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phys: add acceleration-time graph

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eggy 2020-09-28 13:24:50 -04:00
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@ -263,24 +263,22 @@ $s$ is commonly used in IB to represent displacement and $s_{0}$ represents the
<img src="/resources/images/position-time-graph.png" width=700>(Source: Kognity)</img>. <img src="/resources/images/position-time-graph.png" width=700>(Source: Kognity)</img>.
On a position-time or displacement-time graph: The slope of the line in a position-time graph represents that object's velocity. If the slope is not linear, the object is not moving uniformly (at a constant speed).
- The object is not moving if the slope is zero.
- The object is moving at a constant *velocity* if the slope is linear.
- The object is moving with constant acceleration if the slope increases exponentially.
- The object is moving with constant deceleration if the slope decreases exponentially.
- The gradient of the graph at any time is the object's *instantaneous velocity* at that exact time.
A **velocity-time graph** is similar to a position-time graph but replaces the position on the vertical axis with an object's velocity instead. A **velocity-time graph** is similar to a position-time graph but replaces the position on the vertical axis with an object's velocity instead.
<img src="/resources/images/velocity-time-graph.png" width=700>(Source: Kognity)</img> <img src="/resources/images/velocity-time-graph.png" width=700>(Source: Kognity)</img>
On a velocity-time graph, the object is moving with constant acceleration/deceleration if the slope is linear. On a velocity-time graph, the slope represents that object's acceleration. If the slope is not linear, the object is not accelerating uniformly (accelerating at a constant rate)
The area below a velocity-time graph at a given time is equal to the displacement (change in position) at that time, since $ms^{-1}×s=m$. When finding the displacement of an object when it is accelerating, breaking up the graph into a rectangle and a triangle then adding their areas will give their resultant displacement. The area below a velocity-time graph at a given time is equal to the displacement (change in position) at that time, since $ms^{-1}×s=m$. When finding the displacement of an object when it is accelerating, breaking up the graph into a rectangle and a triangle then adding their areas will give the displacement.
<img src="/resources/images/velocity-time-displacement.png" width=700>(Source: Kognity)</img> <img src="/resources/images/velocity-time-displacement.png" width=700>(Source: Kognity)</img>
An **acceleration-time graph** is similar to a velocity-time graph but replaces the velocity on the vertical axis with an object's acceleration instead.
The area below an acceleration-time graph at a given time is equal to the velocity at that time.
## 2.2 - Forces ## 2.2 - Forces
## 2.3 - Work, energy, and power ## 2.3 - Work, energy, and power