![]() ![]() So what about when \(x \left( t \right)\) is curved? The trick to use here is to determine if continuing this curve will eventually cause the graph to go horizontal (i.e. \(t\) graph is a straight line, then the velocity is constant, and the object is neither speeding up nor slowing down. Making this determination from the \(x\) vs. \(t\) graph represent motion in which the object is neither speeding up nor slowing down. Naturally horizontal parts of the the \(v\) vs. \(t\) graph at the point in question is heading closer to the horizontal axis, then its velocity is heading toward zero, and it is slowing down, while if it is heading away, it is speeding up. But there is a simpler, physical way to make this determination: If the \(v\) vs. If they are opposite, then it is slowing down. If they have the same sign, then the acceleration is in the same direction as the velocity, and it is speeding up. \(t\) graph, by comparing the slope of the graph with the value of the graph at the same point. We can determine speeding-up/slowing-down from the \(v\) vs. We therefore cannot determine the answer to this question from the acceleration graph alone, because that graph by itself does not provide the direction of motion (the function \(v \left( t \right)\) associated with this acceleration could be above or below the horizontal axis anywhere). This is because if the object is accelerating in the same direction that it is moving, then it is speeding up, and if it is accelerating in the opposite direction as the direction of motion, then it is slowing down. To make this determination, you actually need two pieces of information – the directions of both the velocity and the acceleration. This is probably the trickiest question of all, because it doesn't have a direct correlation to the value or slope of any of the graphs. Q4: Is the object speeding-up or slowing down? This means that with just the acceleration graph we cannot know where the velocity graph crosses the horizontal axis, and therefore have no idea where the object is coming to rest. ![]() \(t\) graph, but not where it is located up-and-down the vertical axis. If we have the acceleration graph, then integrating it to get the velocity graph leaves an unknown constant (\(v_o\)). ![]() Instead, whether the object is moving or not is a simple matter of whether or not the value of \(v\) is zero. \(t\) graph, we have to be careful not to use the same criterion as we did for the \(x\) vs. Obviously an object that is moving is one whose position is changing, so if the \(x\) value is changing, the object is moving. But we should strive to look at this physically as well. Mathematically, we know that the velocity is the slope of the position graph, so since "at rest" means zero velocity, the object is at rest when the tangent line to the \(x\) vs. Although this is a property of velocity we can answer it using the position graph (we only get unknown constants when we integrate, not when we take derivatives). Q2: Is the object at rest, or is it moving?Īnother seemingly obvious question to answer, but again there are things to keep in mind. ![]()
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