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1-D Kinematics. Kinematics Kinematics is the branch of mechanics that describes the motion of objects without necessarily discussing what causes the motion. 1-Dimensional Kinematics (or 1 - Dimensional motion) refers to motion in a straight line . Distance

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## 1-D Kinematics

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**Kinematics**• Kinematics is the branch of mechanics that describes the motion of objects without necessarily discussing what causes the motion. • 1-Dimensional Kinematics (or 1 - Dimensional motion) refers to motion in a straight line. • Distance • The total length of the path traveled by an object is called distance. • “How far have you walked?” is a typical distance question. • The SI unit of distance is the meter (m).**Displacement (Dx)**• The change in the position of a particle is called displacement. • ∆ is a Greek letter used to represent the words “change in”. ∆x therefore means “change in x”. It is always calculated by final value minus initial value. • “How far are you from home?” is a typical displacement question. • The SI unit for displacement is the meter. Calculation of displacement: ∆x = xf - xi**distance**• Distance vsDisplacement • A picture can help you distinguish between distance and displacement. displacement 100m 50m**Distance = 100m**• Another example distance displacement • Walk east 100m • Turn and walk west 100m Displacement = 100m Distance = 200m Displacement = 0m**Questions**• Does the odometer in your car measure distance or displacement? • Distance • Can you think of a circumstance in which it measures both distance and displacement? • If you go one direction and stop it will give both distance and displacement**If Dx is the displacement of a particle, and d is the**distance the particle traveled during that displacement, which of the following is always a true statement? a) d = |∆x| b) d < |∆x| c) d > |∆x| d) d > |∆x| e) d < |∆x|**A particle moves from x = 1.0 meter to x = -1.0 meter.**• What is the distance d traveled by the particle? • What is the displacement of the particle? 2m -3 m -2 m -1 m 0 m 1 m 2 m 3 m -2m, must show direction**You are driving a car on a circular track of diameter 40**meters. After you have driven around 2 ½ times, how far have you driven, and what is your displacement? C = 2πr C = 2π∙20m 40m C = 125.66m Distance = 2.5∙125.66m = 314.16m Displacement = 40m**Average Speed**• Average speed describes how fast a particle is moving. The equation is: • where: save= average speed d = distance ∆t = elapsed time The SI unit of speed is the m/s save = d/t • Average speed is • always a positive • number.**Average Speed**• Average velocity describes how fast the displacement is changing. The equation is: • where: vave= average velocity ∆x= displacement ∆t = elapsed time The SI unit of speed is the m/s vave = ∆x/ ∆t • Average velocity is • + or – depending on direction.**How long will it take the sound of the starting gun to reach**the ears of the sprinters if the starter is stationed at the finish line for a 100 m race? Assume that sound has a speed of about 340 m/s. ∆t = ? ∆x = 100m vavg = 340 m/s vave = ∆x/ ∆t ∆t= ∆x/ vave ∆t= 100m / 340 m/s = 0.294 s**1.**∆t = ? ∆x = 1km = 1000m vavg = 10 m/s • You drive in a straight line at 10 m/s for 1.0 km, and then you drive in a straight line at 20 m/s for another 1.0 km. What is your average velocity? vave = ∆x/ ∆t ∆t= ∆x/ vave ∆t= 1000m / 10 m/s = 100 s 2. ∆t = ? ∆x = 1km = 1000m vavg = 20 m/s vave = ∆x/ ∆t ∆t= ∆x/ vave ∆t= 1000m / 20 m/s = 50 s 3. ∆x = 2000m vavg = ? m/s ∆t total= 150s vave = ∆x/ ∆t = 2000 m /150s = 13.3 m / s**Qualitative Demonstrations**1) Demonstrate the motion of a particle that has an average speed and an average velocity that are both zero. 2) Demonstrate the motion of a particle that has an average speed and an average velocity that are both nonzero. 3) Demonstrate the motion of a particle that has an average speed that is nonzero and an average velocity that is zero. 4) Demonstrate the motion of a particle that has an average velocity that is nonzero and an average speed that is zero.**s= 0.5 m / s**d = 10m t = ? s • You are a particle located at the origin. Demonstrate how you can move from x = 0 to x = 5.0 and back with an average speed of 0.5 m/s. • What the particle’s average velocity for the above demonstration? • 0 m/s because velocity takes into account the is a sign change average velocity • in the positive x is +0.5 m/s • in the negative x is – 0.5 m/s t= d/save save = d/t Cover the total distance in 20s t= 10 m / 0.5 m/s = 20 s**x (m)**• Graphical Problem 0 m t (s) Demonstrate the motion of this particle**x(m)**• Graphical Problem 0 m t (s) Demonstrate the motion of this particle**B**x (m) A ∆x • Graphical Problem ∆t 0 m t (s) vave = ∆x/ ∆t What physical feature of the graph gives the constant velocity from A to B?**x (m)**B • Graphical Problem: Determine the average velocity from the graph. 2.0 A ∆x 1.0 ∆t 0 t (s) 2.0 4.0 6.0 -1.0 -2.0 vave = 1 m/ 3 s = 1 / 3 m/ s**x (m)**• Graphical Problem 0 m t (s) Demonstrate the motion of these two particle**X (m)**• Graphical Problem 0 m t (s) Demonstrate the motion of these two particle**X (m)**• Graphical Problem 0 m t (s) What kind of motion does this graph represent?**x (m)**A B ∆x • Graphical Problem ∆t 0 m T (s) vave = ∆x/ ∆t Can you determine average velocity from the time at point A to the time at point B from the raph?**x (m)**• Graphical Problem: Determine the average velocity between 1 and 4 seconds. ∆x 2.0 ∆t 1.0 0 t (s) 2.0 4.0 6.0 -1.0 -2.0 vave = 1 m / 4 s = 0.25 m/s vave = ∆x/ ∆t**Instantaneous Velocity**• The velocity at a single instant in time. • If the velocity is uniform, or constant, the instantaneous velocity is the same as the average velocity. • If the velocity is not constant, than the instantaneous velocity is not the same as the average velocity, and we must carefully distinguish between the two.**vins = ∆x/ ∆t**X • Instantaneous Velocity ∆x ∆t 0 m t Draw a tangent line to the curve at B. The slope of this line gives the instantaneous velocity at that specific time.**X**• Determine the instantaneous velocity at 1 second. ∆x 2.0 ∆t 1.0 0 t 2.0 4.0 6.0 -1.0 -2.0 vave = ∆x/ ∆t vave = 1 m/ 1 s = 1 m/s**Acceleration (a)**• Any change in velocity over a period of time is called acceleration. • The sign (+ or -) of acceleration indicates its direction. • Acceleration can be… • speeding up • slowing down • turning**Questions**• If acceleration is zero, what does this mean about the motion of an object? it means the velocity is constant at 0 m/s or some velocity greater than 0 m/s The object has a constant speed, and it is moving in a straight line • Is it possible for a racecar circling a track to have zero acceleration? No because the race car is changing direction. When an object changes direction it changes velocity, so it has an acceleration present.**Uniform (Constant) Acceleration**• In Physics B, we will generally assume that acceleration is constant. • With this assumption we are free to use this equation: • The SI unit of acceleration is the m/s2. • Acceleration in 1-D Motion has a sign! • If the sign of the velocity and the sign of the acceleration is the same, the object speeds up. • If the sign of the velocity and the sign of the acceleration are different, the object slows down. a = ∆v / ∆t**Qualitative Demonstrations**1) Demonstrate the motion of a particle that has zero initial velocity and positive acceleration. 2) Demonstrate the motion of a particle that has zero initial velocity and negative acceleration. 3) Demonstrate the motion of a particle that has positive initial velocity and negative acceleration. 4) Demonstrate the motion of a particle that has negative initial velocity and positive acceleration.**∆v = vf – vi = 180 mph – 0 mph = 180 mph**a = ? mph/s • Practice Problem: A 747 airliner reaches its takeoff speed of 180 mph (80.5 m/s) in 30 seconds. What is its average acceleration? ∆t = tf – ti = 30 s – 0 s = 30 s a = ∆v / ∆t a = 180mph / 30 s = 6mph / s ∆v = vf – vi = 80.5 m / s– 0 m / s = 80.5 m / s a = ? m/s2 ∆t = tf – ti = 30 s – 0 s = 30 s a = ∆v / ∆t a = 80.5 m / s / 30 s = 2.683 m / s2**a = -1.81 m/s2**• A horse is running with an initial velocity of 11 m/s, and begins to accelerate at –1.81 m/s2. How long does it take the horse to stop? ∆v = vf – vi = 0 m /s – 11 m /s = -11 m /s ∆t = ? s ∆t = ∆v /a ∆t = -11 m /s / -1.81 m / s2 = 6.1 s**v (m/s)**• Graphical Problem 0 m T (s) Demonstrate the motion of this particle. Is it accelerating?**v (m/s)**• Graphical Problem 0 m T (s) Demonstrate the motion of this particle. Is it accelerating?**B**v (m /s) A ∆v • Graphical Problem ∆t 0 m t (s) a = ∆v/ ∆t What physical feature of the graph gives the acceleration from A to B?**v (m/s)**• Graphical Problem: Determine the acceleration from the graph. 2.0 ∆v 1.0 ∆t 0 t (s) 2.0 4.0 6.0 -1.0 -2.0 a = 1 m /s / 1 s = 1 m/s2 a= ∆v/ ∆t**Area of a triangle = ½ b ∙h**v (m/s) • Graphical Problem: Determine the displacement of the object from 0 to 4 seconds.. 2.0 h 1.0 0 t (s) b 2.0 4.0 6.0 -1.0 h -2.0 b Atriangle = ½ b ∙h = ½ ∙ 2 ∙2 = 2 m Atriangle = ½ b ∙h = ½ ∙ 2 ∙2 = - 2 m disp = 2 m – 2m = 0 m**Need to do this example below**• 5-14-10 1-D kinematics slide 41.xlsx**Pinewood Derby**• On your graph paper, do the following. a) Draw a position vs time graph for the car. b) Draw tangent lines at three different points on the curve to determine the instantaneous velocity at all three points. c) On a separate graph, draw a velocity vs time graph using the instantaneous velocities you obtained in the step above. d)From your velocity vs time graph, determine the acceleration of the car.**You see the car move faster and faster as time elapses.**• For each successive ∆t, ∆x is bigger. • The velocity vs time graph reflects the increasing velocity • Uniformly Accelerating Objects**What are the signs of velocity and acceleration? How is**speed changing? This object is moving with positive velocity and acceleration. Speed is increasing. This object is moving in the negative velocity and acceleration. Speed is increasing. This object is moving with negative velocity and positive acceleration. Speed is decreasing.**x**v a • Draw Graphs for Stationary Particles 0 m 0 m / s 0 m / s2 t t t position vs time position vs time position vs time**x**v a • Draw Graphs for Constant Non-zero Velocity 0 m 0 m / s 0 m / s2 t t t position vs time position vs time position vs time**v**x a • Draw Graphs for Constant Non-zero Acceleration increasing and positive velocity 0 m 0 m / s 0 m / s2 t t t position vs time position vs time position vs time**v**x a • Draw Graphs for Constant Non-zero Acceleration decreasing and negative velocity 0 m 0 m / s 0 m / s2 t t t position vs time position vs time position vs time**v**x a • Draw Graphs for Constant Non-zero Acceleration increasing and negative velocity 0 m 0 m / s 0 m / s2 t t t position vs time position vs time position vs time**v**x a • Draw Graphs for Constant Non-zero Acceleration decreasing and negative velocity 0 m 0 m / s 0 m / s2 t t t position vs time position vs time position vs time**v = v0 + at**On the AP equation sheet x = x0 + 1/2at2 • Kinematic Equations for uniformly accelerating objects x = v0t + 1/2at2 x = x0 + v0t + 1/2at2 On the AP equation sheet v2 = v02 + 2a(∆x) v2 = v02 + 2a(x – x0) On the AP equation sheet

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