Kinematic formulas and projectile motion

Average velocity for constant acceleration

+:to the right
-:to the left
$$initial \quad velocity:\vec{v_i}=5m/s$$ $$final \quad velocity:\vec{v_f}=\vec{v_i}+(\Delta t)\vec{a_{c}}$$ $$constant \quad acceleration:\vec{a_{c}}=2m/s^2$$ $$time:\Delta t=4s$$ How far does he travel?
final velocity $$\vec{v_{f}}=5\frac{m}{s}+(4s)(2\frac{m}{s^2}) =5\frac{m}{s}+8\frac{m}{s}=13\frac{m}{s}$$ displacement1 (area of blue rectangle) $$\vec{s_{1}}=\vec{v_{i}}*\Delta t = 5\frac{m}{s}*(4s) = 20m$$ displacement2 (area of green triangle) $$\vec{s_{2}}=\frac{1}{2}*(\vec{v_{f}}-\vec{v_{i}})*\Delta t = \frac{1}{2}*(13-5)\frac{m}{s}*(4s) = 16m$$ $$\vec{s}=\vec{s_{1}}+\vec{s_{2}}=20m+16m=36m$$ --*--*--*--*--*--*--*--*--*--*-- $$\vec{s}=\vec{v_{i}}*\Delta t + \frac{1}{2}*(\vec{v_{f}}-\vec{v_{i}})*\Delta t$$ $$\quad = \Delta t*(\vec{v_{i}}+\frac{1}{2}*(\vec{v_{f}}-\vec{v_{i}})) =\Delta t*(\vec{v_{i}}+\frac{1}{2}\vec{v_{f}}-\frac{1}{2}\vec{v_{i}})$$ $$\quad = \Delta t*(\frac{1}{2}\vec{v_{i}}+\frac{1}{2}\vec{v_{f}})$$ $$\quad = \Delta t*\frac{1}{2}(\vec{v_{i}}+\vec{v_{f}})$$
Average Velocity
$$\vec{s}=\Delta t*\frac{1}{2}(\vec{v_{i}}+\vec{v_{f}})$$ $$\quad =4s*\frac{1}{2}(5m/s+13m/s)=4s*9m/s=36m$$

Acceleration of aircraft carrier take off

$$runway = 80m$$ $$initial \quad velocity:\vec{v_{i}} = 0m/s$$ $$final \quad velocity:\vec{v_{f}} = 260\frac{km}{hour} =260*\frac{km}{hour}*\frac{1000}{1}*\frac{m}{km}*\frac{1}{3600}*\frac{hour}{s} =72m/s$$
$$displacement:\vec{s}=\vec{v_{a}}*\Delta t$$ $$constant \quad acceleration:\vec{a_{c}}= \frac{\Delta \vec{v}}{\Delta t}$$ $$change \quad in \quad velocity:\Delta \vec{v}=\vec{v_{f}}-\vec{v_{i}}$$ $$change \quad in \quad time:\Delta t$$ $$average \quad velocity:\vec{v_{a}}=\frac{\vec{v_{f}}+\vec{v_{i}}}{2}$$

$$\vec{a_{c}}= \frac{\Delta \vec{v}}{\Delta t} \quad \rightarrow \quad \Delta t=\frac{\Delta \vec{v}}{\vec{a_{c}}} =\frac{\vec{v_{f}}-\vec{v_{i}}}{\vec{a_{c}}}$$ $$\vec{s}=\vec{v_{a}}*\Delta t =\frac{\vec{v_{f}}+\vec{v_{i}}}{2}* \frac{\vec{v_{f}}-\vec{v_{i}}}{\vec{a_{c}}} =\frac{\vec{v_{f}}^2-\vec{v_{i}}^2}{2\vec{a_{c}}}$$ $$\qquad\qquad\qquad\qquad\downarrow$$ $$\vec{a_{c}}=\frac{\vec{v_{f}}^2-\vec{v_{i}}^2}{2\vec{s}} =\frac{(72m/s)^2-(0m/s)^2}{2*80m} =\frac{5184m^2/s^2}{160m} =32.4m/s^2$$ $$\Delta t=\frac{\vec{v_{f}}-\vec{v_{i}}}{\vec{a_{c}}} =\frac{72m/s}{32.4m/s^2} =2.2s$$

Deriving displacement as a function of time, acceleration and initial velocity

Throw a ball straight up into the air. $$go \quad upward:positive$$ $$go \quad downward:negative$$ $$initial \quad velocity:\vec{v_{i}}=19.6m/s$$ $$gravity \quad of \quad the \quad earth:g=9.8m/s^2$$ Earth's gravity accelerate an object downward to the center of Earth. So.. $$g=-9.8m/s^2$$
$$\vec{F}=m\vec{a} \qquad (\vec{F}:force \quad m:mass \quad \vec{a}:acceleration=g\quad assume \quad constant)$$ $$\vec{F}=G\frac{m_{1}m_{2}}{r^2}$$ $$ \quad (\vec{F}:force$$ $$ \quad m_{1}:mass \quad of \quad one \quad object \quad m_{2}:mass \quad of \quad another \quad object$$ $$ \quad r:distance \quad between \quad 2 \quad objects :radius \quad of \quad Earth)$$

displacement is $$\vec{s}=\vec{v_{avg}} \centerdot \Delta t$$ $$assuming \quad constant \quad \vec{a}$$ average velocity is $$\vec{v_{avg}}=\frac{\vec{v_{i}}+\vec{v_{f}}}{2}$$ final velocity is $$\vec{v_{f}}=\vec{v_{i}}+\vec{a} \centerdot \Delta t$$

$$\vec{s}= \frac{\vec{v_{i}}+\vec{v_{i}}+\vec{a} \centerdot \Delta t}{2} \centerdot \Delta t =(\frac{2\vec{v_{i}}}{2} +\frac{\vec{a}\Delta t}{2})\centerdot\Delta t =\vec{v_{i}} \centerdot \Delta t + \frac{1}{2}\vec{a}(\Delta t)^2$$

Use this equation in following section.
$$\vec{s} =\vec{v_{i}} \centerdot \Delta t + \frac{1}{2}\vec{a}(\Delta t)^2$$

Plotting projectile displacement, acceleration and velocity

displacement is $$\vec{s} =\vec{v_{i}} \centerdot \Delta t + \frac{1}{2}\vec{a}(\Delta t)^2$$ acceleration is $$\vec{a_{g}=-9.8m/s^2}$$ final velocity is $$\vec{v_{f}}=\vec{v_{i}}+\vec{a} \centerdot \Delta t$$
$$\Delta t$$ $$\vec{v_{f}}$$ $$\vec{s}$$
0 19.6 0
1 9.8 14.7
2 0 19.6
3 -9.8 14.7
4 -19.6 0

Projectile height given time

Throw a ball in the air. How high in the air?
Air resistance is negligible. $$\Delta t = 5seconds \quad in \quad the \quad air.$$ time upward = time downward

case upward

$$\Delta t_{up}=2.5seconds$$ $$initial \quad velocity:\vec{v_{i}}$$ $$final \quad velocity:\vec{v_{f}}=0$$ $$gravity \quad acceleration:\vec{a_{g}}=-9.8m/s^2$$ $$change \quad in \quad velocity: \Delta \vec{v}=\vec{v_{f}}-\vec{v_{i}} =(0-\vec{v_{i}}) =-\vec{v_{i}}$$ $$\Delta \vec{v} = \vec{a} \centerdot \Delta t$$

$$\Delta \vec{v} =-9.8m/s^2 \centerdot 2.5s=-24.5m/s$$ $$\Delta \vec{v} =-\vec{v_{i}}=-24.5m/s$$ $$\vec{v_{i}}=24.5m/s$$

$$displacement:\vec{s} = \vec{v_{avg}} \centerdot \Delta t$$

$$\vec{s}=\frac{24.5 + 0}{2}m/s \centerdot 2.5s=30.625m$$

Deriving max projectile displacement given time

$$\Delta t:total \quad time \quad in \quad the \quad air $$ $$\Delta t_{up}= \frac{\Delta t}{2} \quad 時間全体の半分が上昇する時間$$ $$initial \quad velocity:\vec{v_{i}}$$ $$final \quad velocity:\vec{v_{f}}=0$$ $$change \quad in \quad velocity: \Delta \vec{v}=\vec{v_{f}}-\vec{v_{i}} =0-\vec{v_{i}} =-\vec{v_{i}}$$ $$\Delta \vec{v} = \vec{a} \centerdot \Delta t$$ $$average \quad velocity:\frac{\vec{v_{i}}+\vec{v_{f}}}{2}$$ $$displacement:\vec{s} = \vec{v_{avg}} \centerdot \Delta t$$

change in velocity equals to minus initial velocity $$\Delta \vec{v} =-\vec{v_{i}} = -9.8m/s^2 \centerdot \frac{\Delta t}{2}$$ initial velocity is $$\vec{v_{i}}=4.9m/s^2 \centerdot \Delta t$$

displacement is $$\vec{s} = \frac{\vec{v_{i}}+\vec{v_{f}}}{2} \centerdot \frac{\Delta t}{2} \quad =\frac{4.9m/s^2 \centerdot \Delta t+0}{2}\centerdot \frac{\Delta t}{2} \quad =\frac{4.9m/s^2 \centerdot (\Delta t)^2}{4} \quad =1.225m/s^2 \centerdot (\Delta t)^2$$

max displacement
$$\vec{s_{max}}=1.225m/s^2 \centerdot (\Delta t)^2$$

Impact velocity given height

$$$$ $$\vec{v_{i}}= 0$$ $$$$ $$$$ $$$$ $$$$ $$\vec{v_{f}}= ?\quad negative \quad value$$
Throw a rock off to the ground. How fast is it going to be right before it hits to the ground?
$$air \quad is \quad negligible$$ $$height:h \quad meters$$ acceleration of gravity is assumed constant and $$\vec{a_{c}}=-9.8m/s^2$$ $$upward:positive \quad value$$ $$downward:negative \quad value$$ $$initial \quad velocity:\vec{v_{i}}=0m/s$$ $$displacement:\vec{s}=\vec{v_{avg}} \centerdot \Delta t$$

$$average \quad velocity \quad is \quad \frac{\vec{v_{i}}+\vec{v}_{f}}{2} =\frac{0+\vec{v}_{f}}{2} =\frac{\vec{v}_{f}}{2}$$

$$displacement \quad is \quad \vec{s}=\frac{\vec{v}_{f}}{2} \centerdot \Delta t$$

$$\Delta t \quad is \quad \Delta \vec{v}=\vec{a} \centerdot \Delta t \rightarrow \Delta t=\frac{\Delta \vec{v}}{\vec{a}}$$

$$displacement \quad is \quad \vec{s}=\frac{\vec{v}_{f}}{2} \centerdot \frac{\Delta \vec{v}}{\vec{a}}$$

$$change \quad in \quad velocity \quad is \quad \Delta \vec{v}=\vec{v_{f}}-\vec{v_{i}} =\vec{v_{f}}-0 =\vec{v_{f}}$$

$$displacement \quad is \quad \vec{s}=\frac{\vec{v}_{f}}{2} \centerdot \frac{\vec{v_{f}}}{\vec{a}}$$

$$\qquad \qquad \swarrow$$

$$2\vec{a}\vec{s}=(\vec{v_{f}})^2$$ $$\vec{s} = -h \quad A \quad rock \quad falls \quad down, \quad so \quad \vec{s} \quad is \quad negative$$ $$-19.6\frac{m}{s^2} \centerdot -hm = (\vec{v_{f}})^2$$ $$19.6\frac{m^2}{s^2} \centerdot h = (\vec{v_{f}})^2$$ $$final \quad velocity \quad is \quad negative \quad and \quad \vec{v_{f}}=-\sqrt{19.6h}\frac{m}{s}$$

$$h=5m \rightarrow \vec{v_{f}}=-9.9m/s$$

impact velocity

Viewing g as the value of Earth's gravitational field near the surface

$$g=-9.81m/s^2$$ g is the acceleration due to gravity near Earth's surface for an object in free fall to the center of the Earth.