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# Lecture 02 Kinematics in one dimension

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### Lecture 02 Kinematics in one dimension

1. 1. Kinematics in One Dimension<br />
2. 2. The Cheetah: A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds.<br />
3. 3. B<br />d= 20 m<br />A<br />Distance and Displacement<br />Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below:<br />Distancedis a scalar quantity (no direction):<br />Contains magnitude only and consists of a number and a unit.<br />
4. 4. B<br />Δs= 12 m, 20o<br />A<br />q<br />Distance and Displacement<br />Displacement is the straight-line separation of two points in a specified direction.<br />A vector quantity:<br />Contains magnitude AND direction, a number,unit & angle.<br />
5. 5. Δx<br />8 m,E<br />x = +8<br />x = -4<br />12 m,W<br />Distance and Displacement<br />For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W.<br />Net displacement Δxis from the origin to the final position:<br />x<br />Δx= 4 m, W<br />What is the distance traveled?<br />d = 20 m !!<br />
6. 6. d<br />t<br />20 m<br /> 4 s<br />B<br />vs= = <br />A<br />Definition of Speed<br />Speed is the distance traveled per unit of time (a scalar quantity).<br />d= 20 m<br />vs= 5 m/s<br />Not direction dependent!<br />Time t = 4 s<br />
7. 7. s = 20 m<br />B<br />Δx=12 m<br />A<br />20o<br />Time t = 4 s<br />Definition of Velocity<br />Velocity is the displacement per unit of time. (A vector quantity.)<br />= 3 m/s at 200 N of E<br />Direction required!<br />
8. 8. B<br />s = 20 m<br />C<br />A<br />Time t = 4 s<br />Average Speed and Instantaneous Velocity<br /><ul><li>The averagespeed depends ONLY on the distance traveled and the time required.</li></ul>The instantaneousvelocity is the magn-itude and direction of the velocity at a par-ticular instant. (v at point C)<br />
9. 9. s1 = 200 m<br />s2 = 300 m<br />Avg. speed= 8 m/s<br />Example 1.A runner runs 200 m, east, then changes direction and runs 300 m, west. If the entire trip takes 60 s, what is the average speed and what is the average velocity?<br />Recall that average speed is a function only of total distance and total time:<br />start<br />Total distance: s = 200 m + 300 m = 500 m<br />Direction does not matter!<br />
10. 10. t = 60 s<br />x1= +200 m<br />x= -100 m<br />xo = 0<br />Average velocity:<br />Example 1 (Cont.)Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters. <br />xo= 0 m; x= -100 m<br />Direction of final displacement is to the left as shown.<br />Note: Average velocity is directed to the west.<br />
11. 11. Definition of Acceleration<br /><ul><li>The direction of accel- eration is same as direction of force.
12. 12. The acceleration is proportional to the magnitude of the force.
13. 13. An acceleration is the change in velocity per unit of time. (A vector quantity.)
14. 14. A changeinvelocity requires the application of a push or pull (force).</li></ul>A formal treatment of force and acceleration will be given later. For now, you should know that:<br />
15. 15. F<br />a<br />2F<br />2a<br />Acceleration and Force<br />Pulling the wagon with twice the force produces twice the acceleration and acceleration is in direction of force.<br />
16. 16. Force<br />+<br />vo= +8 m/s<br />Example 3 (No change in direction):A constant force changes the speed of a car from 8 m/s to 20 m/s in 4 s. What is average acceleration?<br /> t = 4 s<br />v = +20 m/s<br />Step 1. Draw a rough sketch.<br />Step 2. Choose a positive direction (right).<br />Step 3. Label given info with + and - signs.<br />Step 4. Indicate direction of force F.<br />
17. 17. Example 3 (Continued):What is average acceleration of car?<br />Force<br />+<br /> t = 4 s<br />v= +20 m/s<br />vo= +8 m/s<br />Step 5. Recall definition of average acceleration.<br />
18. 18. Graphical Analysis<br />slope:<br />velocity:<br />acceleration:<br />
19. 19. Position vs time graph (velocity)<br />x, (m)<br />
20. 20. velocity vs time graph (acceleration)<br />Δvor v= width<br />Δt = length<br />Δx= vΔt<br />area=length x width<br />v, (m/s)<br />The area under v against t graph is the distance traveled by the object!<br />
21. 21. Graphical Analysis<br />slope<br />x2<br />Dx<br />Dx<br />Displacement, x<br />x1<br />Dt<br />Dt<br />t2<br />t1<br />Time<br />Average Velocity:<br />Instantaneous Velocity:<br />
22. 22. Uniform Acceleration in One Dimension:<br />Motion is along a straight line (horizontal, vertical or slanted).<br />Changes in motion result from a CONSTANT force producing uniform acceleration.<br />The velocity of an object is changing by a constant amount in a given time interval.<br />The moving object is treated as though it were a point particle.<br />
23. 23. Average velocity for constant a:<br />setting to = 0<br />combining both equations:<br />For constant acceleration:<br />
24. 24. Formulas based on definitions:<br />Derivedformulas:<br />For constant acceleration only<br />
25. 25. vo = 400 ft/s<br />Δx =300 ft<br />v = 0<br />+<br />Example 6:An airplane flying initially at 400 ft/s lands on a carrier deck and stops in a distance of 300 ft. What is the acceleration?<br />Step 1. Draw and label sketch.<br />Step 2. Indicate + direction <br />
26. 26. +<br />Example: (Cont.)<br />vo = 400 ft/s<br />v = 0<br />Δx =300 ft<br />Step 3. List given; find information with signs.<br />Given:vo = 400 ft/s - initial velocity of airplane<br />v = 0 - final velocity after traveling Δx = +300 ft<br />Find:a = ? - acceleration of airplane<br />
27. 27. 0<br />-vo2<br /> 2x<br />-(400 ft/s)2<br /> 2(300 ft)<br />a = = <br />Given:vo = +400 ft/s<br />v = 0<br />Δx= +300 ft<br />Step 4. Select equation that contains aand not t.<br />v 2- vo2= 2aΔx<br />a = - 300 ft/s2<br />Why is the acceleration negative?<br />Because Force is in a negative direction which means that the airplane slows down<br />
28. 28. +<br />Δx<br />-2.0 m/s<br />5.0 m<br />8.0 m/s<br />t = 4.0 s<br />Example 5:A ball 5.0 m from the bottom of an incline is traveling initially at 8.0 m/s. Four seconds (4.0 s)later, it is traveling down the incline at 2.0 m/s. How far is it from the bottom at that instant?<br />Given: d = 5.0 m - distance from initial position of the ball<br />vo= 8.0 m/s - initial velocity<br /> v = -2.0 m/s - final velocity after t = 4.0 s<br />Find: x = ? - distance from the bottom of the incline<br />
29. 29. Given: d = 5.0 m <br />vo= 8.0 m/s - initial velocity<br /> v = -2.0 m/s - final velocity after t = 4.0 s<br />Find: x = ? - distance from the bottom of the incline<br />Solution: <br />where<br />x = 17.0 m<br />
30. 30. W<br />g<br />Earth<br />Acceleration Due to Gravity<br />Every object on the earth experiences a common force: the force due to gravity.<br />This force is always directed toward the center of the earth (downward).<br />The acceleration due to gravity is relatively constant near the Earth’s surface.<br />
31. 31. Gravitational Acceleration<br />In a vacuum, all objects fall with same acceleration.<br />Equations for constant acceleration apply as usual.<br />For an object that is thrown vertically upward, the time in going up is equal to the time in going down.<br />The velocity magnitude is the same at the same height from a certain reference.<br />Near the Earth’s surface:<br />(Air resistance is negligible)<br />a = g = -9.80 m/s2 or -32 ft/s2<br /> Directed downward (usually negative).<br />
32. 32. UP = +<br />Sign Convention:A Ball Thrown Vertically Upward<br />v = 0<br />a = -<br />y = +<br />a = -<br /><ul><li>Displacement is positive (+) or negative (-) based on LOCATION. </li></ul>y = +<br />y = +<br />v = +<br />a = -<br />v = -<br />y = 0<br />y = 0<br />v = -<br />a = -<br />Release Point<br /><ul><li>Velocity is positive (+) or negative (-) based on direction of motion.</li></ul>y = -Negative<br />v= -Negative<br />a = -<br /><ul><li>Acceleration is (+) or (-) based on direction of force (weight).</li></li></ul><li>+<br />a = g<br />Example 7:A ball is thrown vertically upward with an initial velocity of 30.0 m/s. What are its position and velocity after 2.00 s, 4.00 s, and 7.00 s? Find also the maximum height attained<br />Given: a = -9.8 m/s2<br />vo= 30.0 m/s<br />t = 2.00 s; 4.00 s; 7.00 s<br />Find:<br />Δy = ? – displacement <br />v = ? - final velocity<br />After those three “times”<br />Δy = ? – maximum height<br />vo= +30.0 m/s<br />
33. 33. Given: a = -9.8 m/s2; vo= 30.0 m/s<br />t = 2.00 s; 4.00 s; 7.00 s<br />Solutions: <br />For t = 2.00 s:<br />For t = 4.00 s:<br />For t = 7.00 s:<br />
34. 34. Given: a = -9.8 m/s2; vo= 30.0 m/s<br />t = 2.00 s; 4.00 s; 7.00 s<br />Solutions: <br />For t = 2.00 s:<br />For t = 4.00 s:<br />For t = 7.00 s:<br />
35. 35. Given: a = -9.8 m/s2; vo= 30.0 m/s<br />t = 2.00 s; 4.00 s; 7.00 s<br />Solutions: <br />For maximum height, v = 0 (the ball stops at maximum height):<br />