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Teach A Level Maths Displacement and Velocity. Volume 4: Mechanics 1 Displacement and Velocity.

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Presentation on theme: "Teach A Level Maths Displacement and Velocity. Volume 4: Mechanics 1 Displacement and Velocity."— Presentation transcript:

1 Teach A Level Maths Displacement and Velocity

2 Volume 4: Mechanics 1 Displacement and Velocity

3 and then 2 m left. The distance I have moved is 4 m BUT I’m back where I started. My displacement is 0 m Displacement measures the change in position and is a vector. If I move 2 more metres to the left, my displacement from the start is  2 m ( although the distance I’ve moved is now 6 m ). x Start Suppose I move 2 m to the right in a straight line...

4 Using the example from the previous presentation, instead of graphs showing distance and speed against time, we can draw displacement and velocity graphs. Speed is the rate of change of distance and velocity is the rate of change of displacement. Displacement and velocity are both vector quantities so have direction as well as magnitude. When we talk about “rate of change” for distance and displacement we mean how the quantities change with time. Speed is the magnitude of velocity.

5 time (s) displacement (m) time (s) velocity (ms -1 ) Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12

6 time (s) displacement (m) Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 x start

7 time (s) displacement (m) Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 x start

8 time (s) displacement (m) Here I’m back at the start. Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 x start

9 time (s) displacement (m) Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 x start

10 time (s) displacement (m) Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 The negative displacement shows where I am on the left, rather than the right, of the start. x start

11 Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 time (s) velocity (ms -1 ) x start

12 Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 time (s) velocity (ms -1 ) x start

13 Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 time (s) velocity (ms -1 ) x start The negative velocity shows movement in the opposite direction ( towards the left of the start instead of towards the right ).

14 Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 time (s) velocity (ms -1 ) x start

15 Move Speed (m s -1 ) Time (s) 2 m right 12 at rest 02 2 m left 12 at rest 02 2 m left 12 time (s) velocity (ms -1 ) x start I’m still walking to the left.

16 time (s) velocity (ms -1 ) time (s) displacement (m) On a displacement-time graph gradient gives velocity.

17 time (s) displacement (m) On the displacement graph, the two sections with negative gradients give the velocity as  1. 22 2 Tip: When the gradient is negative, as the quantity on the time axis increases... the quantity on the displacement axis decreases. A line with negative gradient falls backwards

18 time (s) displacement (m) Average velocity  total displacement total time e.g.Using the graph, find the average velocity for (a) the entire journey, and (b) the 1 st 6 seconds of the motion Solution: (a) Average velocity 22 10   0·2 m s -1  (b) Average velocity for 1 st 6 seconds 0 6   0 0 Average velocity is zero when the displacement is zero.

19 time (s) velocity (ms -1 ) Decide how the parts below the axis give us displacement. Ans: We find the area... We can find displacement from a velocity-time graph but we have to be careful. e.g. but since area is positive and displacement is negative we must insert a minus sign.

20 time (s) velocity (ms -1 ) 11 Tip: For areas below the time axis the negative velocity reminds us that the displacement is also negative.

21 time (s) velocity (ms -1 ) Area A  2  1  2 The total displacement  Area A – Area B – Area C  The total displacement  2  2  2   2 m Area B  Area C  Area A  2 A BC

22 We can also find the distance travelled from the velocity/time graph. time (s) velocity (ms -1 ) To find the total distance, we just ignore the negative signs on the displacements where the graph lies below the axis. The total distance travelled   6 m The total displacement  2  2  2   2 m

23 SUMMARY  Displacement measures the change in position and is a vector.  Velocity is the rate of change of displacement and is also a vector.  Motion in a straight line can be illustrated on graphs. continued

24 time (s) displacement (m) negative velocity time (s) velocity (ms -1 ) negative displacement 6 6  Areas below the axis give negative displacements. If negative signs for displacement are ignored the area gives distance.  On a velocity-time graph we can find displacement and distance from area.  On a displacement-time graph gradient gives velocity. SUMMARY

25 (i)Constant speed of 4 m s -1 for 4 s from A to C. (ii)Constant speed of 2 m s -1 for 2 s from C to B. (iii)At rest at B for 2 s (b)Use the graph to find the displacement from A after 6 s. (c)Complete the graph if the particle returns to A in a further 2 s. (a)Sketch a velocity-time graph to show the following motion of a particle moving for 10 s in a straight line ABC where B lies between A and C. Your sketch need not use squared paper nor be to scale but you must mark the key values on the axes. EXERCISE

26 Since B is between A and C the particle changes direction so the velocity is negative. (a)(i)Constant speed of 4 m s -1 for 4 s from A to C. (ii)Constant speed of 2 m s -1 for 2 s from C to B. (iii)At rest at B for 2 s. Solution: velocity (ms -1 ) time (s) 22 EXERCISE

27 velocity (ms -1 ) time (s) 4 4 Solution: 86 22 (b)Find the displacement from A after 6 s The displacement after 6 s  16 – 4  12 m A to C C to B EXERCISE

28 velocity (ms -1 ) time (s) 4 4 Solution: 86 2 A to C C to B (c)Complete the graph if the particle returns to A in a further 2 s. 10 66 The displacement of B from A So, the particle must move at 6 m s -1 to return to A in 2 s.  12 m EXERCISE

29 velocity (ms -1 ) time (s) 4 4 Solution: 86 2 A to C C to B (c)Complete the graph if the particle returns to A in a further 2 s. 10 66 We can check this result by noticing that the total displacement is zero. 12 B to A EXERCISE

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31 The slides that follow are in a form suitable for photocopying for example 1 and the summary.

32 21 2 m left 20 at rest 21 2 m left 20 at rest 2 m right Move 21 Time (s) Speed (m s -1 ) Displacement and Velocity - Data and Graphs for e.g.1 time (s) displacement (m) time (s) velocity (ms -1 )

33 Summary DISPLACEMENT AND VELOCITY TEACH A LEVEL MATHS – MECHANICS 1  Displacement measures the change in position and is a vector.  Velocity is the rate of change of displacement and is also a vector.  Motion in a straight line can be illustrated on graphs. time (s) displacement (m) negative velocity time (s) velocity (ms -1 ) negative displacement 6 6  Areas below the axis give negative displacements. If negative signs for displacement are ignored the area gives distance.  On a velocity-time graph we can find displacement and distance from area.  On a displacement-time graph gradient gives velocity.


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