GRAPHICAL REPRESENTATION OF MOTION💖.pptx

Information about, Motion and Distance time graph

1. MOTION GRAPHICAL REPRESENTATION OF MOTION

2. INTRODUCTION: Graphs provide us a convenient way to present basic information about various events. For example: Run rate shown in cricket matches. Used to solve equations in two variables.

3. USAGE OF GRAPHS: W e  can use line graphs to describe motion.  We can show the dependence of one physical quantity on other through line graphs ​. For example: Distance-time graphs   Velocity-time graphs Uniform velocity and uniform speed are equal when the magnitude of distance and displacement is equal.

4. SIMPLE FLOW CHART TO UNDERSTAND THE TOPIC: GRAPHICAL REPRESENTATION OF MOTION DISTANCE-TIME GRAPHS VELOCITY-TIME GRAPHS CALCULATE SPEED USING DISTANCE AND TIME CALCULATE DISTANCE /d /a USING VELOCITY AND TIME

5. DISTANCE-TIME GRAPHS: In a distance-time graph the distance is represented along the “y-axis” and time is represented along the “x-axis”. Using a distance-time graph the change in the position of an object with respect to time can be determined. A distance-time graph helps us find the speed of the object moving. When the object is under going uniform motion(velocity) the graph obtained will always be a straight line graph, as the distance and time are directly proportional to each other When the object is under going non-uniform motion(velocity) the graph obtained will not be a straight line graph. When we are finding the speed we have to use the concept of slope.

6. HOW TO DETERMINE THE SPEED FROM THIS GRAPH: To determine speed using the graph we have to: Take two consecutive points and draw perpendiculars to the x & y axis as shown in the figure. Name the two points ‘A’ & ‘B’. Name the point on the y-axis as S 1 and the point on the x-axis asT 1 (from point A). Name the point on the y-axis as S 2 and the point on the x-axis as T 2 (from point B). Speed  Speed =slope  Slope  S lope  

7. HOW TO SOLVE: Here s 1 =30km, s 2 = 40km; t 1 =60mins,t 2 =80mins. So, according to the formula- Slope Slope  Slope = = = 0.5x60 Therefore, according to this diagram the vehicle is moving with a speed of 30 km/hr.  

8. NON-UNIFORM SPEED(VELOCITY) The nature of this graph shows nonlinear variation of the distance travelled by the car with time. Thus, the graph shown in represents motion with non-uniform speed .

9. VELOCITY-TIME GRAPH: In a velocity-time graph the velocity is represented along the “y-axis” and the time is represented along the “x-axis”. Whenever we fix a velocity on the y-axis its always a straight line parallel line to the x-axis. The velocity is always independent of time i.e. in whichever time we try to find the velocity the velocity remains the same. When the object has an uniform or constant velocity the graph obtained will always be a straight line graph. Whenever the object does not have an uniform velocity the graph obtained will be a curved line graph. When we have to find the distance we cannot use the concept of slope as slope uses the concept of division but [D= SxT ]. When we are finding the distance we have to find the area under the graph.

10. HOW TO DETERMINE DISTANCE FROM THIS GRAPH: To determine distance from this graph we have to: (If the velocity is a constant) draw a straight line parallel to the x-axis. Then take two consecutive points and name them ‘A’ & ‘B’. Then draw two perpendicular lines from point ‘A’ & ‘B’ and name them ‘C’ & ‘D’. Mark the two points ‘C’ & ‘D’ as T 1 & T 2 respectively. We have to find the area of the rectangle ABDC= lxb .

11. HOW TO SOLVE: LxB = ACxCD Here the distance= ACxCD . Where AC=40m/ s,CD =6secs. Therefore, ACxCD =40x6=240m. Which implies that according to this diagram the moving vehicle has covered a distance of 240m.

12. HOW TO DETERMINE DISTANCE USING THIS GRAPH: Scale  y-axis=10m/s for 1 unit; x-axis=2secs for 1 unit. In this figure we can calculate the distance in two ways Area of trapezium ABDE or Area of triangle ABC+Area of rectangle ACDE.

13. HOW TO SOLVE: Using area of trapezium: Area of trapezium So, in the above figure: AE ll BD and AC=height. Therefore, = = = 320 km Thus, the vehicle in this case has travelled a distance of 320 km.  

14. HOW TO SOLVE: Using area of triangle+area of rectangle=1/2xbxh+lxb Area of triangle- pn : Change in colour represents cancellation. 30x4=120 km Area of rectangle- lxb =25x8 25x8=200 km Therefore the total distance=120+200=320 km.  

15. HOW TO DETERMINE DISTANCE USING THIS GRAPH: In this case to find the distance we have to find the area under the graph which is the area of the triangle. Area of triangle= Here we can take a point anywhere on the graph and we have to find the are between the origin and the point taken. We have taken the point as ’A’ on the perpendiculars of 8 and 2. The origin is ‘O’ on zero and the other point on two as ‘C’. So we have to find the area of the triangle AOC. The difference between the previous graph and this graph is just the marking of points, in the 1 st one it’s any 2 points on the graph whereas here in the 2 nd one it only one point on the graph and your other point is nothing but the origin.  

16. HOW TO SOLVE: Area of triangle= = Therefore, the area of the triangle AOC=8m 2 Thus, the vehicle has travelled a distance of 8m.  

17. VELOCITY-TIME GRAPHS OF AN OBJECT IN NON-UNIFORMLY ACCELERATED MOTION: (a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while ( b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time.

18. TYPES OF DISTANCE-TIME GRAPHS: If the graph obtained is a straight line, then the object is said to moving in increasing uniform motion If the graph obtained is a straight line parallel to the time axis, then object is in rest. If the graph obtained is a straight line as shown, then the object is said to moving in uniform motion and is moving towards the initial position

19. If the graph obtained is a curved line as shown, then the object is said to be in increasing non uniform motion If the graph obtained is a curved line as shown, then the object is said to be in decreasing non uniform motion

20. TYPES OF VELOCITY TIME GRAPHS: If the graph obtained is a straight line parallel to the time axis, then the object is said to be with uniform velocity. Therefore the object is having zero acceleration If the graph obtained is a straight line as shown, then the body is said to have uniform acceleration.

21. RETARDATION: NEGATIVE ACCELERATION: If the graph is a straight line as shown, then the object is said to be moving with uniform retardation If the graph is a curved line as shown, then the object is said to be moving with non uniform acceleration . If the graph is a curved line as shown, then the object is said to be moving with non uniform retardation

22. VELOCITY TIME GRAPH In this graph, the object accelerates in AB, moves with constant velocity and zero acceleration till BC and retardation occurs till CD DISTANCE TIME GRAPH In this graph, the object covers distance in uniform speed in AB, stays in rest in BC and then comes back to the initial position.

23. CLUES FOR V-T GRAPHS:

24. SHORT SUMMARY OF THE TOPIC: When an object covers equal distance in equal intervals of time – Uniform velocity. When an object covers unequal distances in equal intervals of time – Non-uniform velocity. From the distance-time graph the speed is given by the slope of the line. From the velocity-time graph the distance is found by the area under the graph or the area enclosed by the figure.

25. NUMERICALS: Find the speed of the vehicle between A and B. Sol: Speed=Slope=S2-S1/ T2-T1 = 4-3/6-4 =1/2 =0.5m/s A B

26. NUMERICALS: Find the area of the shaded region. Sol: Area of trapezium=(Sum of ll sides) x height/2 = (20+50)x3/2 =(70)x3/2 =210/2 =105m

27. NUMERICALS: Find the distance covered by finding the shaded region. Sol: Area of rectangle= lxb =l=30,b=6 = lxb =30x6 =180m

28. NUMERICALS: Find the area of the shaded region to find the distance. Sol: Area of the shaded region=Area of a triangle =1/2xbxh =1/ 2 x 4 x40 =1x2x40 =80m

29. NUMERICALS: Find the speed of the car between the perpendicular of (10,1) and ( 5 0,5). Sol: Slope=Speed=S2-S1/T2-T1 = 5 0-10/5-1 =40/4 =10 m/s

30. MCQ’S: Distance-time graph of two objects A and B are shown below . ​ Which statement is true for the speed of object A and B?​ Speed of object A is greater than object B​ Speed of object A is lesser than object B​ Both have same speed​ Speed of Object A is double the speed of object B.

31. MCQ’S In a distance-time graph, if the line is horizontal, then the object is: ​ Accelerating ​ Speeding up​ At rest​ Slowing down ​

32. MCQ’S On the distance-time graph, the Y-axis should be labelled as:​ Distance​ Displacement ​ Speed ​ Time ​

33. MCQ’S The velocity-time graph of an object moving in a fixed direction is shown in the figure below. What do you conclude about the object? Moves with a constant speed. Moves with varying speeds. Moves with a non zero acceleration. Is at rest.

34. MCQ’S The slope of the distance-time graph is: Distance A cceleration Speed Displacement

35. MCQ’S The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of: Velocity S peed A cceleration Distance

36. MCQ’S For a constant acceleration, the nature of velocity-time graph is: Graph-B​ Graph-D ​ Graph-C ​ Graph-A ​

37. MCQ’S Area under a v – t graph represents a physical quantity which has the unit : m2​ m/s ​ m/s2 ​ m ​

38. MCQ’S The acceleration of an object moving in a straight line can be determined from: ​ T he  area between the distance-time graph and time axis.​ The  area between the velocity-time graph and time axis​ The  slope of the velocity-time graph.​ The  slope of the distance-time graph.​

39. MCQ’S This is a velocity-time graph when the body is under : Non-uniform retardation.​ Uniform acceleration.​ Uniform retardation.​ Non-uniform acceleration. ​ ​

40. DONE BY: DIVYA & SHRIMAYI