# A-level Physics/Forces and Motion/Kinematics

Kinematics is the study of the way objects move. It focuses on describing an object's motion, and doesn't explain how forces affect it.

## Distance and displacement

You may already be familiar with the term distance, as the distance between two points is the length of the path a body takes between those two points.
Distance is a scalar, so if you were to walk 10m North, and then 10m South, you would have covered a distance of 20m.
Displacement, however, is a vector quantity. Displacement, in a sense, is simply the shortest distance between any two points. If a body ends up at the same spot as its initial position after travelling through some distance, we say that the displacement of the body is 0, so the above example would give you a total displacement of 0m.
In the diagram on the right, if the distance covered was 25m, then the displacement would be 10m. You can find the displacement by measuring the length of the line between the start and end points.
A measurement is a displacement if it has a specified direction, otherwise it is a distance.
The symbol for distance is d, and the symbol for displacement is s or x. Be careful not to confuse the s for displacement with s for seconds.

## Speed and velocity

The speed of an object is the distance it moves in a unit of time.
You can find the speed of an object if you know the distance an object moved, and the time it took to move that distance:
$average\ speed={\frac {distance\ moved}{time\ taken}}$, or $s={\frac {d}{t}}$.
Velocity is a vector, and similar to the difference between distance and displacement, velocity is speed in a specified direction.
A vehicle could be moving with constant speed, but have a changing velocity. This happens when the vehicle turns. Imagine a racing car is moving along the track with a speed of 20m s$^{−1}$. If this direction is taken to be positive, then the car's velocity is also 20m s$^{−1}$. Now, if the car was to turn into a "U" bend, its velocity would change. When the car is perpendicular to the first straight, the car will still have a speed of 20 m/s$^{−1}$, but its velocity will now be 0 m/s$^{−1}$. When the car has made the turn and is coming back to the starting point, the speed is still 20 m/s$^{−1}$, but the velocity is -20 m/s$^{−1}$, since the car is now moving in the opposite direction.
The symbol for speed is s, and the symbol for velocity is v. Be careful not to confuse the s for speed with s for displacement or s for seconds.

## Acceleration

Acceleration is the rate of change of velocity. In other words, acceleration is the amount an object's velocity changes in a unit of time.
If you know the change in velocity and the time the change took, you can find acceleration using the formula:
$acceleration={\frac {change\ in\ velocity}{time\ taken\ for\ change}}$, or $a={\frac {\Delta v}{\Delta t}}$.
Alternatively, if you have the initial and final velocities, you can use the formula:
$a={\frac {v-u}{\Delta t}}$, where $u$ is the initial velocity, and $v$ is the final velocity.
$\Delta$ (Delta) means "change in".
Acceleration is a vector, and can slow down objects as well as speed them up. An object will slow down when its acceleration is opposite to its velocity. The object is now decelerating.
Something can be said to be accelerating if it's changing direction. In the example above, the car changes its velocity by turning a corner. Since acceleration is the rate of change of velocity, the car is accelerating.
Acceleration is measured in metres per second per second, or $m/s^{2}$. If something had an acceleration of $10m/s^{2}$, it means that its speed increases by $10m/s$ each second.

## Measuring speed and acceleration
### Light gates

The Light Gate has an infrared transmitter and receiver, mounted in a robust steel housing avoids any misalignment problems. The Light Gate can be used for studying free fall, air track and incline plane experiments.

### Ticker tape timer

The ticker tape timer is used in the measurement of velocity acceleration and general timing. It has a frequency of 50 to 60 Hz (varies according to type) equivalent to that of the mains power supply. It will give good results if operated from a 12V a.c. power supply. The timer uses an electromagnet which activates a striker producing dots via a carbon disk on the ticker tape. At 50 Hz, each dot will represent 0.02seconds.While at 60 Hz, one dot represents 0.01seconds.

### Accelerometer

This device is used for the measurement of acceleration and the unit is in metre per second square (m/s$^2$)

## The equations of motion

From the definitions of position, velocity and acceleration, one can derive the relationships between the three vectors.
* ${\vec {v_{f}}}={\vec {v_{i}}}+{\vec {a}}\Delta t\$
* $\Delta {\vec {s}}={\vec {v_{i}}}\Delta t+{\frac {1}{2}}{\vec {a}}(\Delta t)^{2}$
* ${\vec {v_{f}}}^{2}={\vec {v_{i}}}^{2}+2{\vec {a}}\Delta s\$
Where ${\vec {a}}$ is acceleration in the direction of the velocity, ${\vec {s}}$ is displacement, $t$ is time, ${\vec {v_{i}}}$ is initial velocity and ${\vec {v_{f}}}$ is final velocity. Note that these equations only work when an object has constant acceleration and the direction of motion is linear.

### Using the equations

A car accelerate uniformly from rest at 5m/s². What is it velocity after 5 seconds. Firstly, we gather our data from the question above; Acceleration [a]= 5m/s² Time[∆t]= 0m/s² Initial velocity[u]= 0m/s Final velocity[v]=? Using the data gathered, we go for the equation: v=u+a∆t We then substitute our data for the equation: v=0m/s² + (5m/s²×10s) v=0m/s² + 25m/s v=25m/s The car attains a velocity of 25m/s after five seconds from start.

### Deriving the equations

As stated above, the equations of motion are derived from:
${\vec {v}}={\frac {\Delta {\vec {s}}}{\Delta t}},{\vec {a}}={\frac {\Delta {\vec {v}}}{\Delta t}}$.

#### Deriving $\Delta s=v_{avg}\times \Delta t$

The expression ${\frac {v_{i}+v_{f}}{2}}$ is the average velocity during the time frame. This is allowed only in a linear context; non linear kinematics requires the use of calculus. Because the average velocity accounts for the variations of the velocities during a linear change, the total distance traveled is thus the velocity during the time frame multiplied by the time.

#### Deriving s = ut + 1/2 a(Δt)$^2$

This equation is derived from the equations $v=u+a\Delta t\$ and $\Delta s={\frac {(u+v)}{2}}\cdot \Delta t$.
Substitute $v=u+a\Delta t\$ into the equation $\Delta s={\frac {(u+v)}{2}}\cdot \Delta t$
    $\Delta s\$ = ${\frac {u+u+a\Delta t}{2}}\cdot \Delta t$ = ${\frac {2u\Delta t}{2}}+{\frac {a(\Delta t)^{2}}{2}}$ = $u\Delta t+{\frac {1}{2}}a(\Delta t)^{2}$
An alternative derivation that is more commonly used is done by Calculus. Assuming $v=u+at$
* ${\frac {ds}{dt}}=v$
* $s=\int {vdt}$
* $s=\int {(u+at)dt}$
* $s=ut+{\frac {at^{2}}{2}}+C$
* the boundary condition is that at t = 0, \Delta s = 0 so C = 0. Thus the equation becomes
* $s=ut+{\frac {at^{2}}{2}}$

#### Deriving v$^2$ = u$^2$ + 2aΔs

This equation is derived from the equations $\Delta s=u\Delta t+{\frac {1}{2}}a(\Delta t)^{2}$ and $a={\frac {v-u}{\Delta t}}$.
Multiply $\Delta s=u\Delta t+{\frac {1}{2}}a(\Delta t)^{2}$ by $a={\frac {v-u}{\Delta t}}$
    $a\Delta s={\frac {1}{2}}(v+u)\cdot \Delta t\cdot {\frac {(v-u)}{\Delta t}}$
    $2a\Delta s=(v+u)(v-u)\$
    $2a\Delta s=v^{2}-u^{2}\$
    $v^{2}=u^{2}+2a\Delta s\$

### Deriving the equations in vectors

* All constant are taken in capital letters
* All variable are given in small letter

#### Deriving v(t) = U$_i$ + A (t - T$_i$)

    ${\vec {a}}={\frac {d{\vec {v}}}{dt}}={\frac {d^{2}\ {\vec {s}}}{dt^{2}}}$
so
    ${\vec {v}}(t)=\int _{T_{i}}^{t}{\vec {a}}dt$
for zero or constant acceleration A we have
    ${\vec {v}}(t)={\vec {A}}\int _{T_{i}}^{t}dt$
    ${\vec {v}}(t)={\vec {A}}\,[t]_{T_{i}}^{t}$
    ${\vec {v}}(t)={\vec {A}}\,(t-{T_{i}})+K$
we have one unknown K , we need to consider initial or final condition
* Lets take initially we have ${\vec {v}}(T_{i})={\vec {U}}_{i}$
    ${\vec {U}}_{i}={\vec {A}}\,({T_{i}}-{T_{i}})+K$
i.e.
    $k={\vec {U}}_{i}$
    ${\vec {v}}(t)={\vec {U}}_{i}+{\vec {A}}\,(t-{T_{i}})$
If we take final condition in consideration then i.e.: ${\vec {v}}(T_{f})={\vec {V}}_{f}$
    ${\vec {v}}(t)={\vec {V}}_{f}+{\vec {A}}\,({T_{f}}-t)$
constant acceleration can be found using initial and final condition
    ${\vec {A}}={\frac {{\vec {V}}_{f}-{\vec {U}}_{i}}{T_{f}-T_{i}}}$

#### Deriving s(t) = S$_i$ + U$_i$(t-T$_i$) + (1/2)A (t - T$_i$)$^2$

now we have
    ${\vec {v}}(t)={\frac {d{\vec {s}}}{dt}}={\vec {U}}_{i}+{\vec {A}}\,(t-{T_{i}})$
    ${\vec {s}}(t)=\int _{T_{i}}^{t}{\Big{\vec {U}}_{i}+{\vec {A}}\,(t-{T_{i}}){\Big )}{dt}$
    ${\vec {s}}(t)=({\vec {U}}_{i}-{\vec {A}}\,T_{i})\int _{T_{i}}^{t}{dt}+{\vec {A}}\int _{T_{i}}^{t}t\ {dt}$
    ${\vec {s}}(t)=({\vec {U}}_{i}-{\vec {A}}\,T_{i})\ {\Bigt{\Big ]}_{T_{i}}^{t}+{\vec {A}}{\Bigg{\frac {t^{2}}{2}}{\Bigg ]}_{T_{i}}^{t}+K$
    ${\vec {s}}(t)=({\vec {U}}_{i}-{\vec {A}}\,T_{i})({t}-{T_{i}})+{\frac {\vec {A}}{2}}(t^{2}-{T_{i}}^{2})+K$
for eliminating K we need either final or initial condition i.e. ${\vec {s}}(T_{i})={\vec {S}}_{i}$
    ${\vec {S_{i}}}=({\vec {U}}_{i}-{\vec {A}}\,T_{i})({T_{i}}-{T_{i}})+{\frac {\vec {A}}{2}}({T_{i}}^{2}-{T_{i}}^{2})+K$
    $K={\vec {S_{i}}}$
    ${\vec {s}}(t)={\vec {S_{i}}}+({\vec {U}}_{i}-{\vec {A}}\,T_{i})({t}-{T_{i}})+{\frac {\vec {A}}{2}}(t^{2}-{T_{i}}^{2})$
    ${\vec {s}}(t)={\vec {S_{i}}}+{\vec {U}}_{i}({t}-{T_{i}})-{\vec {A}}({t}T_{i}-{T_{i}}^{2})+{\frac {\vec {A}}{2}}(t^{2}-{T_{i}}^{2})$
    ${\vec {s}}(t)={\vec {S_{i}}}+{\vec {U}}_{i}({t}-{T_{i}})+{\frac {\vec {A}}{2}}(t^{2}-2{t}T_{i}+{T_{i}}^{2})$
    ${\vec {s}}(t)={\vec {S_{i}}}+{\vec {U}}_{i}({t}-{T_{i}})+{\frac {\vec {A}}{2}}(t-T_{i})^{2}$
If we would have taken final condition i.e. ${\vec {s}}(T_{f})={\vec {S}}_{f}$ then
    ${\vec {s}}(t)={\vec {S_{f}}}+{\vec {V}}_{f}({T_{f}}-{t})-{\frac {\vec {A}}{2}}(T_{f}-t)^{2}$

#### Deriving |v(t)|$^2$ = |U$_i$|$^2$ + 2 A . (s(t)-S_i)

    ${\vec {v}}(t)={\vec {U}}_{i}+{\vec {A}}\,(t-{T_{i}})$
    ${\vec {v}}(t)\cdot {\vec {v}}(t)={\big{\vec {U}}_{i}+{\vec {A}}\,(t-{T_{i}}){\big )}\cdot {\big{\vec {U}}_{i}+{\vec {A}}\,(t-{T_{i}}){\big )}$
    ${\vec {v}}(t)\cdot {\vec {v}}(t)={\vec {U}}_{i}\cdot {\vec {U}}_{i}+{\vec {A}}\cdot {\vec {A}}\ (t-T_{i})^{2}+2{\vec {A}}\cdot {\vec {U}}_{i}(t-T_{i})$
    ${\vec {v}}(t)\cdot {\vec {v}}(t)={\vec {U}}_{i}\cdot {\vec {U}}_{i}+{\vec {A}}\cdot {\vec {A}}\ (t-T_{i})^{2}+2{\vec {A}}\cdot {\Bigg{\vec {s}}(t)-{\vec {S}}_{i}-{\frac {\vec {A}}{2}}(t-T_{i})^{2}{\Bigg )}$
    ${\vec {v}}(t)\cdot {\vec {v}}(t)={\vec {U}}_{i}\cdot {\vec {U}}_{i}+{\vec {A}}\cdot {\vec {A}}\ (t-T_{i})^{2}+2{\vec {A}}\cdot ({\vec {s}}(t)-{\vec {S}}_{i})-2{\vec {A}}\cdot ({\frac {\vec {A}}{2}}(t-T_{i})^{2})$
    ${\vec {v}}(t)\cdot {\vec {v}}(t)={\vec {U}}_{i}\cdot {\vec {U}}_{i}+2{\vec {A}}\cdot ({\vec {s}}(t)-{\vec {S}}_{i})$
taking case for final velocity
    ${\vec {V}}_{f}\cdot {\vec {V}}_{f}={\vec {U}}_{i}\cdot {\vec {U}}_{i}+2{\vec {A}}\cdot ({\vec {S}}_{f}-{\vec {S}}_{i})$
or
    $\left|{\vec {V}}_{f}\right|^{2}=\left|{\vec {U}}_{i}\right|^{2}+2{\vec {A}}\cdot ({\vec {S}}_{f}-{\vec {S}}_{i})$