warehouse/content/physics/kinematics.md

title, date

title	date
Vectors and Kinematics	2024-08-19

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.

-> Prime of the Faith

Motion in 1D

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.

Reference Frames and Displacement

• Any measurement about motion is taken in terms of a reference frame.

Note

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.

• In one dimension, only one axis, the $x$-axis of a coordinate plane is used.
• the net distance an object has traveled is known as displacement
• total distance is the overall distance traveled by the object/particle regardless of reference frame or initial/final positions
• change in positions is described using \Delta x

Average Velocity

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.

\text{average speed}=\frac{\text{distance traveled}}{\text{time elapsed}}=\frac{\Delta x}{\Delta t}

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)|

Instantaneous Velocity

If the position equation is defined as x(t), then:

\text{instantaneous velocity} = \tfrac{\mathrm{d}x}{\mathrm{d}t} = v(t) = \lim_{\Delta t \to 0}{\tfrac{\Delta x}{\Delta t}}

Average Acceleration

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.

\text{average acceleration} = \frac{\text{change of velocity}}{\text{time elapsed}}

Or more mathematically:

a=\frac{v_2-v_1}{t_2-t_1}=\tfrac{\Delta v}{\Delta t}

Instantaneous Acceleration

If the velocity function is defined as v(t) (this notation only applies to 1 dimension):

a(t)=\tfrac{\mathrm{d}v}{\mathrm{d}t}=\lim_{\Delta t \to 0}{\tfrac{\Delta v}{\Delta t}}

Please take calculus/study calculus if you want a neuron or two to function during the course of this, well, course.

Motion at Constant Acceleration

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:

\bar{v}=\frac{x-x_0}{t-t_0}=\frac{x-x_0}{t}

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:

a=\frac{v-v_0}{t}

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:

v=v_0+at 

Note

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.

[!faq]- Answer 4.0*6.0=24\mathrm{m}/\mathrm{s}

Next, calculating the position:

x=x_0+\bar{v}t 

Because of the symmetric and uniform growth of the velocity over time:

\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})

This finally results in the formula for position in these scenarios:

x=x_0+tv_0+\tfrac{1}{2}at^2

Objects in free fall

Just remember that g=9.80\mathrm{m}/\mathrm{s}^2

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