====== 2. Motion in one and two dimensions ======
**<nspages courses:phy101:2 -h1 -textPages="" -simpleLineBreak>**

===== - Position ($x$) =====
In order to define the position of a particle at a particular instant in time, we need to first define a coordinate system or reference frame.

{{ :courses:phy101:3.2.3.png?nolink |}}

In the above example, home is the origin of a one-dimensional coordinate system and the person is walking from home in different directions in different time intervals as shown below.

^ Time $t_i$ (min) ^ Position $x_i$ (km) ^ Displacement $\Delta x_i$ (km) ^
| $t_0$ = 0 | $x_0$ = 0 | $\Delta x_0$ = 0 |
| $t_1$ = 9 | $x_1$ = 0.5 | $\Delta x_1$ = $x_1-x_0$ = 0.5 |
| $t_2$ = 18 | $x_2$= 0 | $\Delta x_2$ = $x_2-x_1$ = -0.5 |
| $t_3$ = 33 | $x_3$ = 1.0 | $\Delta x_3$ = $x_3-x_2$ = 1.0 |
| $t_4$ = 58 | $x_4$ = -0.75 | $\Delta x_4$ = $x_4-x_3$ = -1.75 |

The **position** is always measured from the origin. It is positive for right-ward movement and negative for left-ward movement. **Displacement**, on the other hand, is measured with respect to the previous position.

$$ \text{Position } \equiv \text{Current position } - \text{Origin}. $$
$$ \text{Displacement } \equiv \text{Current position } - \text{Previous position}. $$

Displacement is a vector and can be positive or negative, but and the magnitude of the displacement vector is **distance**.

===== - Velocity ($v$) =====
The position $x$ of a particle at different times $t$ is shown in the following $x-t$ graph.

{{ :courses:phy101:2.1.png?nolink |}}

Now the **average velocity** between, for example, times $t_1$ and $t_6$

$$ v_a = \frac{\Delta x}{\Delta t} = \frac{x_6-x_1}{t_6-t_1} $$

where $x_6$ is the position at time $t_6$ and so on. You immediately see that $v_a$ is actually the slope represented by the red line. Now imagine another average velocity between $t_2$ and $t_5$

$$ v'_a = \frac{x_5-x_2}{t_5-t_2} $$

which is represented by the green line in the graph. In the same fashion, if we keep decreasing the time interval, the slope will keep decreasing and when the interval $\Delta t$ is very close to write the equation for the **instantaneous velocity** at a particular instant in time $t_0$

$$ v = \lim_{\Delta t\rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} $$

which is represented by the blue line in the curve. You see that the blue line is a tangent to the curve and it describes the slope at point $t_0$. And this is also the definition the first derivative of the dependent variable $x$ with respect to the independent variable $t$.

$$ \text{First derivative } \equiv \text{ Slope of tangent line}. $$

Velocity is the first derivative of position with respect to time. So velocity is the slope of a tangent line at a particular point on an $x-t$ curve.

Because velocity is the rate of change of position, its unit is meter / second or **m/s**.

===== - Acceleration ($a$) =====
Now velocity can also change with time and the rate of change of velocity is called acceleration.

{{ :courses:phy101:2.2.png?nolink |}}

We can perform a similar thought experiment with the above $v-t$ graphs to define **average acceleration** as

$$ a_a = \frac{\Delta x}{\Delta t} $$

and, similarly, the **instantaneous acceleration**

$$ a = \lim_{\Delta t\rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{dv}{dt}. $$

So acceleration is sole of a tangent line at any point on a $v-t$ graph. And acceleration can also be expressed as a **second derivative** of position with respect to time:

$$ a = \frac{d}{dt}(v) = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}. $$

Geometrically speaking, the second derivative is related to curvature. So on an $x-t$ graph, the slope of a tangent line at any point gives the velocity, and the curvature at that point gives the acceleration.

$$ \text{Second derivative } \equiv \text{ Curvature}. $$

===== - Motion with varying acceleration =====
Now imagine that the position of a particle is given algebraically by the **equation of motion** $x(t)=\sin(t)$.

So at time $t=1.57$ s, the **position** of the particle is $x(t=1.57) = \sin(1.57)=1$ m.

Now we can immediately find the **velocity** and **acceleration** of the particle at any past ($t$ negative), present ($t=0$) or future ($t$ positive) time using differential calculus.

$$ v(t) = \frac{dx(t)}{dt} = \frac{d}{dt}(\sin t) = \cos(t). $$

$$ a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(\cos t) = -\sin(t). $$

The three curves $x(t)$, $v(t)$ and $a(t)$ are represented below by the blue, red and green lines, respectively.

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The tangent at time $t=3.14$ s is also shown for each curve using the blue, red and green straight lines. You can see the tangent at other points by hovering your mouse on the graph.
===== - Motion with constant acceleration =====
$$ x = x_0 + v_{av}t. $$

$$ v = v_0 + at. $$

Add $v_0$ on both sides of the above equation to get $v_{av}=v_0+at/2$ and replace this in the first equation.

$$ x = x_0 + v_0t + \frac{1}{2}at^2. $$



Replace $v_av$ and $t$ in the first equation to get

$$ v^2 = v_0^2 + 2a(x-x_0). $$

{{ :courses:phy101:2.3.png?nolink |}}

===== - Free falling bodies =====
{{ :courses:phy101:2.4.png?nolink |}}

$$ v = v_0 - gt. $$

$$ v^2 = v_0^2 - 2g(y-y_0). $$

$$ y = y_0 + v_0 t - \frac{1}{2} g t^2. $$

===== - Integral calculus =====
$$ x = \int v dt + C_1. $$

$$ v = \int a dt + C_2. $$
Prove $v=v_0+at$ for constant $a$ from this equation.

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===== - Motion in two dimensions =====
==== Position ====
{{ :courses:phy101:2.5.png?nolink |}}

==== Velocity and acceleration ====
{{ :courses:phy101:2.7.png?nolink |}}

===== - Motion of a projectile =====
{{ :courses:phy101:2.9.png?nolink |}}

Along the $x$-axis, there is no acceleration and hence $v_{0x}=v_x$ and

$$ x = x_0 + v_xt. $$

Along the $y$-axis, there is constant acceleration, so

$$ y = y_0 + v_{y,av}t = y_0 + \frac{v_{0y}+v_y}{2}t. $$

$$ v_y = v_{0y} - gt. $$

$$ y = y_0 + v_{0y} t - \frac{1}{2} g t^2. $$

$$ v_y^2 = v_{0y}^2 - 2g(y-y_0). $$

Use the 4th equation to get the **maximum height**

$$ H = \frac{v_{0y}^2}{2g} = \frac{v_0^2 \sin^2\theta}{2g}. $$

Use the 2nd equation to calculate the **time of flight**

$$ t_f = \frac{2 v_{0y}}{g}. $$

For $x_0=0$ m, $x=v_xt$ and, hence, $t = x/(v_0\cos\theta_0)$. Replace this time in the second equation to get the **trajectory**

$$ y = (\tan\theta_0)x - \left[\frac{g}{2(v_0\cos\theta_0)^2}\right]x^2 $$

which is an equation of a parabola ($y=ax+bx^2$).

Putting $y=0$ in the trajectory equation gives the **range**

$$ R = \frac{v_0^2 \sin 2\theta_0}{g}. $$

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