The first thing that I learned was the idea of instantaneous velocity of an object, or the velocity at a certain point in time which an object is traveling at. I found that there is actually two different ways to find the instantaneous velocity of an object: by finding the slope of the tangent on a position versus time graph, or by using the mathematical equation V(final) = acceleration*time + Velocity (initial). Finding the slope of the tangent seems way harder than it actually is; all that's needed is at least three different points on a position versus time graph. Here's an example: I want to find the instantaneous velocity of an object at 6 seconds into the test. So, I'll use points 5 and 7 to test...
V(final)= a*t + V(initial)
V= (change in)X/ (change in) T
122.5 m- 62.5 m / 7 secs - 5 secs =
60 m / 2 secs =
30 meters / sec
That's one way to find the instantaneous velocity of the object. However, if I didn't have a position versus time graph to use, I can always use the mathematical formula V(final)= a*t + V(initial).
This requires me to be using a velocity versus time graph, and simply finding the slope of the graph. I'm able to do this because the slope is also my acceleration.
The next thing that I learned was how to find the displacement or distance an object travels in a given amount of time by using a velocity versus time graph and by using the model x = 1/2 a*t + v. I already know how to find the displacement of an object using other types of graphs, but this gives me a quicker, more efficient way to an answer. When I use the velocity versus time graph, I have to find the area under the curve, which amounts to the displacement. Here's an example problem:
Imagine a invisible triangle connecting the line to the sides of the graph. This space should look like a triangle and a rectangle. This is what you need to find for the displacement.
15 m/s * 6 secs/ 2 = 45 m/s^2+
15 m/s * 6 secs = 90 m/s^2
Displacement = 135 m/s^2
Now, to the heart of this unit; acceleration! I actually have four different ways to find the acceleration of an object, all of them listed out below:
- Finding the slope of a Velocity versus Time graph
- Using the model a = (change in) Velocity/ (change in) Time
- Using 1/2 a * t + Velocity (initial)
- Using Velocity (final)= a * t + Velocity (initial)
I've already showed an example problem involving three of these four solutions, but I haven't touched on when to use 1/2 a *t + Velocity (initial). I should only use this equation whenever I have a Position versus Time Squared graph and need to use it to find the acceleration. However, I can only do this whenever the starting velocity is equal to 0. The trick to using this method is to not forget to remember that the slope is only equal to 1/2 of the acceleration, so I need to always add back another half to find my full acceleration.
A big part of my first unit at the beginning of the year was learning all about graphs. More specifically, position versus time and velocity versus time. In this unit, I expanded on this previous knowledge by adding acceleration versus time graphs. I also found how to graphically represent an object when the surface isn't flat. This in turn changes my position versus time graph from having a straight line with a constant slope to either an exponential growth or decay graph. I developed new language to describe these new graphic ideas. Despite adding all of these new ideas, I can still find things such as the displacement the same way as before; by finding the difference in the distance from the starting point to the ending point.
The same thing applies to my previous knowledge about velocity versus time graphs. I still retained all the same principles and ideas from the first unit, I just added some new knowledge to the mix. However, this time I played with the idea of a velocity versus time graph having a constant slope, meaning that the acceleration (as the slope) is staying constant as well. This meant that my velocity graph was no longer just a straight line; it began to look more like a position versus time graph from the first unit. This took on a different meaning, depending on which way the object is traveling and if it's accelerating or decelerating. Here's an example graph below:
We can tell that the object is accelerating because it's covering more distance in less time as the test runs. However, this acceleration is towards the sensor, because the position from the sensor is getting smaller. My velocity versus time graph would start down below the line, as the object is moving towards the sensor, but be moving from the origin down at a constant pace. The velocity graph is constant because the object's speed is constantly increasing, which is the acceleration (also the slope for velocity versus time). This summary has giving me a clearer picture of what I have learned going into the test, and I hope to revisit this throughout the year for additional help if needed!
The next thing that I learned was how to find the displacement or distance an object travels in a given amount of time by using a velocity versus time graph and by using the model x = 1/2 a*t + v. I already know how to find the displacement of an object using other types of graphs, but this gives me a quicker, more efficient way to an answer. When I use the velocity versus time graph, I have to find the area under the curve, which amounts to the displacement. Here's an example problem:
Imagine a invisible triangle connecting the line to the sides of the graph. This space should look like a triangle and a rectangle. This is what you need to find for the displacement.15 m/s * 6 secs/ 2 = 45 m/s^2+
15 m/s * 6 secs = 90 m/s^2
Displacement = 135 m/s^2
Now, to the heart of this unit; acceleration! I actually have four different ways to find the acceleration of an object, all of them listed out below:
- Finding the slope of a Velocity versus Time graph
- Using the model a = (change in) Velocity/ (change in) Time
- Using 1/2 a * t + Velocity (initial)
- Using Velocity (final)= a * t + Velocity (initial)
I've already showed an example problem involving three of these four solutions, but I haven't touched on when to use 1/2 a *t + Velocity (initial). I should only use this equation whenever I have a Position versus Time Squared graph and need to use it to find the acceleration. However, I can only do this whenever the starting velocity is equal to 0. The trick to using this method is to not forget to remember that the slope is only equal to 1/2 of the acceleration, so I need to always add back another half to find my full acceleration.
A big part of my first unit at the beginning of the year was learning all about graphs. More specifically, position versus time and velocity versus time. In this unit, I expanded on this previous knowledge by adding acceleration versus time graphs. I also found how to graphically represent an object when the surface isn't flat. This in turn changes my position versus time graph from having a straight line with a constant slope to either an exponential growth or decay graph. I developed new language to describe these new graphic ideas. Despite adding all of these new ideas, I can still find things such as the displacement the same way as before; by finding the difference in the distance from the starting point to the ending point.
The same thing applies to my previous knowledge about velocity versus time graphs. I still retained all the same principles and ideas from the first unit, I just added some new knowledge to the mix. However, this time I played with the idea of a velocity versus time graph having a constant slope, meaning that the acceleration (as the slope) is staying constant as well. This meant that my velocity graph was no longer just a straight line; it began to look more like a position versus time graph from the first unit. This took on a different meaning, depending on which way the object is traveling and if it's accelerating or decelerating. Here's an example graph below:
We can tell that the object is accelerating because it's covering more distance in less time as the test runs. However, this acceleration is towards the sensor, because the position from the sensor is getting smaller. My velocity versus time graph would start down below the line, as the object is moving towards the sensor, but be moving from the origin down at a constant pace. The velocity graph is constant because the object's speed is constantly increasing, which is the acceleration (also the slope for velocity versus time). This summary has giving me a clearer picture of what I have learned going into the test, and I hope to revisit this throughout the year for additional help if needed!
