Kinematics

Equations

$$x=x_0+V_{0x}t+0.5A_x{t}^2$$

$$V_x=V_{0x}+A_{x}t$$

$$V_x^2=V_{0x}^2+2A_x(x-x_0)$$

$$x=x_0+\frac{1}{2}(v_x+v_{0x})t$$

Centripetal

$$A_c=V^2/r$$

A stone thrown upwards

Gravity: $A=-9.8m/s^2$
Initial Velocity: $V_0=8m/s$
Initial Position: $x_0=5$
(In Graph, X is time, Y is position)

y=-9.8x^2 + 8x + 5   

$$x=A{t}^2+V0+x_0$$ $$x=-9.8t^2+8t+5$$

Final time

$$\frac{-b+\sqrt{b^2-4ac}}{2a}$$ $$\frac{-8+\sqrt{8^2-4(-9.8)5}}{2(-9.8)}$$ $$t_{final}=1.23084262738$$

Drone Descent

At 2m, -3m/s^2
Constant decel so x=0 when dx/dt = 0

y=-3x+2+(9/8)x^2  

Position Equation

$$x=-3t+2+\alpha t^2$$

Velocity equation

$$v=\partial x=-3t+2+\alpha t^2$$ $$v=-3+2\alpha t$$

Set both Position equation and velocity equation to zero because that’s what we care about, then solve for t.
Velocity:

$$t=\frac{3}{2\alpha }$$

Position:

$$t=\frac{3\pm \sqrt{9-8\alpha }}{2\alpha }$$

Solve as system to find A:

$$\frac{3}{2\alpha }=\frac{3\pm \sqrt{9-8\alpha }}{2\alpha }$$ $$3=3\pm \sqrt{9-8\alpha }$$ $$0=\sqrt{9-8\alpha }$$ $$0=9-8\alpha$$ $$\alpha =9/8$$

$\alpha$ is the coefficient responsible for acceleration in the equation for the drones position

But equation for position in terms of acceleration (derived from Constant acceleration eq 1) (with other factors removed) is:

$$x=\frac{1}{2}A{t}^2$$

Our equation (with other factors removed) is:

$$x=\frac{9}{8}\alpha t^2$$

So for $\alpha$ to equal $A$, we need to factor out the $1/2$ coefficient, resulting in:

$$x=\frac{1}{2}\frac{9}{4}{t}^2$$

Antelope Accelerating

$\Delta X$ = 65
$\Delta t$ = 7.8
$V_{t=7.8}$ = 15.9

Diver at an Angle

V = 8 m/s & 30* above horizontal
Hits water 1.9s later
What was height of diving board
What angle did diver enter water

X axis

$V_0=4\sqrt{3}m/s^2$
$t=1.9s$
$a=-9.8m/s^2$
$x=0$

$$x=x_0+V_{0x}t-48.11$$ $$0=x_0+4\sqrt{3}\cdot 1.9-48.11$$ $$x_0=-4\sqrt{3}\cdot 1.9-48.11$$

__$x_0=34.9464138625$

Y axis

$V_0=4$
$t=1.9$
$a=0$
$Y_0=0$

$$Y_t=4\ast 1.9$$

$Y_t=7.6$

Combine

Car Ramp

Length of Ramp

$X_0=0$
$X=5$
$V_0=0$
$t=0.8$

$$x=x_0+\frac{1}{2}(v_x+v_{0x})t$$ $$5=0+1/2(v_x+0)0.8$$

$V_x=12.5$

Angle = 36.87 degrees up from horizon

Y Axis (Up)

$$V_y=12.5\sin (36.87)$$

$V_y=7.5$
$A_y=-9.81m/s^2$
$Y_0=3$
$Y=0$

$$x=x_0+V_{0x}t+0.5A_x{t}^2$$ $$0=3+7.5t+0.5(-9.81)t^2$$

$t=1.8582$

X Axis (Right)

$$V_x=12.5cos(36.87)$$

$V_x=10$
$t=1.8582$

$$X=1.8582\cdot 10$$

$X=18.52$

Shooting water into tank

Spinning String or Hoop, axis aligned with gravity

$$T_y=G$$ $$T_x=C$$

Work and Energy