Question #8ebbc

1 Answer
Oct 27, 2015

The method is the same as any other graph :
The gradient #=(y_2-y_1)/(x_2-x_1)#

Explanation:

The method is the same as for finding the gradient (or slope) of any other graph:
1. Choose two points that are on the line, point 1 and point 2.
2. Those points will have coordinates (#x_1#, #y_1#) and (#x_2#, #y_2#) respectively. Read off the coordinates.
3. The gradient #=(y_2-y_1)/(x_2-x_1)#

It may be that the question is actually about the analysis of a distance v. time² graph. If so, in order to answer the question the equation that is part of the question needs to be shared (the equation might be #s=ut + 1/2 at^2 #).

If the equation is as above the method of analysis is as follows:
1. Compare the equation that is given to #y=mx+c# (which is the general equation of a straight line).
2. Identify x and y from what is plotted on the x and y axes. In our case (distance v. time² ) it would be distance on the y-axis and time² on the x-axis.
3. Everything multiplied by the "x" term would be equal to the gradient. In our case we can then write: gradient #=1/2 a# .
4. The separate term in the equation is equal to the y-intercept. In our case it would be: intercept #=ut#.
5. Use the gradient and intercept values determined from the graph to solve for the unknowns. In our case those are #a# and #u#.


Check out the image below of an example of my graph work including working out the gradient and analysis to determine exactly acceleration of a trolley down a ramp.
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