66 lines
2.9 KiB
Markdown
66 lines
2.9 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. $v = v_0 + at$ - Use when missing $x$
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2. $x = x_0 + \frac{1}{2}(v_0 + v)t$ - Use when missing $a$
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3. $x = x_0 + v_0 t + \frac{1}{2} a t^2$ - Use when missing $v$
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4. $v^2 = v_0^2 + 2a(x - x_0)$ - Use when missing $t$
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(Actually useful equations)
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$$ t = \frac{v_f - v_i}{a} $$
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$$s = v_it + \frac{1}{2}at^2 $$
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| -------------- | ----- |
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| $t_0$ | $t$ |
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| $v_0$ | $v$ |
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| $x_0$ | $x$ |
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| $a$ (constant) | $a$ |
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## Examples
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> Sally aggressively drives her Alfa Romeo from rest to 50 m/s in 6s. What is her acceleration?
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1. Start:
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- $v_0 = 0 m/s$
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- $t_0 = 0s$
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- $x_0 = 0m$
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2. Final:
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- $v = 50 m/s$
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- $t = 6s$
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- $x = ?$
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3. Using the equation $v = v_0 + at \to a = \frac{(v - v_0)}{t}$, $a$ = $(50 m/s - 0 m/s)/ 6s = 8.3m/s^2$
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4. Assess: do the units make sense? Is the answer reasonable?
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> An object slides until coming to a rest. It traveled 15 meters in 3 seconds. What was the object's acceleration?
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1. The problem is a kinematics problem, refer to the above formula.
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2. Acceleration is a constant
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3. Problem has two unknowns, $v_0$ and $a$, let's solve for $v_0$ first.
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4. $x = x_0 + \frac{1}{2}(v_0 + v) t \to v_0 = 2*(x - x_0)/ t \to 2*(15)/3 = 10$
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5. $v = v_0 + at \to a = (v - v_0) /t \to a = (-10 m/s) / 3s = -3.33m/s^2$
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6. Asses: Units? Answer make sense? Significant figures? Acceleration speeding up or slowing down?
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# Gravity
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- Is the problem 1 dimensional?
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- Could draw axis pointing upwards, call it y axis
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- If not specified, assume acceleration is $g = 9.8 m/s^2$.
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- Direction is important, $g$ is down towards the earth, so it's often negative. The sign depends on your choice of axis. |