88 lines
4.2 KiB
Markdown
88 lines
4.2 KiB
Markdown
---
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title: Vectors and Kinematics
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date: 2024-08-19
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---
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> One must always start a study into the heavily crippled IB editions of the glorious subject of Physics with the initial understanding that the road ahead lead to pain immeasurable.
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>
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> -> Prime of the Faith
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# Motion in 1D
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In this rendering of motion, you will *never* need to use vector quantities to describe movement due to the scalar nature of all quantities discussed.
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## Reference Frames and Displacement
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- Any measurement about motion is taken in terms of a *reference frame*.
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> [!NOTE]
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> **Example:** A train that moves with respect to the ground being held stationary, is not moving with respect to a stationary person inside that train. If a person were to walk, at let's say $5$ km/s toward the back of the train while it moves forward at $80$ km/h, the person is moving at $75$ km/h with respect to the stationary ground.
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- In one dimension, only one axis, the $x$-axis of a coordinate plane is used.
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- the *net* distance an object has traveled is known as *displacement*
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- *total distance* is the overall distance traveled by the object/particle regardless of reference frame or initial/final positions
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- change in positions is described using $\Delta x$
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## Average Velocity
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It is important to note that this equation was derived from the more complicated calculus variant of the velocity equation. There are 2 important terms here.
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$$
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\text{average speed}=\frac{\text{distance traveled}}{\text{time elapsed}}=\frac{\Delta x}{\Delta t}
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$$
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For any form of velocity, the speed is simply calculated by using the magnitude of the velocity vector $||\vec{v}||$, but since we exist in one dimension now, we will use $|v(t)|$
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## Instantaneous Velocity
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If the position equation is defined as $x(t)$, then:
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$$
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\text{instantaneous velocity} = \tfrac{\mathrm{d}x}{\mathrm{d}t} = v(t) = \lim_{\Delta t \to 0}{\tfrac{\Delta x}{\Delta t}}
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$$
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## Average Acceleration
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The acceleration of an object is the rate at which the velocity of said object changes. **Average Acceleration** is defined as the change in velocity from 2 distinct points divided by the change in time between those 2 distinct points.
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$$
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\text{average acceleration} = \frac{\text{change of velocity}}{\text{time elapsed}}
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$$
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Or more mathematically:
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$$
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a=\frac{v_2-v_1}{t_2-t_1}=\tfrac{\Delta v}{\Delta t}
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$$
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## Instantaneous Acceleration
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If the velocity function is defined as $v(t)$ (this notation only applies to 1 dimension):
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$$
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a(t)=\tfrac{\mathrm{d}v}{\mathrm{d}t}=\lim_{\Delta t \to 0}{\tfrac{\Delta v}{\Delta t}}
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$$
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Please take calculus/study calculus if you want a neuron or two to function during the course of this, well, course.
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## Motion at Constant Acceleration
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During this section and any point you use these formulas, always take $t_0$ to be equivalent to $0=t_1$. This allows for the following line of reasoning:
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$$
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\bar{v}=\frac{x-x_0}{t-t_0}=\frac{x-x_0}{t}
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$$
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This clearly makes sense because $x_0$ is taken to be the initial position of the particle in 1D, and $t_0$ is taken to be simply $0$, when the clock starts. Therefore acceleration must be a constant like so:
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$$
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a=\frac{v-v_0}{t}
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$$
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Firstly, it is important to note that all variables without subscripts are taken to be definitions of that variable completely. In contrast, when you saw that $v$ had a little *bar* on top of it, $\bar{v}$ meant to be the ***average*** of $v$. Moving on, if you are required to find the velocity of an object after elapsed time $t$:
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$$
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v=v_0+at
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$$
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> [!NOTE]
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> **Example:** You are given that the acceleration of a motorcycle is constantly $4.0 \mathrm{m}/\mathrm{s}^2$. You need to find the speed of that motorcycle (note that we haven't given you any sense of position, $\therefore$ did not ask for velocity) after elapsed time $t=6.0\mathrm{s}$. The motorcycle starts from total rest, and when $t_1=t_0=0$.
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> [!faq]- Answer
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> $4.0*6.0=24\mathrm{m}/\mathrm{s}$
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Next, calculating the position:
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$$
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x=x_0+\bar{v}t
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$$
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Because of the symmetric and uniform growth of the velocity over time:
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$$
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\bar{v}=(\frac{v_0+v}{2})t=x_0+t(\frac{v_0+v_0+at}{2})=x_0+t(\frac{2v_0+at}{2})
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$$
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This finally results in the formula for position in these scenarios:
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$$
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x=x_0+tv_0+\tfrac{1}{2}at^2
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$$
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## Objects in free fall
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Just remember that $g=9.80\mathrm{m}/\mathrm{s}^2$
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#physics
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