MOTION

Understanding Motion

Reference point and reference frame

• To describe the position of an object we need a reference point or origin. An object may seem to be moving to one observer and stationary to another.
• Example : A passenger inside a bus sees the other passengers to be at rest, whereas an observer outside the bus sees the passengers are in motion.
• In order to make observations easy, a convention or a common reference point or frame is needed. All objects must be in the same reference frame. 

Distance and Displacement

Distance and displacement

The magnitude of the length covered by a moving object is called distance. It has no direction.-

- Displacement is the shortest distance between two points or the distance between the starting and final positions with respect to time. It has magnitude as well direction.

Distance VS Displacement

- Displacement can be zero, but distance cannot.

Magnitude and direction

• Magnitude is the size or extent of a physical quantity. In physics, we have scalar and vector quantities.
• Scalar quantities are only expressed as magnitude. E.g: time, distance, mass, temperature, area, volume
• Vector quantities are expressed in magnitude as well as direction of the object. E.g: Velocity, displacement, weight, momentum, force, acceleration etc.

Time, Average Speed and Velocity

Time and speed

• Time is the duration of an event that is expressed in Seconds. Most physical phenomena occur with respect to time. It is a scalar quantity.
• Speed is the rate of change of distance. If a body covers a certain distance in a certain amount of time, its speed is given by $\frac{Distance}{Time}$- 
• Average speed $=\frac{TotalDistancetravelled}{TotalTimetaken}$

Uniform motion and non-uniform motion

• When an object covers equal distances in equal intervals of time it is in uniform motion.
• When an object covers unequal distances in equal intervals of time it is said to be in non-uniform motion.

Velocity

• Rate of change of displacement is velocity. It is a vector quantity. Here the direction of motion is specified.  Velocity is $\frac{Displacement}{Time}$ 
• Average velocity = $\frac{Initialvelocity+Finalvelocity}{2}=\frac{u+v}{2}$

Acceleration

Acceleration

• The rate of change of velocity is called acceleration it is a vector quantity.
• In non-uniform motion
  velocity varies with time, i.e change in velocity is not 0.
• $Acceleration=\frac{Changeinvelocity}{Time}$
• $Or,a=\frac{v-u}{t}$   

Motion Visualised

Distance-time graphs

• Distance-Time graphs show the change in position of an object with respect to time.
• Linear variation = uniform motion & non-linear variations imply non- uniform motion
• Slope gives us speed

Distance - Time Graph

• OA implies uniform motion with constant speed as slope is constant
• AB implies body is at rest as slope is zero
• B to C is non-uniform motion

Velocity-time graphs

• Velocity-Time graphs show the change in velocity with respect to time.
• Slope gives acceleration
• Area under curve gives displacement
• Line parallel to x-axis implies constant velocity- 

Velocity - Time Graph

OA = constant acceleration, AB = constant velocity , BC = constant retardation

Equations of Motion

Equations of motion

The motion of an object moving at uniform acceleration can be described with the help of three equations, namely
     (i) $v=u=at$
     (ii) $v^2-u^2=2as$
     (iii) $s=ut+\frac{1}{2}a{t}^2$

Derivation of velocity-time relation by graphical method

Velocity - Time Graph

• A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a .
• From the graph BC = v (final velocity)... DC = u (initial velocity).....(eq 1)
• BD = BC - DC… (eq 2)
• We know acceleration a = slope = BD/AD or AD = OC = t (time taken to reach point B)
• $\therefore$ BD = at…(eq 3)
• Substitute everything in (we get : at = v - ueq 2)
• Rearrange to get v = u + at

Derivation of position-time relation by graphical method

Velocity - Time Graph

• A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a 
• Area under the graph gives Displacement $=A\left(\Delta ABD\right)+A\left(\square OADC\right)=(\frac{1}{2}AD\times BD)+OA\times OC$.....(1)
• OA = u , OC = t and BD = at
• Substituting in (1) we get $s=ut+\frac{1}{2}a{t}^2$

Derivation of position-velocity relation by graphical method

Velocity - Time Graph

• A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a 
• Displacement covered will be area under the curve which is the trapezium OABC.
• We know area of trapezium $=s=\frac{(OA+BC)}{2\times OC}$
• OA = u and BC = v and OC = t
• $\therefore s=\frac{(v+u)}{2\times t}$ ....(eq1)
• We also know that $t=\frac{(v-u)}{a}$..... (eq 2)
• Substitute (eq 2) in (eq 1), and arrange to get 
• $v^2-u^2=2as$

Uniform Circular Motion

Uniform circular motion

• If an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
• Velocity is changing as direction keeps changing.
• Acceleration is constant