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Further Explanation on Question 16 of Chemical and Physical Section of GS-Free
- Thread starter agileduck
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If you know some calculus, dx/dt=v (velocity=derivative of displacement) so dx=v*dt ->integrate-> x=integral(v*dt)
An integral can be thought of as the area between a curve and x axis, so when the net area=0, displacement=0
Without calculus, you can memorize that the displacement is the area between a curve and x axis of a graph of v vs. t
The times the net area=0 in that plot are at the origin and at c, where the net area = (areaof+trapezoid)+(areaof-trapezoid)
example plot image:
An integral can be thought of as the area between a curve and x axis, so when the net area=0, displacement=0
Without calculus, you can memorize that the displacement is the area between a curve and x axis of a graph of v vs. t
The times the net area=0 in that plot are at the origin and at c, where the net area = (areaof+trapezoid)+(areaof-trapezoid)
example plot image:
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It is more simple than that. It starts with understanding what is going on in a velocity vs. time graph conceptually.
Any time the line is above the x-axis the object has some positive velocity. This would move the object in some direction away from the origin/start.
So, although point A may "look" like the object has returned to its starting point, note that during the second downward-sloped (negative) portion of the graph (next to A) the object still has POSITIVE velocity, so it is still moving FURTHER away from the origin. The slope is only negative because its velocity is getting lower in magnitude (the slope is acceleration).
In your car it would be like you...took off with a certain acceleration for x period of time, then STOPPED accelerating and traveled at constant velocity for a time, then began decelerating for x time again (negative slope). After all that...where are you? You aren't back where you started, you are a maximum distance away. However, past point A the velocity becomes negative. This indicates the object is moving opposite in direction to what it was in the first segment. Because the two segments are symmetrical, we know the object ends up exactly at the origin at approximately point C (if C is taken to be on the x-axis, it is kind of floating there).
Any time the line is above the x-axis the object has some positive velocity. This would move the object in some direction away from the origin/start.
So, although point A may "look" like the object has returned to its starting point, note that during the second downward-sloped (negative) portion of the graph (next to A) the object still has POSITIVE velocity, so it is still moving FURTHER away from the origin. The slope is only negative because its velocity is getting lower in magnitude (the slope is acceleration).
In your car it would be like you...took off with a certain acceleration for x period of time, then STOPPED accelerating and traveled at constant velocity for a time, then began decelerating for x time again (negative slope). After all that...where are you? You aren't back where you started, you are a maximum distance away. However, past point A the velocity becomes negative. This indicates the object is moving opposite in direction to what it was in the first segment. Because the two segments are symmetrical, we know the object ends up exactly at the origin at approximately point C (if C is taken to be on the x-axis, it is kind of floating there).
