From e0a4ea7fdbc565200b3d325e3160f232b7fbeb5e Mon Sep 17 00:00:00 2001 From: arc Date: Wed, 3 Sep 2025 13:47:38 -0600 Subject: [PATCH] vault backup: 2025-09-03 13:47:38 --- education/physics/PHYS2210/Unit 1.md | 8 +++++++- 1 file changed, 7 insertions(+), 1 deletion(-) diff --git a/education/physics/PHYS2210/Unit 1.md b/education/physics/PHYS2210/Unit 1.md index 2de2417..c19a618 100644 --- a/education/physics/PHYS2210/Unit 1.md +++ b/education/physics/PHYS2210/Unit 1.md @@ -22,4 +22,10 @@ $$ v_{\text{instant}} = v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \fr - $a(t)$ -> **slope** of velocity-vs-time (derivative of $v(t)$) # Acceleration To find the instantaneous acceleration, we can apply the formula: -$$a_{\text{instant}} = a = \frac{dv}{dt} = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}$$ \ No newline at end of file + +$$a_{\text{instant}} = a = \frac{dv}{dt} = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}$$ +## Equations of Motion for Constant Acceleration +1. $v = v_0 + at$ +2. $x = x_0 + \frac{1}{2}(v_0 + v)t$ +3. $x = x_0 + v_0 t + \frac{1}{2} a t^2$ +4. $v^2 = v_0^2 + 2a(x - x_0)$