# physics

## Acceleration

### Acceleration

Acceleration: Change in velocity over time.

A car start to move to the east.
Its velocity will be 90km/h in 3 seconds.
How much is the acceleration?

$$\vec{a}:acceleration$$ $$\Delta \vec{v}:change \quad in \quad velocity=(90-0)km/h$$ $$\Delta t:time=3seconds$$

$$\vec{a}=\frac{\Delta \vec{v}}{\Delta t}=\frac{(90-0)km/h}{3sec} =\frac{30km/h}{sec}$$

※Velocity increases 30km/h every second.

$$\vec{a}=\frac{30kmph}{sec}=\frac{30\frac{km}{hour}}{sec} =30\frac{km}{hour*sec}*\frac{1}{3600}*\frac{hour}{sec} =\frac{1}{120}\frac{km}{sec^2}$$ $$\quad=\frac{1}{120}\frac{km}{sec^2}*\frac{1000}{1}*\frac{meters}{km}=8.33\frac{m}{sec^2} =\frac{8.33\frac{m}{sec}}{sec}$$

※Velocity increases 8.33m/s every second.

### Airbus A380 take-off time

$$take \quad off \quad velocity=280 km/hour$$ $$acceleration=1.0\frac{m/s}{s}=1.0\frac{m}{s^2}$$

How long take off last?

$$\vec{v} = 280\frac{km}{hour} = 280\frac{km}{hour}*\frac{1000}{1}*\frac{m}{km}*\frac{1}{3600}*\frac{hour}{s} =78\frac{m}{s}$$

$$\vec{a}=\frac{\Delta\vec{v}}{\Delta t} \rightarrow \Delta t =\frac{\Delta\vec{v}}{\vec{a}}$$ $$\Delta t = \frac{78\frac{m}{s}}{1.0\frac{m}{s^2}}=78s$$

### Airbus A380 take-off distance

How long is a runway needed to take off?

$$velocity:\vec{v}=78\frac{m}{s}$$ $$acceleration\vec{a}=1.0\frac{m}{s^2}$$

if acceleratin is constant,

$$Average \quad velocity = \frac{final \quad velocity + initial \quad velocity}{2}$$

$$\vec{v_{a}}=\frac{(78+0)\frac{m}{s}}{2} =39\frac{m}{s}$$

$$\vec{v}_{a}=\frac{\vec{s}}{\Delta t}\rightarrow \vec{s}=\vec{v}_{a}\centerdot \Delta t$$ $$\vec{s}=39\frac{m}{s}\centerdot 78s=3042m$$

### Why distance is area under velocity-time line

$$velocity:\vec{v}=5\frac{m}{s}\quad right \quad constant$$ $$time:t=5seconds$$

$$\vec{v}=\frac{\vec{s}}{t}\rightarrow \vec{s}=\vec{v}*t=5\frac{m}{s}*5s=25m$$

$$acceleration:\vec{a}=1\frac{m}{s^2} \quad constant$$ $$initial\quad velocity:\vec{v_{i}}=0$$ $$time:t=5seconds$$

Velocity increases 1m/s every second.

$$\vec{s}= \frac{1}{2}*t*\vec{v} =\frac{1}{2}*5s*5\frac{m}{s}=12.5m$$