Criado por shania catania
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Finding the gradient of a straight-line graph It is often useful or necessary to find out what the gradient of a graph is. For a straight-line graph, pick two points on the graph. The gradient of the line = (change in y-coordinate)/(change in x-coordinate) . In this graph, the gradient = (change in y-coordinate)/(change in x-coordinate) = (8-6)/(10-6) = 2/4 = 1/2 We can of course use this to find the equation of the line. Since the line crosses the y-axis when y = 2, the equation of this graph is y = ½x + 2 . Finding the gradient of a curve To find the gradient of a curve, you must draw an accurate sketch of the curve. At the point where you need to know the gradient, draw a tangent to the curve. A tangent is a straight line which touches the curve at one point only. You then find the gradient of this tangent. Example: Find the gradient of the curve y = x² at the point (3, 9). Note: this method only gives an approximate answer. The better your graph is, the closer your answer will be to the correct answer. If your graph is perfect, you should get an answer of 6 for the above question.
The Equation of a Straight Line Equations of straight lines are in the form y = mx + c (m and c are numbers). m is the gradient of the line and c is the y-intercept (where the graph crosses the y-axis). NB1: If you are given the equation of a straight-line and there is a number before the 'y', divide everything by this number to get y by itself, so that you can see what m and c are. NB2: Parallel lines have equal gradients. The above graph has equation y = (4/3)x - 2 (which is the same as 3y + 6 = 4x). Gradient = change in y / change in x = 4 / 3 It cuts the y-axis at -2, and this is the constant in the equation. Graphs of Quadratic Equations These are curves and will have a turning point. Remember, quadratic equations are of the form: y = ax² + bx + c (a, b and c are numbers). If 'a' is positive, the graph will be 'U' shaped. If 'a' is negative, the graph will be 'n' shaped. The graph will always cross the y-axis at the point c (so c is the y-intercept point). Graphs of quadratic functions are sometimes known as parabolas. Example: Drawing Other Graphs Often the easiest way to draw a graph is to construct a table of values. Example: Draw y = x² + 3x + 2 for -3 £ x £ 3 x -3 -2 -1 0 1 2 3 x² 9 4 1 0 1 4 9 3x -9 -6 -3 0 3 6 9 2 2 2 2 2 2 2 2 y 2 0 0 2 6 12 20 The table shows that when x = -3, x² = 9, 3x = -9 and 2 = 2. Since y = x² + 3x + 2, we add up the three values in the table to find out what y is when x = -3, etc. We then plot the values of x and y on graph paper. Intersecting Graphs If we wish to know the coordinates of the point(s) where two graphs intersect, we solve the equations simultaneously. This can be done using the graphs. Simultaneous Equations You can solve simultaneous equations by drawing graphs of the two equations you wish to solve. The x and y values of where the graphs intersect are the solutions to the equations. Example: Solve the simultaneous equations 3y = -2x + 6 and y = 2x -2 by graphical methods. From the graph, y = 1 and x = 1.5 (approx.). These are the answers to the simultaneous equations. Solving Equations Any equation can be solved by drawing a graph of the equation in question. The points where the graph crosses the x-axis are the solutions. So if you asked to solve x² - 3 = 0 using a graph, draw the graph of y = x² - 3 and the points where the graph crosses the x-axis are the solutions to the equation. If you are asked to draw the graph of y = x² - 3x + 5 and then are asked to use this graph to solve 3x + 1 - x² = 0 and x² - 3x - 6 = 0, you would proceed in the following way: 1) Make a table of values for y = x² - 3x + 5 and draw the graph. 2) Make the equations you need to solve like the one you have the graph of. So for 3x + 1 - x² = 0: i) multiply both sides by -1 to get x² - 3x -1 = 0 ii) add 6 to both sides: x² - 3x + 5 = 6 Now we know that y = x² - 3x + 5 and x² - 3x + 5 = 6, therefore, y = 6. Find out what x is when y = 6 and these are the answers (you should get two answers). Try solving x² - 3x - 6 = 0 yourself using your graph of y = x² - 3x + 5. You should get a answers of
Travel Graphs Bookmark this page Speed, Distance and Time The following is a basic but important formula which applies when speed is constant (in other words the speed doesn't change): Speed = distance time Remember, when using any formula, the units must all be consistent. For example speed could be measured in m/s, distance in metres and time in seconds. If speed does change, the average (mean) speed can be calculated: Average speed = total distance travelled total time taken Example: If a car travels at a speed of 10m/s for 3 minutes, how far will it travel? Firstly, change the 3 minutes into 180 seconds, so that the units are consistent. Now rearrange the first equation to get distance = speed × time. Therefore distance travelled = 10 × 180 = 1800m = 1.8km Units In calculations, units must be consistent, so if the units in the question are not all the same (e.g. m/s, m and s or km/h, km and h), change the units before starting, as above. The following is an example of how to change the units: Example: Change 15km/h into m/s. 15km/h = 15/60 km/min (1) = 15/3600 km/s = 1/240 km/s (2) = 1000/240 m/s = 4.167 m/s (3) In line (1), we divide by 60 because there are 60 minutes in an hour. Often people have problems working out whether they need to divide or multiply by a certain number to change the units. If you think about it, in 1 minute, the object is going to travel less distance than in an hour. So we divide by 60, not multiply to get a smaller number. Velocity and Acceleration Velocity is the speed of a particle and its direction of motion (therefore velocity is a vector quantity, whereas speed is a scalar quantity). When the velocity (speed) of a moving object is increasing we say that the object is accelerating. If the velocity decreases it is said to be decelerating. Acceleration is therefore the rate of change of velocity (change in velocity / time) and is measured in m/s². Example: A car starts from rest and within 10 seconds is travelling at 10m/s. What is its acceleration? Acceleration = change in velocity = 10 = 1m/s² time 10 Distance-time graphs: These have the distance from a certain point on the vertical axis and the time on the horizontal axis. The velocity can be calculated by finding the gradient of the graph. If the graph is curved, this can be done by drawing a chord and finding its gradient (this will give average velocity) or by finding the gradient of a tangent to the graph (this will give the velocity at the instant where the tangent is drawn). Velocity-time graphs/ speed-time graphs: A velocity-time graph has the velocity or speed of an object on the vertical axis and time on the horizontal axis. The distance travelled can be calculated by finding the area under a velocity-time graph. If the graph is curved, there are a number of ways of estimating the area (see trapezium rule below). Acceleration is the gradient of a velocity-time graph and on curves can be calculated using chords or tangents, as above. On travel graphs, time always goes on the horizontal axis (because it is the independent variable). Trapezium Rule This is a useful method of estimating the area under a graph. You often need to find the area under a velocity-time graph since this is the distance travelled. Area under a curved graph = ½ × d × (first + last + 2(sum of rest)) d is the distance between the values from where you will take your readings. In the above example, d = 1. Every 1 unit on the horizontal axis, we draw a line to the graph and across to the y axis. 'first' refers to the first value on the vertical axis, which is about 4 here. 'last' refers to the last value, which is about 5 (green line).] 'sum of rest' refers to the sum of the values on the vertical axis where the yellow lines meet it. Therefore area is roughly: ½ × 1 × (4 + 5 + 2(8 + 8.8 + 10.1 + 10.8 + 11.9 + 12 + 12.7 + 12.9 + 13 + 13.2 + 13.4)) = ½ × (9 + 2(126.8)) = ½ × 262.6 = 131.3 units²
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