Presentation on theme: "9/15 do now Finish chapter 1 test Homework: 2.1, 2.3, 2.5, 2.7"— Presentation transcript:
19/15 do now Finish chapter 1 test Homework: 2.1, 2.3, 2.5, 2.7 No Post Tomorrow
2Motion Along a Straight Line Chapter 2Motion Along a Straight Line
3Goals for Chapter 2 To study motion along a straight line To define and differentiate average and instantaneous linear velocityTo define and differentiate average and instantaneous linear accelerationTo explore applications of straight-line motion with constant accelerationTo examine freely falling bodiesTo consider straight-line motion with varying acceleration
5Displacement: change in position, it is a vector quantity Displacement: change in position, it is a vector quantity. Its direction is from start to end.∆x = x2 – x1Time: change in time∆t = t2 – t1Average x-velocity: the displacement, ∆x, divided by the time interval ∆t .
6Distance and Average Speed Distance: length of the path, it depends on the path. It is a scalar quantity. It has no direction.Average x-speed: the distance traveled ∆s divided by the time interval ∆t. It is a scalar.Average speed vs. average velocityWhen Alexander Popov set a world record in 1994 by swimming m in sec, his average speed was (100.0 m) / (46.74 s) = m/s. but because he swam four lengths in a 25 meter pool, he started and ended at the same point and he had zero total displacement and zero average velocity!
7example What is the average velocity of the car? Position at t2 = 4.0 sPosition at t1 = 1.0 s+displacementx2 = 277 mx1 = 19 mWhat is the average velocity of the car?vav-x = (277 m – 19 m) / (4.0 s – 1.0 s) = 86 m/sThe average velocity is positive because it is moving in the positive direction.Note: you can choose any way as +.
8P-T graph of the car ∆x ∆t Slope = vav-x t (s) X (m) x1 = 19m
9Check your understanding 2.1 Each of the following automobile trips takes one hour. The positive x-direction is to the east.A travels 50 km due east.B travels 50 km due westC travels 60 km due east, then turns around and travels 10 km due westD travels 70 km due east.E travels 20 km due west, then turns around and travels 20 km due east.Rank the five trips in order of average x-velocity from most positive to most negative.Which trips, if any, have the same average x-velocity?For which trip, if any, is the average x-velocity equal to zero?4, 1, 3, 5, 21, 35
10Practice 2.2In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin in the nest and extend the +x-axis to the release point, what was the bird’s average velocity in m/sFor the return flight?For the whole episode, from leaving the nest to returning?-4.42 m/s0 m/s
11Practice 2.4Starting from a pillar, you run 200 m east (the +x-axis) at an average speed of 5.0 m/s, and then run 280 m west at an average speed of 4.0 m/s to a post. CalculateYour average speed from pillar to post,You average velocity from pillar to post.4.4 m/s-0.72 m/s
12Practice 2.6Two runners start simultaneously from the same point on a circular 200 m track and run in the same direction. One runs at a constant speed of 6.20 m/s, and the other runs at a constant speed of 5.50 m/s.When will the fast one first “lap” the slower one and how far from the starting point will each have run?When will the fast one overtake the slower one for the second time, and how far from the starting point will they be at that instant?286 s, 1770 m, 1570 m572 s, 3540 m, 3140 m
13Practice 2.8A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t) = αt2 – βt3, where α = 1.50 m/s2 and β = m/s3. Calculate the average velocity of the car for each time interval:t = 0 to t = 2.00 s;t = 0 to t = 4.00 st = 2.00 s to t = 4.00 s.
14exampleA cat runs along a straight line (the x-axis) from point A to point B to point C, as shown. The distance between points A and C is 5.00 m, the distance between points B and C is 10.0 m, and the positive direction of the x-axis points to the right. The time to run from A to B is 20.0 s, and the time from B to C is 8.00 s.BCAWhat is the average speed of the cat between points A and C?What is the average velocity of the cat between points A and C?
15Example - Walking 1/2 the time vs. Walking 1/2 the distance Tim and Rick both can run at speed vr and walk at speed vw, with vw < vr. They set off together on a journey of distance D. Rick walks half of the distance and runs the second half. Tim walks half of the time and runs the other half.a) Draw a graph showing the positions of both Tim and Rick versus time.b) Write two sentences explaining who wins and why.c) How long does it take Rick to cover the distance D?d) Find Rick's average speed for covering the distance D.e) How long does it take Tim to cover the distance?
16solutionTim wins because he takes short time to cover the same distance as Rick.a.txDD/2tTim½ tTimtRickc.d.e.
179/16 do now – on a new sheet Homework questions? Vectors V1 and V2 shown above have equal magnitudes. The vectors represent the velocities of an object at times t1 and t2, respectively. The average acceleration of the object between time t1 and t2 wasZeroDirected northDirected westDirected north of eastDirected north of westHomework questions?Homework: 2. 9, 2.11 and work sheet
182.2 Instantaneous velocity Instantaneous velocity is defined as the velocity at any specific instant of time or specific point along the path.Instantaneous velocity is a vector quantity, its magnitude is the speed, its direction is the same as its motion’s direction.How long is an instant?In physics, an instant refers to a single value of time.
19P2P1To find the instantaneous velocity at point P1, we move the second point P2 closer and closer to the first point P1 and compute the average velocity vav-x = ∆x / ∆t over the ever shorter displacement and time interval. Both ∆x and ∆t become very small, but their ratio does not necessarily become small.In the language of calculus, the limit of ∆x / ∆t as ∆t approaches zero is called the derivative of x with the respect to t and is written dx/dt.The instantaneous velocity is the limit of the average velocity as the time interval approaches zero; it equals the instantaneous rate of change of position with time.
20Example 2.1A cheetah is crouched 20 m to the east of an observer’s vehicle. At time t = 0 the cheetah charges an antelope and begins to run along a straight line. During the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation x = 20 m + (5.0 m/s2)t2.Find the displacement of the cheetah between t1 = 1.0 s and t2 = 2.0 sFind the average velocity during the same time interval.Find the instantaneous velocity at time t1 = 1.0 s by taking ∆t = 0.1 s, then ∆t = 0.01 s, then ∆t = s.Derived a general expression for the instantaneous velocity as a function of time, and from it find vx at t = 1.0 s and t = 2.0 s
219/17 do nowVectors A and B are shown. Vector C is given by C=B−A. The magnitude of vector A is 16.0 units, and the magnitude of vector B is 7.00 units. What is the magnitude of C?Homework: 2.9, 2.11, worksheet, ,
23Average and instantaneous velocities in x-t graph Secant line – average velocitytangent line – instantaneous velocity
24exampleThe automobiles make a 5 hour trip over a total distance of 200 km.Which car starts later?When does A & B pass each other?Which car reaches 200 km first?Calculate average speed of A and B.
25The Derivative…aka….The SLOPE! Suppose an eccentric pet ant is constrained to move in one dimension. The graph of his displacement as a function of time is shown below.tx(t)t + Dtx(t +Dt)ABAt time t, the ant is located at Point A. While there, its position coordinate is x(t).At time (t+Dt), the ant is located at Point B. While there, its position coordinate isx(t + Dt)
26The secant line and the slope Suppose a secant line is drawn between points A and B. Note: The slope of the secant line is equal to the rise over the run.tx(t)t + Dtx(t +Dt)ABThe slope of the secant line is average velocity
27The “Tangent” line READ THIS CAREFULLY! If we hold POINT A fixed while allowing Dt to become very small. Point B approaches Point A and the secant approaches the TANGENT to the curve at POINT A.tx(t)t + Dtx(t +Dt)ABBx(t +Dt)Ax(t)tt + DtWe are basically ZOOMING in at point A where upon inspection the line “APPEARS” straight. Thus the secant line becomes a TANGENT LINE. The slope of the tangent line is ____________________ velocity.
28The derivative Mathematically, we just found the slope! Lim stand for “__________" and it shows the ∆t approaches zero. As this happens the top numerator approaches a finite #.This is what a derivative is. A derivative yields a NEW function that defines the rate of change of the original function with respect to one of its variables. The above example shows the rate of change of "x" with respect to time.
29In most Physics books, the derivative is written like this: Mathematicians treat as a SINGLE SYMBOL which means find the derivative. It is simply a mathematical operation.The bottom line: The derivative is the slope of the line tangent to a point on a curve.
30exampleConsider the function x(t) = 3t +2; What is the time rate of change of the function (velocity)?This is actually very easy! The entire equation is linear and looks like y = mx + b . Thus we know from the beginning that the slope (the derivative) of this is equal to 3.We didn't even need to INVOKE the limit because the ∆t is cancel out.Regardless, we see that we get a constant.
31ExampleConsider the function x(t) = kt3, where k = proportionality constant.What happened to all the ∆t's ? They went to ZERO when we invoked the limit!What does this all mean?
32The MEANING?For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 meters.The derivative, however, tell us how our DISPLACEMENT (x) changes as a function of TIME (t). The rate at which Displacement changes is also called VELOCITY. Thus if we use our derivative we can find out how fast the object is traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12 m/s
37Addition and Subtraction Rule The derivative of the sum (or difference) of two or more functions is the sum (or difference) of the derivatives of the functions.Examplex =2t5 + 3t-137
38Chain ruleIf x is a function of f, and f is a function of t, so indirectly, x is a function of t: x(f(t))Example
39Class work Find the derivatives (dx/dt) of the following function x = t3x = 1/t = t-1x = (6t3 + 2/t)-2x = 16t2 – 16t + 4
40Average velocity vs. instantaneous velocity Example A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t) = αt2 – βt3, where α = 1.50 m/s2 and β = m/s3.Calculate the average velocity of the car for the time interval: t = 0 to t = 4.00 s;Determine the instantaneous velocity of the car at t = 2.00 s and t = 4.00 s.
41exampleAn object is moving in one dimension according to the formula x(t) = 2t3 – t2 – 4. find its velocity at t = 2 s.
42exampleThe position of an object moving in a straight line is given by x = (7 + 10t – 6t2) m, where t is in seconds. What is the object’s velocity at 4 seconds?
43exampleIf s is distance and t is time, what must be the dimensions of C1, C2, C3, and C4 in each of the following equations?S = C1tS = ½ C2t2S = C3sinC4tHint: the argument of any trigonometric function must be dimensionless.
44ExampleAn object moves vertically according to y(t) = 12 – 4t+ 2t3. what is its velocity at t = 3 s?92 m/s
45exampleAn object moves in one dimension such that x(t) is proportional to t5/2. this means v2 will be proportional tot3/2t7/2t7tt3
46exampleAn object is forced to move along the x axis in such a way that is displacement is given by x = t – 15t2 where x is in m and t is in s.Find expressions for the velocity of the object.At what time and distance from the origin is the velocity zero?At what time and location is the velocity -50 m/st = 0.67 s; x = 36.7 mt = 2.3 s; x = m
47exampleThe position of a vehicle moving on a straight track along the x-axis is given by the equation x(t) = 3t3 – 2t2 + t, where x is in meters and t is in seconds. What is its velocity at time t = 3s?70 m/s
48ExampleThe position of a vehicle moving on a straight track along the x-axis is given by the equation x(t) = t2 + 3t + 5, where x is in meters and t is in seconds. What is its velocity at time t = 5 s?13 m/s
49exampleThe position of a particle moving along the x-axis is given by the equation x(t) = 2 + 6t2, where x is in meters and t is in seconds. What is the average velocity during the interval t = 0 to t = 0.5 s?3 m/s
50exampleAn object’s motion is given by the equation x(t) =2 + 4t3. what is the equation for the object’s velocity?v(t) = 12t2
51Follow the motion of a particle The motion of the particle may be described from x-t graph.
52QuestionsThe graph above shows velocity v versus time t for an object in linear motion. Which of the following is a possible graph of position x versus time t for this object?
53Test your understanding 2.2 According to the graphRank the values of the particle’s x-velocity vx at the points P, Q, R, and S from most positive to most negative.At which points is vx positive?At which points is vx negative?At which points is vx zero?Rank the values of the particle’s speed at the points P, Q, R, and S from fastest to slowest.PRQ, SR, P, Q = S
54Example 2.10A physics professor leaves her house and walks along the side walk toward campus. After 5 min it starts to rain and she returns home. According to the graph, at which of the labeled points is her velocityZero?Constant and positive?Constant and negative?Increasing in magnitude?Decreasing in magnitude?IVIVIIIII
55exampleIn which of the following is the rate of change of the particle’s momentum zero?In which of the following is the particle’s acceleration constant?I, III, II, IIItdIIIIII
56exampleIn which of these cases is the rate of change of the particle’s displacement constant?IItvIIIIII
57example Which pair of graphs represents the same 1-dimensional motion? B.C.D.
58exampleThe graph represents the relationship between distance and time for an object. What is the instantaneous speed of the object att = 5.0 seconds?t = 2.0 seconds?1.5 m/s
59exampleThe graph represents the relationship between the displacement of an object and its time travel along a straight line. What is the average speed of the object during the first 4.0 seconds?2 m/s
60exampledtoAccording to the graph, the velocity of the object must beZeroConstant and positiveConstant and negativeIncreasingdecreasing
61exampledtoAccording to the graph, the acceleration of the object must beZeroConstant and positiveConstant and negativeIncreasingdecreasing
62Examplex(m)15The graph of an object’s motion (along a line) is shown. Findthe instantaneous velocity of the object at points A and B.the object’s average velocity.its acceleration.10B0.5 m/s0.5 m/s5A1020t (s)
63Examplex(m)15Refer to the graph. The object’s motion is represented by the curve. Findthe instantaneous velocity at point F.the instantaneous velocity at point D.the instantaneous velocity at point C.the instantaneous velocity at point E.D10CF5Et (s)1020
64Example 6.7 m/s Distance east (m) B 40 A -10 m/s 20 C t min 5 10 15 20 A girl walks along an east-west street, and a graph of her displacement from home is shown in the graph. Find herAverage velocity for the whole time intervalInstantaneous velocity at A, B and C6.7 m/sDistance east (m)Average velocity for the time interval t = 7 min to t = 14 minThe instantaneous velocity at t = 13.5 min and t = 15 minB40A-10 m/s20Ct min5101520-20--14 m/s; 6 m/s
65exampleA car with an initial positive velocity slows to a stop with a constant acceleration.Which graph best represents its position vs. time graph?Which graph best represents the velocity vs. time graph?tttABCttDE
66examplevAn object moves with a velocity vs. time graph as shown. The position vs. time graph for the same time period would betxxxABCtttxxttDE
67exampleAn object is moving in a straight line (the x-axis). The graph shows the x-coordinate of this object as a function of time. Which one of the following statements about this object is correct?Between points A and B, both the x-component of its average velocity and its average speed are greater than 0.75 m/s.Between points A and B, both the x-component of its average velocity and its average speed are less than 0.75 m/s.Between points A and B, the x-component of its average velocity is 0.75 m/s, but its average speed is greater than 0.75 m/s.Between points A and B, both the x-component of its average velocity and its average speed are equal to 0.75 m/s.
682.3 average and instantaneous acceleration The average acceleration of the particle as it moves from P1 to P2 is a vector quantity, whose magnitude equals to the change in velocity divided by the time interval.Velocity describes how fast a body’s position change with time.Acceleration describes how fast a body’s velocity change, it tells how speed and direction of motion are changing.
69Example 2.2An astronaut has left an orbiting spacecraft to test a personal maneuvering unit. As she moves along a straight line, her partner on the spacecraft measures her velocity every 2.0 s, starting at time t = 1.0 s:Find the average x-acceleration, and describe whether the speed of the astronaut increases or decreases, for each of these time intervals:t1 = 1.0 s to t2 = 3.0st1 = 5.0 s to t2 = 7.0 st1 = 9.0 s to t2 = 11.0 st1 = 13.0 s to t2 = 15.0 stvx1.0 s0.8 m/s9.0 s-0.4 m/s3.0 s1.2 m/s11.0 s-1.0 m/s5.0 s1.6 m/s13.0 s-1.6 m/s7.0 s15.0 s-0.8 m/s0.2 m/s2; speed increases-0.2 m/s2; speed decreases-0.3 m/s2; speed increases0.4 m/s2; speed decreases
70exampleA racquetball strikes a wall with a speed of 30 m/s. the collision takes 0.14 s. If the average acceleration of the ball during collision is 2800 m/s/s. what is the rebound speed?
71Instantaneous acceleration The instantaneous acceleration is the limit of average acceleration as the time interval approaches zero.
72Average and instantaneous acceleration Example 2.3 Suppose the x-velocity vx of a car at any time t is given by the equation: vx = 60 m/s + (.50 m/s2)t2Find the change in x-velocity of the car in the time interval between t1 = 1.0 s and t2 = 3.0 s.Find the average x-acceleration between t1 = 1.0 s and t2 = 3.0 s.Derive an expression for the instantaneous x-acceleration at any time, and use it to find the x-acceleration at t= 1.0 s and t = 3.0 s.4.0 m/s2.0 m/s2a = (1.0 m/s3)t; 1.0 m/s2; 3.0 m/s2
73example x(t) = at3 – bt2 + ct - d, 42 m/s2 16 m/s2 The position of an object as a function of time is given byx(t) = at3 – bt2 + ct - d,where a = 3.6 m/s3, b = 5.0 m/s2; c = 6 m/s; and d = 7.0 m(a) Find the instantaneous acceleration at t = s.(b) Find the average acceleration over the first seconds.42 m/s216 m/s2
74exampleThe position of a vehicle moving on a straight track along the x-axis is given by the equation x(t) = t2 + 3t + 5 where x is in meters and t is in seconds. What is its acceleration at time t = 5 s?(2 m/s2 )
75ExampleA particle moving along the x-axis has a velocity given by v = 4t – 2.50t2 cm/s for t in seconds. Find its acceleration att = 0.50 st = 3.0 s1.50 cm/s2-11.0 cm/s2
76exampleAn object moves vertically according to y(t) = 12 – 4t + 2t3. What is its acceleration at t = 3 s?36 m/s2
77ExampleAn object beginning at the origin moves in one dimension according to v(t) = 12/(6 + 7t). Determine the acceleration of the object.
78exampleThe equation of the position of an object moving along the x-axis is given by x(t) = 1.5t3 – 4.5t2 + .5t, where x is in meters and t is in seconds. What is the object’s displacement when its instantaneous acceleration is equal to zero?-2.5 m
79exampleThe velocity of a particle moving along the x-axis is given by the equation v(t) = 1 + 5t + 2t2, where v is in m/s and t is in seconds. What is the average acceleration during the interval t = 0 to t = 2?9 m/s2
80exampleThe position of a particle moving along te x-axis is given by the equation x(t) = 1 + 2t2 + 3t3, where x is in meters and t is in seconds. What is the average acceleration during the interval t = 0 to t = 1?13 m/s2
81exampleAn object’s motion is given by the equation x(t) = 4t + 4t3. what is the equation for the object’s acceleration?a(t) = 24t
82exampleA truck moving along a straight road at 30 m/s applies its breaks such that its velocity is given by the equation v(t) = 30 – 2t, where v is in m/s and t is in seconds. What is the truck’s acceleration at t = 1 s?-2 m/s2
833.59 (3000)An objects is forced to move along the x axis in such a way that ist displacement is given by x = t – 15t2 where x is in m and t is in s.Find expressions for the velocity and acceleration. Is the acceleration constant?What are the initial position and the initial velocity of the object?at what time and distance from the origin is the velocity zero?at what time and location is the velocity -50 m/sv = (20 – 30t) m/sa =– 30 m/s2xo = 30 m; vo = 20 m/st = 2.33s; x = -5.0 m
843.63 (3000)A mass at the end of a spring vibrates up and down according to the equation y = 8sin(1.5t) cm, where t is the time in seconds and the complete argument (angle) of the sine function, 1.5t is in radian.What is the velocity of the mass at t = 0.75 s?At t = 3.0 s?What is the maximum velocity of the mass?5.2 cm/s-2.5 cm/sv = (+ or -) 12.0 cm/s
85ExampleThe velocity of a car traveling on a straight track along the y-axis is given by the equation v(t) = -12t2 + 6t + 2, where x is in meters and t is in seconds.What is its velocity at time t = 3 s?What is its acceleration at time t = 3 s?-88 m/s-66 m/s/s
86Finding acceleration on a vx-t graph and ax-t graph Average acceleration can be determined by v-t graph
87Finding the acceleration on v-t graph A graph of and t may be used to find the acceleration.Average acceleration: the slope of secant line.Instantaneous acceleration: the slope of a tangent line at point.
88Caution: The sign of acceleration and velocity a is in the same direction as va is in the opposite direction as vv: posa: neg.v: posa: pos.v: neg.a: pos.v: neg.a: neg.
89We can obtain an object’s position, velocity and acceleration from it v-t graph pointxvaAGiven 0BCDENeg.Pos.Pos.Pos.
90Finding acceleration on a x-t graph On a x-t graph, the acceleration is given by the curvature of the graph.Curves up from the point: acceleration is positivestraight or not curves up or down: acceleration is zeroCurves down: acceleration is negative
91pointxvaABCDEFinding x, v, a in x-t graphNeg.pos.pos.pos.
92ExampleThe figure is graph of the coordinate of a spider crawling along the x-axis. Graph its velocity and acceleration as function of time.Vta
93Check your understanding 2.3 Refer to the graph,At which of the points P, Q, R, and S is the x-acceleration ax positive?At which points is the x-acceleration ax negative?At which points does the x-acceleration appear to be zero?At each point state whether the speed is increasing, decreasing, or not changing.SQP, RP: v is not change;Q: v is zero, changing from pos. to neg., first decrease in pos. then increase in neg.,R: v is neg., constant;S: v is zero, changing from neg. to pos., first decrease in neg. then increase in pos.,
94example1. At which of the labeled points is the magnitude of the velocity greatest?CpositionBDDAE2. At which of the labeled points is the velocity zero?timeC, E3. At which of the labeled points is the magnitude of the acceleration greatest?E
95exampleA child standing on a bridge throws a rock straight down. The rock leaves the child's hand at t= 0. Which of the graphs shown here best represents the velocity of the stone as a function of time?vtvvABCttvvDEtt
96examplexWhich of the following graphs best describes the x-component of the velocity as a function of time for this object?ttvtvtvABCtvtvDE
97exercise 2.35Two cars, A and B, move along the x-axis. The figure is a graph of the positions of A and B versus time.At what time(s), if any, do A and B have the same position?1 s, 3 ssketch velocity versus time for A & B .At what time(s), if any, does A pass B?At what time(s), if any, does B pass A?3 svv1 sttAB
982.3 motion with constant acceleration Given:(assume t0 = 0)derive:vx = vx0 + axtx = x0 + vx0 + ½ axt2vx2 – vx02 = 2ax(x – x0)Homework – extra credit
101a-t graph The area indicate the change in velocity during ∆t A horizontal line indicate the slope = 0, a = 0Since ax = ∆v / ∆t; ∆v = ax∙∆t which is represented by the area.The area indicate the change in velocity during ∆t
103Example 2.4A motorcyclist heading east through a small Iowa city accelerates after he passes the signpost marking the city limits. His acceleration is a constant 4.0 m/s2. At time t = 0 he is 5.0 m east of the signpost, moving east at 15 m/s.Find his position and velocity at time t = 2.0 s.Where is the motorcyclist when his velocity is 25 m/s?
104Example 2.5A motorist traveling with a constant speed of 15 m/s passes a school crossing corner, where the speed limit is 10 m/s. Just at the motorist passes, a pplice officer on a motorcycle stopped at the corner starts off in pursuit with constant acceleration of 3.0 m/s2.How much time elapses before the officer catches up with the motorist?What is the officer’s speed at that point?What is the total distance each vehicle has traveled at that point?
105Test your understanding 2.4 Four possible vx-t graphs are shown for the two vehicles in example 2.5. which graph is correct?
106exampleAt what time after t = 0 does the object again pass through its initial position?9 sDuring which interval does the particle have the same average acceleration as 12 s < t < 14 s?v (m/s)209s < t < 11s2s < t < 5s0s < t < 3s3s < t < 7s5s < t < 11s1051014-10-20
107Example 2.22 a. b. 1 mi/hr = 0.447 m/s 173 mi/h = 77.3 m/s The catapult of the aircraft carrier USS Abraham Lincoln accelerates an F/A-18 Hornet jet fighter from rest to a takeoff speed of 173 mi/h in a distance of 307 ft. Assume constant acceleration.Calculate the acceleration of the fighter in m/s2Calculate the time required for the fighter to accelerate to takeoff speed.1 mi/hr = m/s 173 mi/h = 77.3 m/s1 ft = m 307 ft = 93.6 ma.ax = 32.0 m/s2b.t = 2.42 s
108ExamplevtThe graph above shows velocity v versus time t for an object in linear motion. Sketch a graph of position x versus time t for this object?xt
1092.5 Free Falling Bodies a = -g If we ignore air friction and the effects due to the earth’s rotation, all objects fall and rise at the constant acceleration.The constant acceleration of a freely falling body is called the acceleration due to gravity, and we use letter g to represent its magnitude. Near the earth’s surface g = 9.81 m/s/s = 32 ft/s/sOn the surface of the moon, g = 1.6 m/s/sOn the surface of the sun, g = 270 m/s/sa = -g
110Example 2.6A one-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest and falls freely. Compute its position and velocity after .1.0 s. 2.0 s, and 3.0 s.
113Example 2.7You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.80 m/s2. findThe position and velocity of the ball 1.00 s and 4.00 s after leaving your handThe velocity when the ball is 5.00 m above the railingThe maximum height reached and the time at which it is reachedThe acceleration of the ball when it is at its maximum height.
115Example 2.8Find the time when the ball in Example 2.7 is 5.00 m below the roof railing.
116Check your understanding 2.5 If you toss a ball upward with a certain initial speed, it falls freely and reaches a maximum height h at time t after it leaves your hand.If you throw the ball upward with double the initial speed what new maximum height does the ball reach?If you throw the ball upward with double the initial speed, how long does it take to reach its maximum height?4h2t
117ExampleA rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. If it falls the second half of the distance in t2 seconds, what is the value of t2/t1? (ignore air resistance)
118ExampleAn object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the object travels during the first 6 s of its fall?176 m
119ExampleThe position of an object is given by the equating x = 3.0t t + 4.5, where x is in meters and t is in seconds. What is the instantaneous acceleration of the object at t = 3.00 s?6 m/s2
1202.6 velocity and position by integration Finding v(t) and x(t) when given a(t) In the case of straight-line motion, if the position x is a known function of time, we can find vx = dx/dt to find x-velocity. And we can use ax = dvx/dt to find the x-acceleration as a function of timeIn many situations, we can also find the position and velocity as function of time if we are given function ax(t).
121The “AREA”In v-t graph, the area under the line represent displacement.v (m/s)v (m/s)x = areax = areat(s)t(s)t1t2t1t2v (m/s)v (m/s)v (m/s)v (m/s)How ever, if acceleration is not constant, how can we determine x(t)?v(t)t(s)t1t1t1t2
122Zoom inWe have learned that the rate of change of displacement is defined as the VELOCITY of an object. Consider the graph below∆Area = v(t)∙dtt1t2v (m/s)t(s)tv(t)dtTOTAL DISPLACEMENT = ∑∆Area = ∑v(t)∙dt
123The “Integral” – the area The temptation is to use the conventional summation sign “S" . The problem is that you can only use the summation sign to denote the summing of DISCRETE QUANTITIES and NOT for something that is continuously varying. Thus, we cannot use it.When a continuous function is summed, a different sign is used. It is called an Integral, and the symbol looks like this:When you are dealing with a situation where you have to integrate, realize:• WE ARE GIVEN: the derivative already• WE WANT: The original function x(t)So what are we basically doing?WE ARE WORKING BACKWARDS!!!!! OR FINDING THE ANTI -DERIVATIVE
124Example These are your LIMITS! 45 m An object is moving at velocity with respect to time according to the equation v(t) = 2t.a) What is the displacement function? Hint: What was the ORIGINAL FUCNTION BEFORE the “derivative? was taken?b) How FAR did it travel from t = 2s to t = 7s?These are your LIMITS!45 mYou might have noticed that in the above example we had to find the change(D) over the integral to find the area, that is why we subtract. This might sound confusing. But integration does mean SUM. What we are doing is finding the TOTAL AREA from 0-7 and then the TOTAL AREA from 0-2. Then we can subtract the two numbers to get JUST THE AREA from 2-7.
125In summary…So basically derivatives are used to find SLOPES and Integrals are used to find AREAS.
126Example Here is a simple example of which you may be familiar with: Assume we know the circumference of a circle is 2pr, where r is the radius. How can we derive an expression for the area of a circle whose radius is R?We begin by taking a differential HOOP of radius "r" and differential thickness “dr” as shown.If we determine the area of JUST OUR CHOSEN HOOP, we could do the calculation for ALL the possible hoops inside the circle.Having done so, we would then SUM up all of those hoops to find the TOTAL AREA of the circle. The limits are going to be the two extremes, when r = R and when r = 0
127Backward power rule ∫xn dx = + C ∫xndx = - ∫ ∫ dx = x + C d (xn+1) = (n+1) xn∫xn dx = Cxn+1n+1∫xndx =bn+1n+1aban+1General expression, no limitsEvaluation with given limits∫abdx = b - aSpecial case: n = 0∫ dx = x + Cor
128Addition/subtraction rule constant multiplier rule ∫ (u v) dx = ∫ u dx ∫ v dx+-+-∫ au dx = a ∫ u dx
129Example 2.9Sally is driving along a straight highway in her classic 1965 Mustang. At time t = 0, when Sally is moving at 10 m/s in the positive x-direction, she passes a signpost at x = 50 m. her x-acceleration is a function of time:find her x-velocity and position as functions of timeWhen is her x-velocity greatest?What is the maximum x-velocity?Where is the car when it reaches the maximum x-velocity?
131Example 2.10Use Eqs.To find vx and x as functions of time in the case in which the acceleration is constant.
132ExampleA particle moving in one dimension has a position function defined as: x(t) = 6t4-2tAt what point in time does the particle change its direction along the x-axis?b) In what direction is the body traveling when its acceleration is 12 m/s/s?a)t = sb)t = sv = m/s
133Example 2.50The acceleration of a bus is given by ax(t)=αt, where α = 1.2 m/s3.If the bus’s velocity at time t = 1.0 s is 5.0 m/s, what is its velocity at time t = 2.0 s?If the bus’s position at time t = 1.0 s is 6.0 m, what is its position at time t = 2.0 s?Sketch ax-t, vx-t and x-t graphs for the motion.
134exampleAn object initially at rest experiences a time-varying acceleration given by a = (2 m/s3)t for t >= 0. How far does the object travel in the first 3 seconds?9 ma = (2 m/s3)tvx = (1 m/s3)t2vx = vox + (2 m/s3)½ t2vox = 0vx = (1 m/s3)t2x = xo + (1 m/s3)(1/3)t3x(3 s) – x (0 s) = (1 m/s3)(1/3)(3 s)3x(3 s) – x (0 s) = 9 m
135Check your understanding 2.6 If the x-acceleration ax is increasing with time, will the vx-t graph beA straight line,Concave upConcave down