2Chapter Objectives Describe accelerated motion Use graphs and equations to solve problems involving moving objectsDescribe the motion of objects in free fall (separate presentation - not on this test)
3Section 3.1 Acceleration Define acceleration Relate velocity and acceleration to the motion of an objectCreate velocity-time graphs
4Uniform Motion Nonuniform Motion Moving at a constant velocityIf you close your eyes, you feel as though you are not moving at allMoving while changing velocityCan be changing the rate or the directionYou feel like you are being pushed or pulled
5Changing VelocityConsider the following motion (particle model) diagramConstant VelocityNot movingIncreasing VelocityDecreasing Velocity
6Changing velocity You can indicate change in velocity by the motion diagram spacingthe magnitude (length) of the velocity vectors.If the object speeds up, each subsequent velocity vector is longer.If the object slows down, each vector is shorter than the previous one.
7Velocity-Time GraphsDistance being covered is longer, thus the runner is speeding up.Distance being covered is longer, thus the runner is speeding up.
8Velocity-Time Graphs Time (s) Velocity (m/s) 1 5 2 10 3 15 4 20 25 Slope???Time (s)Velocity (m/s)1521031542025Area???
9Change in Position → can’t get position Velocity-Time GraphsAnalyze the unitsSlope = rise over runm = ∆y / ∆xSlope = m/s/ s = m/s^2m/s^2 is the unit for accelerationArea = ½ b hb = s * m/s = mm is the unit for displacementThe slope of a velocity-time graph is the ACCELERATION and the area is DISPLACEMENT.Change in Position → can’t get position
11Velocity – Time Graphs How Fast something is moving at a given time? Average AccelerationUse the information on the x & y axis to plug into the equationa = ∆v / tSlopeInstantaneous AccelerationFind the slope of the line (straight line)Find the slope of the tangent (curve)DisplacementFind the area under the curveYou do not know the initial or final position of the runner, just the displacement.
12Velocity-Time Graphs Describe the motion of each sprinter. A Constant velocityZero AccelerationPositive displacementBConstant AccelerationStarts from RestPositive displacementDescribe the motion of each sprinter.
13Velocity-Time Graphs D Constant Acceleration Positive Acceleration Comes to a StopZero displacementEConstant VelocityZero AccelerationNegative displacementCConstant AccelerationNegative AccelerationComes to a StopPositive displacement
14Sample QuestionWhat is the average acceleration of the car shown on the graph below?A. 0.5 m/s2B m/s2C. 2 m/s2D. -2 m/s2
15Acceleration The rate at which an object’s velocity changes Variable: aUnits: m/s2It is the change in velocity which measures the change in position. Thus, it is measuring a change of a change, hence why the square time unit.When the velocity of an object changes at a constant rate, it has constant acceleration
16Motion Diagrams & Acceleration In order for a motion diagram to display a full picture of an object’s movement, it should contain information about acceleration by including average acceleration vectors.The vectors are average acceleration vectors because motion diagrams display the object at equal time INTERVALS (intervals always mean average)Average acceleration vectors are found by subtracting two consecutive velocity vectors.
17Average Acceleration Vectors You will have: Δv = vf - vi = vf + (-vi).Then divide by the time interval, Δt. The time interval, Δt, is 1 s. This vector, (vf - vi)/1 s, shown in violet, is the average acceleration during that time interval.
18Average Acceleration Vectors vi = velocity at the beginning of a chosen time intervalvf = velocity at the end of a chosen time interval.∆v = change in velocity* Acceleration is equal to the change in velocity over the time interval** Since the time interval is 1s, the acceleration is equal to the change in velocity***Anything divided by 1 is equal to itself…
19Average vs. Instantaneous Acceleration Average AccelerationChange in velocity during some measurable time interval divided by the time intervalFound by plugging into the equationInstantaneous AccelerationChange in velocity at an instant of timeFound by calculating the slope of a velocity-time graph at that instant𝐚= ∆𝒗 𝒕
20Velocity & Acceleration How would you describe the sprinter’s velocity and acceleration as shown on the graph?
21Velocity & Acceleration Sprinter’s velocity starts at zeroVelocity increases rapidly for the first four seconds until reaching about 10 m/sVelocity remains almost constant
22Average vs. Instantaneous Acceleration What is the acceleration for the first four seconds?Refers to average acceleration because there is a time intervalSolve using the equation a = ∆v /tvi = 0 m/s; vf = 11 m/s; t = 4sa = (11m/s – 0 m/s)/ 4sa = 2.75 m/s2
23Average vs. Instantaneous Acceleration What is the acceleration at 5s?Refers to instantaneous acceleration because it is looking for acceleration at an instantNeed to find the slope of the line to solve for accelerationSlope is zero; thus instantaneous acceleration is zero at the instant of 5s.
24Instantaneous Acceleration Solve for the acceleration at 1.0 sDraw a tangent to the curve at t = 1sThe slope of the tangent is equal to the instantaneous acceleration at 1s.a = rise / run
25Instantaneous Acceleration The slope of the line at 1.0 s is equal to the acceleration at that instant .
26Positive & Negative Acceleration These four motion diagrams represent the four different possible ways to move along a straight line with constant acceleration.
27Object is moving in the positive direction Displacement is positiveThus, velocity is positiveObject is getting fasterAcceleration is positive
28Object is moving in the positive direction Displacement is positiveThus, velocity is positiveObject is getting slowerAcceleration is negative
29Object is moving in the negative direction Displacement is negativeThus, velocity is negativeObject is getting fasterAcceleration is negative
30Object is moving in the negative direction Displacement is negativeThus, velocity is negativeObject is getting slowerAcceleration is positive
31Positive & Negative Acceleration When the velocity vector and acceleration vector point in the SAME direction, the object is INCREASING SPEEDWhen the velocity vector and acceleration vector point in the OPPOSITE direction, the object is DECREASING SPEED
32+ UP - Down Displacement Velocity Acceleration Speeding UP Or Displacement & Velocity always have the same signDisplacementVelocityAccelerationSpeeding UPOrSlowing Down+UP-DownUp = sameDown = Different
33Sample QuestionHow can the instantaneous acceleration of an object with varying acceleration be calculated?A. by calculating the slope of the tangent on a distance versus time graphB. by calculating the area under the graph on a distance versus time graphC. by calculating the area under the graph on a velocity versus time graphD. by calculating the slope of the tangent on a velocity versus time graph
35Segmentt(s)vi (m/s)vf(m/s)∆vavg. a(m/s2)ins. AXi(m)Xf∆XA0 BCDE**Can not assume position on graph. Velocity time graphs can only be used to figure out displacement. You must be given an initial position.
363.2 Motion with Constant Acceleration Interpret position-time graphs for motion with constant accelerationDetermine mathematical relationships among position, velocity, acceleration, and timeApply graphical and mathematical relationships to solve problems related to constant acceleration.
37Constant acceleration: x-t Graphs Velocity is constantly increasing, which means more displacement.Slope must be getting steeper.Results in a curve that is parabolic.
38Constant acceleration: x-t Graphs x (m)t (s)Concave UP = +at (s)Concave UP = + ax (m)x (m)t (s)Concave Down = -aConcave Down = -ax (m)t (s)
39Kinematics EquationsThree equations that relate position, velocity, acceleration, and time.First two are derived from a v-t graph and the third is a substitution.Total of five different variables.Δx (displacement), vi (initial velocity), vf (final velocity), a (acceleration), and t (time).Must know any three in order to solve for the other two.
40First Kinematics Equation Remember that the slope of a v-t graph is the average acceleration.Rearranging the equation, gives us the first kinematics equation.Replace tf – ti with tvf = vi + at
41Second Kinematics Equation We remember that area of a v-t graph equals displacementBreak into two known shapes (rectangle & triangle).Total Area = Area of rectangle + area of triangleΔx = vit + ½ (vf –vi)tvf – vi = at(substitute)Δx = vit + ½ at2
42Third Kinematics Equation First equation substituted into the second to cancel out the time variable.vf = vi + at t = (vf – vi) / aΔx = vit + ½ at2Δx = vi((vf – vi)/a) + ½ a ((vf – vi)/a) 2Δx = vivf – vi2 + ½ a (vf2 – 2 vivf + vi2 )/a22a Δx = 2 vivf - 2 vi2 + vf2 – 2 vivf + vi22a Δx = - vi2 + vf2 (rearrange)SimplifyMultiply by 2a to get rid of fractionCombine like termsvf2 = vi2 + 2a Δx
43No displacement (position) KinematicsEquationUse when:NO ∆xNo displacement (position)No vfNo final velocityNo tNo timevf2 = vi2 + 2a ΔxΔx = vit + ½ at2vf = vi + atMust have 3 of the 5 variables to start. (∆x, vi, vf, a, t)Must always know acceleration.If acceleration is not given, then must solve for it first.
44Sample Problem Given Unknown Formula Substitution Answer An automobile starts at rest and speeds up at 3.5 m/s2 after the traffic light turns green. How far will it have gone when it is traveling at 25 m/s?vi = 0 m/svf = 25 m/sa = 3.5 m/s2∆x = ? mvf2 = vi2 + 2a ∆x(25m/s)2 = (0 m/s)2 + 2 (3.5 m/s2)(∆x )∆x = 89.3 mGivenUnknownFormulaSubstitutionAnswer
45Sample ProblemYou are driving a car, traveling at constant velocity of 25 m/s, when you see a child suddenly run onto the road. It takes you 0.45 s to react and apply the brakes. As a result, the car slows with a steady acceleration of 8.5 m/s2 and comes to a stop. What is the total distance that the car moves before it stops?ReactionStoppingt =0.45 sv = 25 m/svi = 25 m/svf = 0 m/sa = -8.5 m/s2∆xreaction = ? m∆xstopping = ? m𝑣= ∆𝑥 𝑡25 𝑚 𝑠 = ∆𝑥 0.45𝑠(0 m/s)2 = (25 m/s)2 + 2(-8.5m/s2)(∆x)∆x = m∆xstopping = mvf2 = vi2 + 2a Δxxtotal = m