Kinematics - (describing how things move)

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Kinematics - (describing how things move). Change in position (How far you travel in a given direction). How far you travel. How fast you travel (in a given direction). How fast you travel. Rate of change of velocity. Describing Motion. - PowerPoint PPT Presentation

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  • Kinematics - (describing how things move)How far you travelChange in position(How far you travel in a given direction)How fast you travelHow fast you travel (in a given direction)Rate of change of velocity

    Scalar (no direction) Vector (w/ direction)Distance (d)Displacement (d)

    Speed (s)Velocity (v)

    Acceleration (a)

  • Describing MotionThere are lots of different ways to describe motion.WordsSketchesTime elapsed photographsPhysical Expressions (Equations)Graphical Representation

  • Kinematics Equations that Make Sense! Average speed: sav = d / change in t SI unit: m/ssav = d / t = d / tf - tiAverage velocity: vav = d / t SI unit: m/svav = (df - di) / tAverage acceleration: aav = v / t SI unit: m/s/s = m/s2aav = (vf - vi) / tdf = di + vav tvf = vi + aav tNote: if the time intervals are very small we call these quantities instantaneous

  • Using Split Times!3.12.11.71.41.2(5)Determine the average velocity for each distance intervalDetermine average velocity of the object over the time recordedDetermine the average acceleration over the time recordedvav = df - di / t= 25m - 0m / 14.5 s= 1.7 m / sa = vf - vi / t= (1.25 m/s - 3.1m/s) / 14.5 s= - 0.13 m / s2Note: the a is negative because the change in v is negative!!

    Position (m)0-55-1010-1515-2020-25Split Time (s)1.62.43.03.54.0Av. Velocity (m/s)

  • Reference FramesAll measurements are made relative to a frame of reference..Sitting at your desk right now you dont seem to be moving but you are infact revolving at 30 km/s (67,000 mi/h) around the sun The earth also has a rotational speed about its axis. People on earth experience different tangential speeds depending on their latitude. The space shuttle at Kennedy Space Center is already traveling at 410 m/s (917 mi/h) before it even gets off the ground. A geosynchronous satellite that is in a stationary orbit over the earth is traveling at over 3080 m/s (6890 mi/h)Someone moving south at 5 km/h on a train which itself is moving south at 80 km/h will be moving at 85 km/h relative to the earth. If the person had been moving north on the train their velocity would have been 75 km/h relative to the earth.

  • Solving Kinematics Problems

    Assign a coordinate system Define which directions are positive and negative.

    Write down your known variables and show unknowns with a question mark.

    Write down the kinematics expression that will allow you to solve for one variable. All the others in your expression should be known. Rearrange if necessary.

    Substitute numbers and units into your physical expression.

    Solve the equation for your unknown and include the correct units.

    Check your answer.

    Does the magnitude of your answer make sense? Do the units come out right? Can you use another expression to check your answer?

  • Example 1 Distance Run by a Jogger

    How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m/s?t = 5400s sav = 2.22 m/s d = ?sav = d / t So d = sav t = (2.22 m/s)(5400s)d = 12000 m Check: Units come out right (m) when multiplied

  • Example 2 The Worlds Fastest Jet-Engine Car

    Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997. To establish such a record, the driver makes two runs through the course, one in each direction,to nullify wind effects. From the data, determine the averagevelocity for each run.a) t = 4.740 s x = +1609m vav = ?vav = x / t = (+1609m) / (4.740 s) vav = + 339.5 m/s b) t = 4.695 s x = -1609m vav = ?vav = x / t = (-1609m) / (4.695 s) vav = - 342.7 m/s

  • Example 3 Acceleration and Increasing Velocity

    Determine the average acceleration of the plane.vi = 0 km/h vf = 260 km/h ti = 0s tf = 29saav = (vf - vi) / (tf ti)aav = (+260 km/h 0 km/h) / (29s 0s)aav = + 9.0 km/h /s

  • Graphical Representation of MotionKinematics Relationships Through Graphing:1. The slope of a d-t graph at any time tells you the av. velocity of the object.2. The slope of a v-t graph at any time tells you the av. acceleration of the object.The area under a v-t graph tells you the displacement of the object during that time.4. The area under a a-t graph tells you the change in velocity of the object during that time

  • Constant MotionOn the d-t graph at any point in time vav = d / t vav = (50 - 0)m / (5 - 0)svav = 10 m/sThe slope is constant on this graph so the velocity is constantOn the v-t graph at any point in time aav = vf - vi / taav = (10 - 10)m/s / (5 - 0)saav = 0 m/s2Looking at the area between the line and the x-axis.Area of rectangle = b x hArea = 5s x 10 m/s = 50 mWhich is of course displacementOn the a-t graph the area between the line and the x-axis is.Area of rectangle = b x hArea = 5s x 0 m/s2 = 0 m/sThe area thus represents.v = aav tChange in velocity

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  • Changing MotionOn the d-t graph at any point in time vav = d / t The slope is constantly increasing on this graph so the velocity is increasing at a constant rateThe slope of a tangent line drawn at a point on the curve will tell you the instantaneous velocity at this positionOn the v-t graph at any point in time aav = vf - vi / taav = (20 - 0)m/s / (5 - 0)saav = 4 m/s2Looking at the area between the line and the x-axis.Area of triangle = 1/2 (b x h)Area = 1/2 (5s x 20 m/s) = 50 mWhich is of course displacementOn the a-t graph the area between the line and the x-axis is.Area of rectangle = b x hArea = 5s x 4 m/s2 = 20 m/sThe area thus represents.Change in velocity

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  • To determine the velocity at any point in time you need to find the slope of the distance-time graph. This means that you need to find the slope of the tangent line drawn at the point of interest. By selecting two points spaced evenly on either side of the point of interest, a line can be drawn between them that has the same slope as the tangent. (shown below).Slope between 1s and 3s shows the velocity at 2sThe velocity at 2s is p/t = (18m - 2m)/ (3s - 1s) = 8m/sThe acceleration is given by the slope of the velocity-time graph. Therefore: a = v / t = 20m/s / 5s = 4m/s2

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  • Example ProblemA student is late for the school bus. She runs east down the road at 3 m/s for 30s, then thinks that she has dropped her calculator so stops for 10s to check. She jogs back west at 2 m/s for 10s, stops for 5 s then accelerates uniformly from rest to 4 m/s east over a 10 second period. Sketch the velocity-time graph of the students motionDetermine the total distance and displacement of the student during this timeDetermine the students average velocity during this time

  • Velocity-Time Graph of the Students Motion

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  • Total distance traveled by the student is.dtotal = d1 + d2 + d3 + d4 + d5dtotal = s1t1 + s2 t2+ s3 t3 + s4 t4 + s5 t5dtotal = (3m/s)(30s) + (0m/s)(10s) + (2m/s)(10s) + (0m/s)(5s) + (1/2(4m/s)(10s)dtotal = 130 mTotal displacement by the student is.dtotal = d1 + d2 + d3 + d4 + d5dtotal = v1t1 + v2 t2+ v3 t3 + v4 t4 + v5 t5dtotal = (3m/s)(30s) + (0m/s)(10s) + (-2m/s)(10s) + (0m/s)(5s) + (1/2(4m/s)(10s)dtotal = + 90 m (East)+ east - west

  • Average velocity of the student is..vav = dtotal / ttotal= + 90m East / 65svav = 1.4 m/s East+ 90 m- 20 m+ 20 m

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  • More Kinematics Equations that Make Sense!df = di + vav tdf = di + (vi + vf) /2 t but vf = vi + aav tdf = di + (vi + (vi + aav t) /2 tdf = di + vi t + 1/2 aavt2 df = di + (vi + vf) /2 t but t = (vf - vi ) / aavdf = di + (vi + vf) /2 (vf - vi) /aavd = (vi + vf) /2 (vf - vi) /aavSo d = (vf2 - vi2 ) /2aavandvf2 = vi2 + 2aav dd = vi t + 1/2 aavt2 orbut vav = (vi + vf) / 2

  • Free FallFree fall describes the motion of an object which is only under the influence of gravity. I.e. a ball thrown upwards or droppedKinematics equations can be used for solving free fall problems by replacing aav in the expressions with g where g is 9.8 m/s2 downwardsAn object in free fall experiences a constant uniform acceleration of 9.8 m/s2 in the downwards directionvf = vi + g tdv = vi t + 1/2 g t2 vf2 = vi2 + 2g dvvi is positive+-g is negativeIf you define up as the positive direction, g must be negative because the velocity gets less positive over time-+vi is negativeg is positiveIf you define up as the negative direction, g must be positive because the velocity gets less negative over timeAir resistance limits the time of free fall. Eventually a falling object will reach a constant velocity downwards known as its terminal velocity

  • Freely Falling BodiesExample 1 A Falling Stone

    A stone is dropped from the top of a tall building. After 3.00sof free fall, what is the displacement y of the stone?

  • Freely Falling Bodies?- 9.8 m/s20 m/s3.00 s

    dvavfvit

  • Freely Falling Bodiesdv = vi t + g t2= (0 m/s)(3.00s) + (-9.8 m/s2)(3.00s)2= - 44.1 m

    dvavfvit?-9.80 m/s20 m/s3.00 s

  • Freely Falling BodiesExample 2 How High Does it Go?

    The referee tosses the coin upwith an initial speed of 5.00m/s.In the absence if air resistance,how high does the coin go aboveits point of release?

  • Freely Falling Bodies?- 9.80 m/s20 m/s+ 5.00 m/s

    dvavfvit

  • Freely Falling Bodiesvf2 = vi2 + 2 g dv= 1.28 m

    dvavfvi t?-9.80 m/s20 m/s+5.00 m/s

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