description about kinematics

Published on: **Mar 3, 2016**

Source: www.slideshare.net

- 1. KINEMATICs OF RECTILINEAR MOTION/motion in one dimension1. DISTANCE and DISPLACEMENT2. SPEED and VELOCITY A. AVERAGE SPEED and AVERAGE VELOCITY B. INTANTANEOUS VELOCITY and INSTANTANEOUS SPEED3. ACCELERATION4. RECTILINEAR MOTION A. DESCRIBING RECTILINEAR MOTION WITH EQUATION 1. UNIFORM RECTILINEAR MOTION 2. ACCELERATED UNIFORM RECTILINEAR MOTION 3. VERTICAL MOTION: Upward Vertical Motion Downward Vertical Motion Free Fall Motion B. DESCRIBING RECTILINEAR MOTION WITH GRAPH 1
- 2. 1. DISTANCE AND DISPLACEMENT • DISTANCE: Length of path travelled by object during a motion • DISPLACEMENT: The change position of an object at a certain time 2
- 3. SAMPLE PROBLEM 11. An object moves in line with x axis direction as shown in the figure below. A B C D X (m) - - - - - - - - - - - - - - -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 Find: a. Distance and displacement of object moves from C to A b. Distance and displacement of object moves from B to A and back to C c. Distance and displacement of object moves from C to D and back to A2. Sprinter running to the east as far as 80 m and then turn to the north direction as far as 60 m before he finally stops. Find displacement and distance travelled by sprinter.3. A particle moves in circular path with radius 14 m. Determine the object’s distance and displacement when moves: a. once round b. half round c. quarter round d. 1/6 round 3
- 4. 2. SPEED AND VELOCITY “speed is scalar quantity, velocity is vector quantity” A. AVERAGE SPEED and AVERAGE VELOCITY total distance travelled Average speed total time taken s v s = distance (m) t = time (s) t v = speed (m/s) displacement Average velocity total time taken s Δs = displacement (m) v t = time interval(s) t v = average velocity (m/s) 4
- 5. SAMPLE PROBLEMS1. A motorist drives north for 30 minutes at 72 km/h and then stops for 30 minutes. He then continues north, traveling 130 km in 2 hours and then he turn to south travelling 50 km in 1 hour a. What is his total displacement? b. What is his average velocity? c. What is his average speed?2. A car moved from A to B in 10 minutes, and continues to C in 5 minutes. Look at the picture! A 8 km B Find: a. Average speed of car 6 km b. Average velocity of car C3. A toy car moves from A to D through B and C in 10 s . B 12 m A Find : a. The average speed b. The average velocity 6m C 9m D 5
- 6. B. INSTANTANEOUS VELOCITY and INSTANTANEOUS SPEED ® INSTANTANEOUS VELOCITY is the velocity measured at a particular moment or average velocity in time interval towards zero Instantaneous velocity is gradient of the s-t graph s ds v lim v lim t 0 t 0 t dt Instantaneous is derivation of position (s) against time (t) ® INSTANTANEOUS SPEED is the magnitude of the instantaneous velocity 6
- 7. 3. ACCELERATION “the change velocity divided by time interval” A. AVERAGE ACCELERATION change of velocity average acceleration time interval : average acceleration (m/s2) v vt v0 a a vt : final velocity (m/s) v0 : initial velocity (m/s) t tt t 0 tt : final time (s) t0 : initial time (s) Speed= 0 4 m/s 8 m/s 12 m/s 16 m/s 20 m/s 7
- 8. B. INSTANTANEOUS ACCELERATION is the acceleration measured at a particular moment or average acceleration in time interval towards zero v dv a lim a lim t 0 t 0 t dt NOTice: if an object moves with constant velocity, average velocity is the same as instantaneous velocity if an object moves with constant acceleration, average acceleration is the same as instantaneous acceleration 8
- 9. 4. RECTILENIER MOTION UNIFORM RECTLINEAR MOTION Velocity/speed = constant Acceleration = zero ACCELERATED UNIFORM RM RECTILINEAR MOTION Velocity/speed = change at certain time interval Acceleration = constant VERTICAL MOTION: Velocity/speed = change at certain time interval Acceleration = constant = gravitational acceleration (g) • UPWARD VERTICAL MOTION • DOWNWARD VERTICAL MOTION • FREE FALL MOTION 9
- 10. A. DESCRIBING RECTILINEAR MOTION WITH EQUATIONS 1. Uniform Rectilinear Motion o To begin, consider a car moving with a constant rightward (+) velocity, say of +10 m/s. Attention to the animation below ! s v v = velocity/speed (m/s) s = displacement/distance (m) t = time (s) t Example Problem 10
- 11. 2. Accelerated Uniform Rectilinear Motion Speed= 0 4 m/s 8 m/s 12 m/s 16 m/s 20 m/s vt v0 at v v 2as 2 t 2 0 a vt : acceleration (m/s2) : final velocity/speed (m/s) v0 : initial velocity/speed (m/s) s v0t 1 2 at 2 t : time interval(s) s : displacement/distance (m) v0 vt s t 2 11
- 12. 3. Vertical Motion a. Upward Vertical Motion the highest point reached (H) vt v0 gt H vH 0 vt v0 vt2 v0 2 gs 2 tH g s v0t 1 2 gt 2 2 V0 v v0 vt sH 0 s t 2g 2 ground g = gravitational acceleration (m/s2) vH = velocity at point H (m/s) sH = the highest point reached (m) tH = time to reach highest point (s) 12
- 13. b. Downward Vertical Motion v0 vt v0 gt vt2 v0 2 gs 2 s v0 t 1 2 gt 2 v 0 vt s t 2 vtc. Free Fall Motion (v0 = 0) ground vt gt g = gravitational acceleration (m/s2) vt2 2 gs vt 2 gs vt = final velocity/velocity at certain t time (m/s) 2s v0 = initial velocity (m/s) s 1 2 gt t 2 g s = displacement/distance(m) t = time interval(s) vt s t 13 2
- 14. B. DESCRIBING RECTILINEAR MOTION WITH GRAPH 1. Describing Motion with Position/displacement vs. Time Graphs (s - t graph) o To begin, consider a car moving with a constant, rightward (+) velocity - say of +10 m/s. If the position-time data for such a car were graphed, then the resulting graph would look like the graph at the right. 14
- 15. o Now consider a car moving with a rightward (+), changing velocity, a car that is moving rightward but speeding up or accelerating. If the position-time data for such a car were graphed, then the resulting graph would look like the graph at the right 15
- 16. The position vs. time graphs for the two types of motion -constant velocity (URM) and changing velocity/accelerated(AURM) are depicted as follows. A. For Constant Velocity (URM) Rightward, v(+) s s fast t position time t s s t position slow 16 time t
- 17. Leftward, v(-) s s t fast time position time t s s t position slow time t 17
- 18. B. For Changing Velocity/Accelerated (AURM) s s time Slow to fast/ position position t accelerated, a(+) t timeRightward s s t position position Fast to slow/ decelerated, a(-) t time s s position Slow to fast/ t position accelerated, a(+) t timeLeftward s s position t position Fast to slow/ decelerated, a(-) t time 18
- 19. .2. Describing Motion with Velocity vs. Time Graphs (v - t graph) Consider a car moving rightward with a constant velocity of +10 m/s If the velocity-time data for such a car were graphed, then the resulting graph would look like the graph at the right Now consider a car moving with a rightward (+), changing velocity. 19
- 20. The velocity vs. time graphs.A. For Constant Velocity (URM) Rightward, v(+) Leftward, v(-) v v t tB. For Changing Velocity (AURM) Slow to fast/accelerated, a(+) Slow to fast/decelerated, a(-) rightward leftward rightward leftward v v v v t t t t 20
- 21. Speeding up (accelerated) Slowing down (decelerated)2. Describing Motion with Acceleration vs. Time Graphs (a- t graph) for Constant Acceleration (AURM) Slow to fast/accelerated, a(+) Slow to fast/decelerated, a(-) a a t t 21
- 22. Determining Velocity on a Position vs. Time. Graph (s-t Graph) “The slope of the line on a position versus time graph is equal to the velocity of the object” s v = the slope of the line ∆y on a s-t graph θ ∆x t y y 2 y1 rise slope tan x x2 x1 run Consider the position versus time graph below! v? for example take points (1, 10) and (3, 30) y 30 10 v 10 m/s x 3 1 22
- 23. Animasi GLB Arah Gerak ke Kanan 23
- 24. Animasi GLB Arah Gerak ke Kiri 24
- 25. Animasi GLBB ke Kanan Dipercepat 25
- 26. Animasi GLBB ke Kiri Dipercepat 26
- 27. Animasi GLBB ke Kanan Diperlambat 27
- 28. Animasi GLBB ke Kiri Diperlambat 28
- 29. Determining Distance, Displacement , and Acceleration on a Velocityvs. Time Graph (v-t Graph) The area bound by the line and the axes on a velocity versus time graph represents the displacement/distance • The slope of the line on a velocity versus time graph is equal to the acceleration of the object”Consider the velocity versus time graph below! v (m/s) B C 6 D F A t(s) 2 4 7 9 12 -4 E 29
- 30. a. The displacement and distance at first 2 s ? •The displacement = the distance = the area of triangle =1/2 x 2 x 6 = 6 m/sb. The displacement and distance in time interval t= 2 s until t= 12 s? •The displacement = the area of trapezoid - the area of triangle = {½ x (5 + 2) x 6 } – { ½ x 5 x 4} = 21 – 10 = 11 m •The distance= the area of trapezoid + the area of triangle = {½ x (5 + 2) x 6 } + { ½ x 5 x 4} = 21 +10 = 21 mc. The acceleration at t= 1 s? 60 •The acceleration = the slope of line AB 3 m/s2 20d. The acceleration at t= 6 s? 06 •The acceleration = the slope of line CD - 2 m/s2 74e. The acceleration at t= 10 s? 0 ( 4 ) •The acceleration = the slope of line EF 4/3 m/s2 12 9 30
- 31. f. Average acceleration in time interval t= 2 s until t= 9 s? v -4 - 6 -10 a -1.4 m/s 2 t 9-2 7Determining velocity on acceleration vs.time graph (a – t graph) The area bound by the line and the axes on a acceleration versus time graph represents the velocity a(m/s2) 2 4 t(s) 6 -1Velocity at time v = The Area Rectangle 1 - The Area Rectangle 2 = (2 x 4) - (1 x 2)t= 6 s ? If v0= 0 31 = 6 m/s
- 32. DETERMINING KINEMATICS QUANTITIES WITH GRAPHS–t To determine v v= The SlopeGraph Position (s) To determine s v= The Areav–t The Slope The AreaGraph To determine a a= The Slope velocity (v)a–tGraph To determine v v= The Area acceleration (a) 32
- 33. General Problem Solving Strategy Step 1: Write down any information given in the problem. These pieces of information are called your “knowns”. Also write down those things you don’t know. These are the “unknowns.” Step 2: Convert the situation into an equation to be solved. Step 3: Solve the problem.1. A car moves on a linear path with constant speed 72 km/hour. Determine: a.The distance travelled by the car in 5 minutes b. time to travel distance 360 km Step 1: Knowns and unknowns v= 72 km/hour a. s= ?, t= 5 minutes = 1/12 hour b. t= ?, s= 360 km Step 2: Mathematical representation s v t 33
- 34. Step 3: Solution a. s = v. t = 72 . 1/12 = 6 km s b. t v 360 t 72 5hour 34
- 35. Jelaskan Animasi Berikut! 35
- 36. 36
- 37. Jelaskan Animasi Berikut! 37
- 38. SAMPLE PROBLEMS1. A car moves on a linear path with constant speed 72 km/hour. Determine: a.The distance travelled by the car in 5 minutes b. time to travel distance 360 km2. How the speed of the object in 15 minutes traveling the distance of 20 km?3. An object moves and expressed by the following position-time graph x (m) 14 6 t (s) 4 6 Determine: a. Velocity at time t = 5 s b. Average velocity in time interval t= 0 s until t= 6 s 38
- 39. 3. An object moves and expressed by the following displacement-time graph S (m) 4 1 4 9 11 14 20 t (s) -6 a. Describing motion of the object! b. Find the velocity of the object at time t= 3 s, t= 11 s, and t= 15 s c. Find the average velocity at time interval t= 9 s until t= 14 s4. A car moves from rest and accelerated to a speed of 30 m/s in 6 seconds. How the distance travelled by the car?5. How long does it take for a car to change its velocity from 10 m/s to 25 m/s if the acceleration is 5 m/s2? 39
- 40. 6. The speed versus time graph below represents the motion of a car. Approximately how far did the car travel during the first 5 seconds?7. A body moves as described by the following v-t graph a. Describe the motion. b. What is the distance travelled during the motion? c. What is instantaneous speed at time t= 2 s? d. What is the average speed for the motion?. e. What is the average acceleration for the motion?. f. What is the average acceleration for time interval t= 2 s until t= 8s g. Instantaneous acceleration at time t= 6 s 40
- 41. 8. A car moves in 36 km/h and braked with constant acceleration and stopped after 5 m. How long does it take to stop?9. A car moves with speed of 25 m/s then braked and required distance of 40 m change its speed becomes 15 m/s . Find total distance travelled until the car was stopped.10. A car is moving with the speed of 72 km/h. Because at its front as far as 200 m a cat is crossing the road, then driver brakes the car with deceleration 5 m/s2. Whether the cat was hit by a car?11. Two cars, R and B are separated each other at a distance 600 m travelling in the opposite direction. Car R and B move with constant velocity of 20 m/s and 10 m/s respectively. When and where do car A and B meet if: a.both of them departs at the same time. b.Car B departs 5 s earlier 600m12. Two cars, A and B are separated each other at a distance 600 m travelling in the opposite direction. Car A with constant velocity of 10 m/s and B depart from rest with acceleration 2 m/s2. When and where do car A and B meet if both of them departs at the same time.13. When and where do car B overtakes car R? t= 0 s, v = 0 m/s, a = 2 m/s2 t= 0 s, v = 20 m/s (constant) 41
- 42. 14. When and where do car R overtakes car B if: a. Car R and B move with constant velocity of 10 m/s and 5 m/s respectively and both of them departs at the same time b. Car R and B move with constant velocity of 10 m/s and 5 m/s respectively and car B departs 5 s earlier c. Car R move with acceleration 2 m/s2 from rest and B move with constant velocity of 10 m/s 100 m15. An object fall freely from the rooftop of a building, and then hit the ground in time interval of 2 seconds. Calculate: a. The velocity to hit the ground b. The high of building16. A boy shoot a projectile vertically upward with initial velocity of 200 m/s. Find: a. The maximum high b. Time to reach high of 500 m c. Time to reach the highest point d. Velocity of projectile has moved at 10th seconds e. Velocity of projectile in high of 2 km17. Someone falls a stone with initial velocity of 20 m/s from peak of tower which has height 225 m above the earth. Determine: a. Time required by the stone to reach the earth b. Velocity of the stone when reaching the earth 42 c. Height of stone from earth when the velocity is 30 m/s
- 43. 18. The diagram below shows a position-time graph of an object which travelling vertically y (m) - - 12 - ● B ●C - 10 - - 8 - - A● 6 - - 4 - - 2 - - t (s) - - - - - - - - - - - - - - - - - - - 5 10 15 Calculate: a. Average velocity at time interval 0.5 s – 7.5 s b. Average velocity at time interval 0 s – 14 s c. Instantaneous velocity at t= 1.5 s d. Instantaneous velocity at t= 7.5 s 43 e. Instantaneous velocity at t= 10 s