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Motion is spelled K-I-N-E-M-A-T-I-C-S Rachael Kate Ana

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After reading this chapter you should be able to… Represent constant and non constant motion. Understand the difference between distance, displacement, speed, and velocity. Be able to create multiple representations of motion. Understand acceleration The goal of this chapter is to be able to represent motion in a variety of ways!

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How do we represent motion? Mechanics is the study of motion Kinematics-the description of motion Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations. Dynamics- the explanation of motion What can we measure? Distance (ruler) Time (stopwatch) Speed (distance/time)

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What is distance, displacement, speed, and velocity? Distance is the change in position, while displacement is when the ending position is different than the beginning position. Speed is the amount of distance traveled in a certain amount of time, while velocity is the displacement in a given time.

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What is distance, displacement, speed, and velocity? Cont’d Below is a list of key terms needed to understand this lesson: Position: where something is Clock reading: what the stopwatch says Distance: how much ground an object has covered during its motion. Time interval: difference in clock readings Displacement: change in position Velocity= displacement time Speed= distance time

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Differing between displacement and distance Let us test our understanding of the distinction between distance and displacement using the diagram below. This diagram shows that a person walks 6 meters East, 4 meters South, 6 meters West, and 4 meters North. It is evident that the person did walk a total of 20 meters, however his displacement is 0 meters. Throughout his travels he covered 20 meters of ground (the distance is 20 meters). On the other hand, when the person has finished his walk his starting position is no different then his beginning (the displacement has not changed).

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Let us take a look on how to represent constant motion There are several ways to represent motion: Motion diagram Position vs. time graph Velocity vs. time graph Mathematical representation Picture

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Types of constant motion There are five different types of constant motion. Moving slowly in the positive direction Moving slowly in the negative direction Moving quickly in the positive direction Moving quickly in the negative direction Not moving (standing still)

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How to draw motion diagrams Draw a dot to represent the object at the start and end of the total motion which you are observing. Draw a dot to represent the object at several equally spaced times. If the object is traveling at a constant speed, these will be equally spaced. Draw an arrow over each dot representing the velocity at that point. The arrow will point in the direction of motion and its length will represent the speed of the object at that point. If the object turns around, you may want to make two motion diagrams; one from the start until the point that the object turns around, and the other from the point it turns around to the end.

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Drawing motion diagrams cont’d The above motion diagram indicates a car traveling at a constant speed in the positive direction. If the car were traveling at a faster speed the arrows would be longer and more spread out. If it were traveling at a slower speed the arrows would be smaller in length and closer together. T=0 t = 0 The motion diagram for an object moving with a constant velocity.

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Constant motion chart WordsMotion Diagram Position vs. Time Graph Velocity vs. Time Graph Mathematical Representation Picture Moving slowly in the positive direction........ x = vt + x i v = v Moving slowly in the negative direction........ x = -vt + x i v = -v Moving fast in the positive direction.. x = vt + x i v = v Moving fast in the negative direction.. x = -vt + x i v = -v Not moving x = x i v = 0 x x x x x t t t t t t t t t t v v v v v t = 0 + – + –

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Experiment 1: Constant Motion Materials needed: stopwatch Meter stick Duct tape Someone to roller skate

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Experiment 1: Steps 1. Before beginning this experiment each group must choose who will be roller skating. Tell the roller skater not to speed up or slow down as they participate in the experiment. 2. To set-up make several marks down a sidewalk (not slanted) at intervals of 5 m. You should only need to go up to 30 m. 3. As the roller skater begins, someone must time how long he/she takes to get to each ticker mark. 4. This experiment may want to be repeated several times do to human error.

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Safety Precautions: The most important objective is not to injure the roller skater, so make sure to wear a helmet and plenty of pads. Make sure not to do this experiment on a busy sidewalk or any where there are cars because you do not want distractions. Make sure the skater knows how to stop!

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Experiment: Constant Motion cont’d 5 m10 m15 m20 m25 m30 m Measure the time that it takes for the skater to get to each of the marked sections and then record in a notebook. You want to draw a position vs. time graph.

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What to find during experiment: Position vs. time graph Time(s)Position (m) 10 15 20 25 30 5 Questions to Answer: 1.Write a mathematical model to describe the position vs. time graph. 2.Was the skater’s speed constant over the entire interval? How do you know?

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How can you represent non constant motion? Acceleration is the rate at which an object changes its velocity. Acceleration values are expressed in units of velocity/time. Typical acceleration units include the following: m/s/s mi/hr/s km/hr/s If an object is slowing down, then its acceleration is in the opposite direction of its motion.

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Acceleration motion diagrams To draw motion diagrams for an object that is accelerating, you would draw it very similar to the velocity motion diagrams. You would draw another arrow above the velocity arrows representing the acceleration. If the object is speeding up, the arrows will get progressively further apart. If it is slowing down, they will get progressively closer together.

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Motion diagrams cont’d T=0 This motion diagram shows acceleration in the positive direction.

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Drawing position vs. time graphs The slope of any position vs. time graph is the velocity of the object. Below is an example of accelerated motion. Observe that the line on the graph is curved. A curved line on a position vs. time graph signifies an accelerated motion. The initial small slope indicates a small initial velocity and the final large slope indicates a large final velocity. The negative slope of the line indicates a negative velocity.

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Velocity vs. time graphs In a velocity vs. time graph a diagonal line signifies an accelerated motion. If a free falling object undergoes an acceleration of 9.8 m/s/s, then one would expect the velocity vs. time graph to be a diagonal line. The graph to the left shows an object moving in the negative direction and speeding up, giving it a negative acceleration. The slope of any velocity vs. time graph is the acceleration of the object. Below is an example of a velocity vs. time graph: For velocity vs. time graphs, the area bounded by the line and the axes represents the distance traveled.

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Acceleration vs. time graphs Acceleration vs. time graphs give us information about acceleration and about velocity. In this graph, an ant is accelerating at 1 m/s 2 from t = 2 to t = 5 and is not accelerating between t = 6 and t = 7; that is, between t = 6 and t = 7 the ant’s velocity is constant. Acceleration vs. time graphs tell us about an object’s velocity in the same way that velocity vs. time graphs tell us about an object’s displacement. The change in velocity in a given time interval is equal to the area under the graph during that same time interval. Be careful: the area between the graph and the t-axis gives the change in velocity, not the final velocity or average velocity over a given time period.

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Acceleration chart Written description Motion diagram Position vs. time Velocity vs. time Acceleration vs. time Pos. direction, speeding up T=0..... Neg direction, speeding up..... Pos direction, slowing down..... Neg direction slowing down..... Turning around....... T=0

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Understanding mathematical representations of acceleration D=f(Vi,t,a) D=Vi*t+[1/2t(vf-vi)] D=Vi*t+1/2t(at) D=Vi*t+1/2at 2 Galileo’s Formula A=VF-VI t v

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Deriving new formulas Acceleration=Vf-Vid= Vit+1/2at 2 t T=Vf-Vi t D=ViVf –Vi 2 +1/2a(Vf 2 -2VfVi+Vi 2 ) a a 2ad=-Vi 2 +Vf 2 Vf 2 =Vi 2 +2ad (Final velocity) 2 = Initial velocity 2 +2(acceleration)(distance) Substitute t’s and a

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Math descriptions of Acceleration 1) D=vit+½at 2 2) Vf 2 =vi 2 +2ad 3) A=∆v/t D=distance Vi=initial velocity A=acceleration T=time Vf=final velocity Vi=initial velocity A=acceleration D=distance A=acceleration ∆v=final velocity- initial velocity (difference in velocity) T=time Now use these equations in the sample problems included to find the variable.

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Example Problem A runner that is moving 4m/s over a 5 second period, increases their speed to 5.5m/s. What is the distance in which the runner increases their speed over?

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Answer: example First, you need to find the acceleration of the runner by using the equation A=∆v/t, where you would get A=5.5m/s-4m/s divided by 5 seconds. Their acceleration would be 0.3m/s/s. After you find their acceleration, you use another equation, either D=vit+½at 2 or Vf 2 =vi 2 +2ad. Let’s use Vf 2 =vi 2 +2ad. (5.5m/s) 2 =(4.5m/s) 2 +2(0.3m/s)D 30.25=16+0.6D 14.25=0.6D D=23.75m

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Experiment 2: Accelerated Motion Things you need: 1 ramp 1 Ball 1 Ruler Stopwatch Pencil and notebook to record your information

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Experiment: cont’d 1) Measure distance from top of ramp to bottom of ramp 2) Split the ramp up into different sections of 20 cm each and mark each section 3) Hold the ball at the top of the ramp, and using your stopwatch, record how long it takes the ball to move from 0cm-20cm on the ramp. Repeat the experiment 3 times for the first section. 4) Repeat step 3, starting the ball at the top of the ramp and measuring how long the ball takes in each section as it rolls and picks up speed down the ramp.

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Acceleration experiment: safety precautions! Wear closed shoes! If the ball falls off the ramp and lands in your foot, you could hurt yourself! Tie your hair back (if you have long hair).

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Experiment: cont’d / 0cm 20cm 40cm 60cm 80cm 1 2 3 4 First you find the time from 0cm-20cm (section 1), then 20cm-40cm (section 2), then from 40cm-60cm (section 3), and then finally from 60cm-80cm(section 4). Start the ball from the top of the ramp every time.

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What to record during experiment: Position (cm) Time (s) Intial Velocity Final velocity (m/s) Acceleration (m/s/s) 0-20 20-40 40-60 60-80

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Chapter Summary As this chapter comes to a close you should now have a basic understanding of what Kinematics is and how it works. Kinematics is the motion of objects described using words, diagrams, numbers, graphs, and equations. (physicsclassroom.com) The basic way to solve a problem should include but not be limited to the following steps: 1. READ the problem. 2. VISUALIZE or DRAW the situation using motion diagrams, position vs. time graphs, velocity vs. time graphs, acceleration vs. time graphs, mathematical representations and pictures. (If this step is skipped the problem may not be properly understood) 3. What information is GIVEN? 4. What is problem asking for? What is NEEDED? 5. THINK! How are you going to approach the problem now? (this is usually the ‘time-consuming’ step… the more you practice the quicker this step becomes, hence the need for homework and problem drills) 6. Consider which EQUATION you will use. Is it APPLICABLE in this situation? 7. CALCULATE. 8. Think about it, does your answer make SENSE? (make sure that your answer fits with what the problem is asking) 9. Don’t forget UNITS!

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Summary You should now have a good understanding of the derivation of the following equations and know what each is used for: D=distance V i =initial velocity V f =final velocity A=acceleration T=time

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Summary cont’d The chart below shows what the three times of graphs discussed in this chapter can be used to find.

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Practice Problems

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1. An airplane accelerates down a runway at 3.20 m/s 2 for 32.8 s until it finally lifts off the ground. Determine the distance traveled before take-off. 2. A car accelerates from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled by the car. 3. A man runs at an accelerating speed of 7.10 m/s through a distance of 35.4 m. What is the acceleration of the man?

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Practice Problems Cont’d 4.The graph at right shows a velocity vs. time graph. Describe the motion of the object. 5. Find the acceleration of the graph to the right.

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Practice Problems Cont’d 6. The graph to the left shows a velocity vs. time graph. Determine the total distance traveled. 7. A stone is dropped into a well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

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Practice Problems Cont’d 8. Joe is driving through town at 25.0 m/s and begins to accelerate at a constant rate of –1.0 m/s 2. Eventually Joe comes to a complete stop. a. Represent Joe’s accelerated motion by sketching a velocity-time graph. Use the velocity-time graph to determine the distance traveled while decelerating. b. Use kinematics equations to calculate the distance which Joe travels while negatively accelerating. 9. Timmy travels 30.0 m/s for 10 seconds. He then accelerates at 3.00 m/s 2 for 5 seconds. a. Construct a velocity-time graph for Timmy’s motion. Use the graph to determine the total distance traveled. b. Divide Timmy’s motion into two time segments and use kinematics equations to calculate the total displacement.

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Practice Problems Cont’d 10. An athlete runs four laps of a 400 m track. What is the athlete’s total displacement? 11. A car accelerates from rest to 35m/s in 6 seconds. What is the acceleration of their car? 12. A person roller-skating starts moving at 7m/s at the top of a hill, and as they start moving down the hill, they start to accelerate. It takes them 4 seconds to skate 50m down the hill. What is their acceleration? 13. If a rocket is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the distance the rocket travels and the acceleration of the rocket?

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Practice Problems Cont’d 14. Kim is approaching a stoplight moving with a velocity of 30.0 m/s. The light turns yellow, and Kim applies the brakes and skids to a stop. If Kim's acceleration is -8.00 m/s 2, then determine the displacement of the car during the skidding process. 15. A motorist drives for 35 minutes at 85 km/hr and then stops for 15 minutes. He then continues driving North. His total trip is 130 km. in 2 hrs. a) What is the motorists average velocity? 16. A woman is speed walking for exercise at 2m/s. She has walked for 45 minutes. How far has she traveled?

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Practice Problems Cont’d 17. A caterpillar is ice skating (frictionless) with his girlfriend. He skates for 20 seconds over 15 meters. She skates the same distance, but it takes her 30 seconds. What is the difference of their velocities? 18. A girl is blowing a paper across the table. The paper is moving 0.5m/s. If the width of the table she is blowing it across is 2m long, would the paper fall off the table after 5 seconds of her blowing?

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Practice Problems Cont’d 19. At a school dance, a boy with a broken foot is excited that his favorite song is on. He starts to limp onto the dance floor at 0.8m/s. If he has 35m to go, and the song is 3 minutes long, will he get onto the dance floor while the song is still playing?

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Solutions

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Solutions to sample problems 1. Given a = 3.20 m/s 2 t = 32.8 s vi = 0 m/s d = vi*t + 0.5*a*t 2 d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2 d = 1720 m 2. acceleration = Change in velocity/time taken a = (46.1 m/s - 18.5 m/s)/(2.47 s) a = 11.2 m/s2 d = vi*t + 0.5*a*t2 d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s) 2 d = 45.7 m + 34.1 m d = 79.8 m 3. vf 2 = vi 2 + 2ad (7.10 m/s) 2 = (0 m/s) 2 + 2(a)(35.4 m) 50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a (50.4 m 2 /s 2 )/(70.8 m) = a a = 0.712 m/s 2 32.8 t 3.20

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Solutions 4. The object is moving in the positive direction with a constant velocity and no acceleration at letter A. At letter B, the object slows down while moving in the positive direction. It has a negative acceleration and then it stops. At letter C the object moves in the negative direction and speeds up with a negative acceleration. 5. The acceleration would be 4m/s 2. To do this problem pick 2 points on the graph and divide the change in y by the change in x.

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Solutions 6. Area = b * h Area = (6 s) * (30 m/s) Area = 180 m So 180 meters is the total distance traveled. 7. d = vi*t + 0.5*a*t 2 d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2 d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 ) d = -57.0 m deep (it is negative because it is in the negative direction)

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Solutions b) Given:vi = 25.0 m/s vf = 0.0 m/s a = –1.0 m/s2 Find:d = ?? vf 2 = vi 2 + 2*a*d (0 m/s) 2 = (25.0 m/s) 2 + 2 * (-1.0 m/s 2 )*d 0.0 m 2 /s 2 = 625.0 m 2 /s 2 + (-2.0 m/s 2 )*d 0.0 m 2 /s 2 - 625.0 m 2 /s 2 = (-2.0 m/s 2 )*d (-625.0 m 2 /s 2 )/(-2.0 m/s 2 ) =d d = 313 m The distance traveled can be found by calculating the area between the line on the graph and the axes. Area = 0.5*b*h Area = 0.5*(25.0 s)*(25.0 m/s) Area = 313 m 8. SAME!

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Solutions 9. a)Area = area of triangle + area of rectangle 1 + area of rectangle 2 Area = 0.5*bt*ht + b1*h1 + b2*h2 Area = 0.5*(5.0 s)*(15.0 m/s) + (10.0 s)*(30.0 m/s) + (5.0 s)*(30.0 m/s) Area = 37.5 m + 300 m + 150 m Area = 487.5 m

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Solutions b) First we want to find the distance for the first 10 seconds. d = v i *t + 0.5*a*t 2 d = (30.0 m/s)*(10.0 s) + 0.5*(0.0 m/s 2 )*(10.0 s) 2 d = 300 m + 0 m d = 300 m Now, we want to find the distance traveled for the last 5 seconds: d = v i *t + 0.5*a*t 2 d = (30.0 m/s)*(5.0 s) + 0.5*(3.0 m/s 2 )*(5.0 s) 2 d = 150 m + 37.5 m d = 187.5 m For the 15 seconds of motion: distance = 300 m + 187.5 m distance = 487.5 m

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Solutions 10. At the end of four laps, the athlete will be back at the starting line for the track, so the athlete’s total displacement will be zero. 11. You would use the equation A=∆v/t to find the car’s acceleration. A=35m/s-0m/s 6 seconds After finding the answer, you would round your answer to the nearest hundredth, which would make your answer A=5.83m/s2 12. You would use the equation D=vit+½at 2 to find the skaters acceleration. The equation you would end up with to solve the problem is: 50m=7m/s(4s)+½a(4s 2 ) 50m=28+8a 22=8a A=2.75m/s/s

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Solutions 13. First, to find the acceleration of the rocket, you would use the equation A=∆v/t, where A=444m/s-0m/s divided by 1.8s. The acceleration is rounded to 246.67m/s/s Then, using the acceleration you just found, find the distance that the rocket travels by using the equation D=vit+½at 2 D=½(246.67)(1.8) D=399.60m

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Solutions 14. Use the equation (0 m/s) 2 = (30.0 m/s) 2 + 2*(-8.00 m/s 2 )*d 0 m 2 /s 2 = 900 m 2 /s 2 + (-16.0 m/s 2 )*d (16.0 m/s 2 )*d = 900 m 2 /s 2 - 0 m 2 /s 2 (16.0 m/s 2 )*d = 900 m 2 /s 2 d = (900 m 2 /s 2 )/ (16.0 m/s 2 ) d = 56.3 m The car will skid a distance of 56.3 meters.

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Solutions 15. Let’s begin this problem by drawing a position vs. time graph. 35 60 90 120 49.6 T(s) 130 D(km) Now let’s draw a velocity vs. time graph. 85 km/hr 35 50 120 v T(s) Average Velocity= 130 km 2hrs = 65 km/hr

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Solutions 16. To find how far she has traveled, you would use the equation D=VT, where D is distance, V is velocity, and T is time. D=(2m/s)(45min) But you should change minutes to seconds, so The 1 mins cancel out, so your left with Finally, you need to multiply 45 and 60, so your left with 2700 seconds So, D=VT D=(2m/s)(2700s) D=5400s

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Solutions 17. To figure out his speed, you need to use the equation V=D/T, and after you plug in the known numbers, you get V=(15m)/20secs V=0.75m/s To figure out her speed, you would also use the equation V=D/T V=(15m)/(30secs) V=0.5m/s He is moving 0.25m/s faster than she is

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Solutions 18. First, you need to figure out how far the paper went using the equation D=VT. D=(0.5m/s)(5s) D=2.5m The distance of the table is 2m, and if the paper went 2.5m, then the paper would fall off the table. 19. First, find out how many seconds are in 3 minutes. 3 minutes(60)=180seconds Then you would use the equation T=D/V to find out how long it would take him to get onto the dance floor. T=(35m)/(0.8m/s) T=43.75s It will take him 43.75 seconds to get onto the dance floor, which leaves him with 136.25 seconds to enjoy his favorite song!

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