32 lines
1.4 KiB
Markdown
32 lines
1.4 KiB
Markdown
Motion in a straight line is one dimensional.
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Kinematics or the physics of motion has 4 noteworthy parameters: time ($t$), position ($x$), velocity ($v$), and acceleration ($a$).
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Kinematic problems have a start and end of motion.
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# Displacement
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Displacement is calculated with the formula:
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$$\Delta x = \text{x-value of final position} - \text{x-value of initial posiion}$$
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# Velocity
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Average velocity over a time interval $\Delta t$ is defined to be: **the displacement** (net change in position), **divided by** **the time taken**.
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$$ \bar{v} = \dfrac{\text{final position-initial position}}{\text{final time - initial time}} = \dfrac{x_2 - x_1}{t_2 - t_1} = \dfrac{\Delta x}{\Delta t}$$
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**Speed (m/s)** is defined to be the total distance traveled divided by the time taken. Speed and velocity are *not the same*.
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$$ v_{\text{instant}} = v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$
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- $x(t)$ -> **position** as a function of time
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- $v(t)$ -> **slope** of position-vs-time (derivative of $x(t)$)
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- $a(t)$ -> **slope** of velocity-vs-time (derivative of $v(t)$)
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# Acceleration
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To find the instantaneous acceleration, we can apply the formula:
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$$a_{\text{instant}} = a = \frac{dv}{dt} = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}$$
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## Equations of Motion for Constant Acceleration
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1. $v = v_0 + at$
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2. $x = x_0 + \frac{1}{2}(v_0 + v)t$
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3. $x = x_0 + v_0 t + \frac{1}{2} a t^2$
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4. $v^2 = v_0^2 + 2a(x - x_0)$
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