Motion In One Dimension |

 Introduction

If a body is in uniform motion (velocity = constant or body has acceleration and velocity acting at 0° or 180°, then path followed is straight line and motion is rectilinear motion.

Uniform motion

$\to$ Velocity is constant and acceleration is zero.

$\to$ Body covers equal displacement in equal time interval.

$\to$ Equation for uniform motion s = ut .

$\to$ time-displacement graph is straight line .

$\to$ Time velocity graph is straight line parallel to time axis

$\to$ Time acceleration graph is straight line coincide it to time axis.

Variable motion at constant acceleration

$\to$ Velocity changes at constant rate and acceleration is constant.

$\to$ Body may cover equal distances in equal time intervals but never covers equal displacements in equal time interval (a and v at 180°)

Distance and Displacement

$\to$ The length of actual path followed by a moving body is called distance. It is a scalar quantity.

$\to$ The shortest distance between initial and final position with direction for a moving body is called displacement . It is a vector quantity.

$\to$ Displacement covered by a moving body in certain time interval may be zero, negative or positive but distance never be zero or negative.

$\to$ Distance ≥ Displacement

$\to \dfrac{\text{Distance}}{\text{Displacement}} ≥ 1$

$\to$ For a moving body, the distance travelled always increases with time displacement may increase or decrease with time.

Speed and Velocity

Speed
$\to$ The time rate of change in distance for a moving body is called speed. It is a scalar quantity.
$\to$ Speed =$\dfrac{\text{distance}}{\text{time taken}}$
$\to$ The speed of a body can be zero or positive but never negative.
$\to$ A body is said to be moving with uniform speed if it covers equal distances in equal intervals of time.
$\to$ A body is said to be moving with a variable speed if it covers equal distances in unequal intervals of time or unequal distances in equal intervals of time.
$\to$ Average speed = $\dfrac{\text{distance covered in given time}}{\text{time taken}}$
$V_{av} = \dfrac{(x-x_0)}{(t-t_0)} = \dfrac{∆x}{∆t}$
Instantaneous speed is limiting value of average speed as time interval tends to zero.
$V_{av} = \displaystyle{\lim_{∆t \to 0}} \dfrac{∆x}{∆t} = \dfrac{dx}{dt}$
It is defined for a point of time .

Velocity

$\to$ The time rate of change in displacement tor a moving body is called velocity . It is a vector quantity .

$\to$ Velocity=$\dfrac{\text{displacement}}{\text{time taken}}$

$\to$ Average velocity = $\dfrac{\text{displacement covered in given time}}{\text{time taken}}$

$V_{av} = \dfrac{(s-s_0)}{(t-t_0)} = \dfrac{∆s}{∆t}$

It is defined for period of time.

Instantaneous speed is limiting value of average speed as time interval tends to zero.

$V= \displaystyle{\lim_{∆t \to 0}} \dfrac{∆s}{∆t} = \dfrac{ds}{dt}$

It is defined for a point of time .

Average velocity and Average speed

$\to$ Average velocity for a moving body in certain time interval may be zero, +ve or -ve but average speed can never be zero or –ve .

$\to$ Average speed ≥ Average velocity

$\to \dfrac{\text{Average speed}}{\text{Average velocity}} ≥1$

Average velocity

When a body moving along straight line covers distances $s_1 , s_2 , . . . s_n$ at different velocities $v_1 , v_2, ... v_n$ then

Average velocity = $\dfrac{\text{Total displacement}}{\text{Total time}}$

$V_{av}= \dfrac{(s_1 +s_2+ , . . .+ s_n)}{(t_1 + t_2+ .. . . .+ t_n)}$

$V_{av}= \dfrac{(s_1 +s_2+ , . . .+ s_n)}{(\dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}+ .. . . .+ \dfrac{s_n}{v_n})}$

$\dfrac{(s_1 +s_2+ , . . .+ s_n)}{V_{av}} = \dfrac{s_1}{v_1} + \dfrac{s_2}{v_2}+ .. . . .+ \dfrac{s_n}{v_n}$

Equal Distance

If $s_1=s_2 . . . . . =s_n$ when body covers equal distances then ,

$\dfrac{(ns)}{V_{av}} = \dfrac{s}{v_1} + \dfrac{s}{v_2}+ .. . . .+ \dfrac{s}{v_n}$

$\dfrac{(n)}{V_{av}} = \dfrac{1}{v_1} + \dfrac{1}{v_2}+ .. . . .+ \dfrac{1}{v_n}$

Average velocity is harmonic mean of individual velocities .

For two equal distance

$V_{av} = \dfrac{2v_1v_2}{v_1+v_2}$

Equal Time

If $t_1=t_2 . . . . . =t_n$ when body covers equal distances then ,

$V_{av}= \dfrac{(s_1 +s_2+ , . . .+ s_n)}{(t_1 + t_2+ .. . . .+ t_n)}$

$V_{av}= \dfrac{(v_1t_1 +v_2t_2+ , . . .+ v_nt_n)}{(t_1 + t_2+ .. . . .+ t_n)}$

$V_{av}= \dfrac{(v_1 + v_2 + ...+ v_n)}{(n)}$

For two equal time.

$V_{av} = \dfrac{v_1+v_2}{2}$

When a moving body covers two equal parts of distances with two different constant velocities $v_1$ and $v_2$ in same or any directions, then

Average speed (v) is given by

$= \dfrac{2v_1v_2}{v_1+v_2}$

But average velocity is given by

$0 ≤ \text{average speed} ≤ \dfrac{2v_1v_2}{v_1+v_2}$

When a moving body travels at two different constant velocities $v_1$ and $v_2$ for equal intervals of time, then Average speed =$\dfrac{v_1+v_2}{2}$

$\dfrac{v_1-v_2}{2} ≤ \text{average speed} ≤ \dfrac{v_1+v_2}{2}$

Acceleration

The rate of change of velocity with respect to time is called acceleration. It is a vector quantity.

$\to$ Acceleration=$\dfrac{\text{velocity}}{\text{time taken}}$

$\to$ Average acceleration = $\dfrac{\text{change in velocity}}{\text{time taken}}$

$\overrightarrow{a_{av}} = \dfrac{(\overrightarrow{v}-\overrightarrow{v_0})}{(t-t_0)} = \dfrac{∆\overrightarrow{v}}{∆t}$

It is defined for a period of time.

Instantaneous acceleration is limiting the value of average speed as the time interval tends to zero.

$\overrightarrow{a} = \displaystyle{\lim_{∆t \to 0}} \dfrac{∆\overrightarrow{v}}{∆t} = \dfrac{d\overrightarrow{v}}{dt}$

It is defined for a point of time.

Acceleration

$\to$ If the change in velocity is constant with time. It is either increasing or decreasing by an equal amount in equal interval of time. It is a uniform acceleration.

$\to$ If the change in velocity is different with time. It is either increasing or decreasing by an unequal amount in equal interval of time. It is a non-uniform acceleration.

If $\overrightarrow{s} = \overrightarrow{s}(t)$

$\to$ velocity(v) = $\dfrac{d\overrightarrow{s}}{dt}$

$\to$ acceleration(a) = $\dfrac{d\overrightarrow{v}}{dt} = \dfrac{d^2\overrightarrow{s}}{dt^2}$

$\to s \propto t^0 \to \text{Rest}$

$\to s \propto t^1 \to \text{Uniform velocity}$

$\to s \propto t^2 \to \text{Uniform acceleration}$

$\to s \propto t^3 \to \text{Variable acceleration}$

If $\overrightarrow{v} = \overrightarrow{v}(t)$

$\to$ displacement(s) = $\int \overrightarrow{v} dt$

$\to$ acceleration(a) = $\dfrac{d\overrightarrow{v}(t)}{dt}$

If $\overrightarrow{a} = \overrightarrow{a}(t)$

$\to$ velocity(v) = $\int \overrightarrow{a} dt$

$\to$ displacement(s) = $\int \overrightarrow{v} dt$

Motion with Constant Acceleration

When the motion of the body starts from rest and moves with constant acceleration 'a'. The distance traveled and velocity in nth second is given by:

$s_n = \dfrac{a}{2}(2n-1)$

$v_n = at$

$\to$ The distance traveled in successive equal interval of time are in the ratio of $1:3:5:7 ...: (2n-1)$

$i.e.$ $s_1:s_2:s_3:...:s_n = 1:3:5:...:(2n-1)$

$\to$ The distance after the end of the successive equal interval of time are in the ratio of $1:4:9:16 ...:n^2$

$i.e.$ $s_1:(s_1+s_2):(s_1+s_2+s_3):...(s_1+s_2+...+s_n) = 1:4:9:...:n^2$

$\to$ The velocity after the end of the successive equal interval of time are in the ratio of $1:2:3:4 ...:n$

$i.e.$ $v_1:v_2:v_3:...:v_n= 1:2:3:...:n$

$\to$ The time taken to reach the end of successive equal intervals of time are in the ratio of $1:\sqrt{2}:\sqrt{3}:\sqrt{4} ...:\sqrt{n}$

$i.e.$ $t_1: (t_1+t_2):(t_1+t_2+t_3):...:(t_1+t_2+t_3+...+t_n)=1:\sqrt{2}:\sqrt{3}:\sqrt{4} ...:\sqrt{n}$

$\to$ The time taken to cover successive equal intervals of time are in the ratio of $1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}:\sqrt{4}-\sqrt{3} ...:\sqrt{n}-\sqrt{n-1}$

$i.e.$ $t_1: t_2:t_3:...:t_n=1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}:\sqrt{4}-\sqrt{3} ...:\sqrt{n}-\sqrt{n-1}$

When the motion of the body starts from rest and moves with constant acceleration 'a'. If $v_1,v_2,v_3,...,v_n$ be the velocities at successive points at equal separation x, then

$\to$ $v_2^2 -v_1^2 = v_3^2 - v_2^2 = v_4^2 - v_3^2 =...= v_n^2 - v_{n-1}^2= 2ax$

When the motion of the body starts from rest and moves with constant acceleration 'a'. If $v_1,v_2,v_3,...,v_n$ be the velocities at equal time interval t, then

$\to$ $v_2 -v_1= v_3- v_2 = v_4- v_3 =...= v_n - v_{n-1}= at$

When a particle accelerates from rest for time $t_1$ over distance $x_1$ and at the acceleration $a_1$ and then retards to rest at the rate of $a_2$ over distance $x_2$ in time $t_2$, then

$\to \dfrac{a_1}{a_2} = \dfrac{t_2}{t_1} = \dfrac{x_2}{x_1}$

$\to$ Net acceleration, (a) = $\dfrac{a_1a_2}{a_1+a_2}$

$\to$ Maximum velocity ($v_{max}$) = $a_1t_1= a_2t_2 = \sqrt{2a_1x_1} = \sqrt{2a_2x_2} = \sqrt{\dfrac{2a_1a_2(x_2+x_2)}{a_1+a_2}} = \dfrac{a_1a_2}{a_1+a_2} \times \text{total time}$

$\to$ Average velocity $v_{av} = \dfrac{a_1a_2}{2(a_1+a_2)} \times (t_1+t_2)$

Motion Under Gravity

A body is released from 'H' under the influence of gravity takes time t to reach the ground. Then,

$\to$ The time taken to reach the ground is $t = \sqrt{\dfrac{2H}{g}}$

$\to$ The velocity with which it hits ground is $t = \sqrt{2gH} = gt$

$\to$ The distance covered in last second $h_t = \dfrac{g}{2}(2t-1)$

$\to$ The velocity of body 'h' height above ground $v_h = \sqrt{2g(H-h)}$

$\to$ The average velocity for entire motion $v_{av} = \dfrac{gt}{2} = \sqrt{\dfrac{gh}{2}}$

$\to$ The distance traveled in successive equal interval of time are in the ratio of $1:3:5:7 ...: (2n-1)$

$\to$ The distance after the end of the successive equal interval of time are in the ratio of $1:4:9:16 ...:n^2$

When a body is thrown vertically upward with initial velocity u under gravity and returns to the thrower's hand. then

$\to$ Maximum height attached:

$H = \dfrac{u^2}{2g}$

$\to$ Time taken to reach the maximum height

$t= \dfrac{u}{g} = \sqrt{\dfrac{2H}{g}}$

$\to$ Time of flight

$T= 2t = 2 \times \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{8H}{g}}$

$\to$ Velocity of the body at height 'h' above the ground

$v_h = \sqrt{2g(H-h)} = u \sqrt{1 - \dfrac{h}{H}}$

$\to$ Average speed for entire trip in upward motion: $v_{av} = \dfrac{u}{2} = \sqrt{\dfrac{gH}{2}}$

$\to$ Distance covered in last second of upward motion is: $h_t = u- \dfrac{g}{2}(2t-1) = \dfrac{g}{2}$

The body passes a point at height twice after time $t_1$ and $t_2$ from starting then

$\to$ Time to reach maximum height $t = \dfrac{t_1+t_2}{2}$

$\to$ Total time of flight $T =t_1+t_2$

$\to$ Height of that point from the ground, $h = \dfrac{g}{2} t_1t_2$

$\to$ Maximum height attained, $H = \dfrac{g}{8} (t_1+t_2)^2$

$\to$ Initial velocity, $(u) = \dfrac{g}{2}(t_1+t_2)$

When a body of mass 'm' is released from rest from height 'h' along a smooth inclined plane having inclination angle $\theta$ and length 'l', then

$\to$ Velocity of the body at the bottom of inclined plane $v = \sqrt{2gh} = \sqrt{2gl \sin \theta}$

$\to$ Acceleration down the plane $a = g \sin \theta$

$\to$ Time taken to fall down the plane $t = \sqrt{\dfrac{2h}{g}} \cosec \theta$

$\to$ Time taken to slide down along different smooth inclined planes of different inclination from same height are in ratio $t_1:t_2:... = \cosec \theta_1: \cosec \theta_2: ...$

$\to$ Time taken to slide down along different smooth inclined planes of equal length of different inclination are in the ratio $t_1:t_2:... = \sqrt{\cosec \theta_1}: \sqrt{\cosec \theta_2}: ...$

Relative Velocity

Introduction

Let velocity of body A be $\overrightarrow{v_A}$ and the velocity of body B be $\overrightarrow{v_B}$.

$\to$