body’s motionUsed to describe the extent of a body’s motion
DistanceDistance – length of the path that a body follows – length of the path that a body follows when moving from one location to the nextwhen moving from one location to the next
DisplacementDisplacement – length of (imaginary) line joining – length of (imaginary) line joining starting position and finishing position and noting starting position and finishing position and noting directiondirection. (i.e. as the crow flies.). (i.e. as the crow flies.)
Therefore, distance is a Therefore, distance is a scalarscalar quantity, and quantity, and displacement is a displacement is a vector vector quantity.quantity.
Example:Example: Boston Marathon Boston Marathon
Hopkinton to Downtown BostonHopkinton to Downtown Boston
42.2 km
38.6 km
HopkintonHopkinton
Downtown BostonDowntown Boston
Dist = 42.2 km (26 miles, 385 yds). Displacement = 38.6 km ENE
Ironman triathlon:Dist = 3.87 km+180.65 km+42.2 km = 226.72km Displ = nearly zero
Speed and VelocitySpeed and Velocity
The rate analogs of distance and displacementThe rate analogs of distance and displacement
• Speed = distance /change in time Speed = distance /change in time
• Velocity = (displacement /change in time) & directionVelocity = (displacement /change in time) & direction
Units are m/s, km/h, miles/hrUnits are m/s, km/h, miles/hr Examples: Examples:
100m sprint in 10.32s: 100m sprint in 10.32s: S = avg. speed = 100m/10.32s = 9.69 m/sS = avg. speed = 100m/10.32s = 9.69 m/s
V = avg. velocity = 100m / 10.32s = 9.69 m/s in an easterly V = avg. velocity = 100m / 10.32s = 9.69 m/s in an easterly direction or 90direction or 90oo from due north from due north
NoteNote: Speed is simply the magnitude of velocity when : Speed is simply the magnitude of velocity when motion is in a straight linemotion is in a straight line
Instantaneous speedInstantaneous speed: average speed over a very : average speed over a very small distance or unit of time so that the speed small distance or unit of time so that the speed does not have time to changedoes not have time to change
Av speed 100m/9.79s = 10.21 m/s 100m/9.92s = 10.08 m/s
ACCELERATIONACCELERATION
The rate of change of velocityThe rate of change of velocity
āā = = VVFinalFinal – V – VInitialInitial = = ∆V∆V ∆ ∆T T ∆T ∆T
0 – 60 mph in 6.6 seconds 60 mph = 27 m/s0 – 60 mph in 6.6 seconds 60 mph = 27 m/s
ā = ā = VVF F – V– VII 27 m/s – 0 m/s27 m/s – 0 m/s ∆ ∆T T 6.6 s 6.6 s
= = 4.1 m/s4.1 m/s22 or 4.1 ms or 4.1 ms-2-2
Negative and Positive AccelerationNegative and Positive Acceleration
Velocity changes during the running strideVelocity changes during the running stride
+ 9.7 m/s + 9.5 m/s + 9.7 m/s
∆t =0.06 s ∆t =0.06 s
ā = (9.5 m/s – 9.7 m/s)/0.06 s
= (- 0.2 m/s)/0.06 s = - 3.3 m/s2
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/s2
Increasing speed in a positive direction: positive acceleration
Weight-acceptance Push-off
- 9.7 m/s - 9.5 m/s - 9.7 m/s
∆t =0.06 s ∆t =0.06 s
Push-off Weight-acceptance
ā = (- 9.7 m/s – (-9.5 m/s))/0.06 s = (-0.2 m/s)/0.06 s = -3.3 m/s2
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/s2
Decreasing speed in a negative direction: positive acceleration
Vertical acceleration in the vertical jumpVertical acceleration in the vertical jump
1 2 3 4 5
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
+
-
- 9.81 m/s2
+
-
āā== ∆V/∆t
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